Current Electricity - Class 12 Medical Physics - Extra Questions
A cell of emf 'E' and internal resistance r is connected across a variable resistor 'R'. Plot a graph showing variation of terminal voltage 'V' of the cell versus the current 'I'. Using the plot, show how the emf of the cell and its internal resistance can be determined.
Given: A cell of emf $$'E'$$ and internal resistance $$r$$ is connected across a variable resistor $$'R'$$.
To find: Plot a graph showing variation of terminal voltage $$'V'$$ of the cell versus the current $$'I'$$. Using the plot, show how the emf of the cell and its internal resistance can be determined.
Solution:
The relation between $$V$$ and $$I$$ is given by:
$$V=E-Ir$$
Thus the graph between $$I$$ and $$V$$ is ass shown in fig(i).
Emf is given by the intercept on the vertical axis, i.e., the $$V$$-axis.
Internal resistance is given by the slope of the line i.e., slope of $$V$$ vs $$I$$ graph.
A voltage $$V_{0}$$ is applied to a potentiometer whose sliding contact is exactly in the middle. A voltmeter $$V$$ is connected between the sliding contact and one fixed end of the potentiometer. It is assumed that the resistance of the voltmeter is not very high in comparison to the resistance of the potentiometer wire. What voltage will the voltmeter show : higher than, less than, or equal to $$V_{0}/2$$?
$$R = $$ Resistance of whole potentiometer wire AB
$$R_v = $$ Resistance of voltmeter
$$R_{AC} = $$ Total Resistance across $$AC$$
$$R_{AC} = \dfrac{R_v (R/2)}{R_v + (R/2)} = \dfrac{R}{2 (1 + R/2 R_v)} < \dfrac{R}{2}$$
The resistance of section CB is equal to $$\dfrac{R}{2}$$
So the reading of voltmeter will be less than $$\dfrac{V_0}{2}$$
Primary circuit of potentio meter is as shown in the diagram find potential gradient.
Given,
$$V=2v,R_{AB}=10\Omega,L=10m,,R_1=1\Omega,R_2=20\Omega$$
Current through the circuit =$$V*R$$= $$\dfrac{2}{31}$$
So the PD across AB is $$\dfrac{2}{31}\times10=0.64$$
Where $$PD=\dfrac{v}{R}\times l$$
$$R=R_1+R_2+R_{AB}$$
Therefore, The potential gradient is=$$V*L$$= $$\dfrac{0.64}{10}=0.064v/m$$
Two cell of emf $$6V$$ and $$4V$$ having internal resistance $$3\Omega$$ and $$2\Omega$$ respectively are connected in parallel so all to send a current through extend resistance of $$8\Omega$$ in the same direction. Find the current through the cells and the potential difference across $$8\Omega$$ Resistor.
$$\cfrac{E_{eff}}{r_{eff}}=\cfrac{E_1}{r_1}+\cfrac{E_2}{r_2}$$
$$r_{eff}=\cfrac{r_1r_2}{r_1+r_2}=\cfrac{(3)(2)}{3+2}=6/5\Omega$$
$$\cfrac{E_{eff}}{6/5}=\cfrac{6}{3}+\cfrac{4}{2}$$
$$\cfrac{5}{6}E_{eff}=2+2=4$$
$$E_{eff}=\cfrac{24}{5}$$
$$I=\cfrac{\cfrac{24}{5}}{\cfrac{6}{5}+8}$$ $$=\cfrac{\cfrac{24}{5}}{\cfrac{46}{5}}=\cfrac{24}{46}$$
$$I_8=\cfrac{24}{46}\times 84$$
$$=\cfrac{96}{23}V$$
A potentiometer wire has an length 4 m and resistance $$4\Omega$$. What resistance must be connected in series with the wire and an accumulator of e.m.f. 2V, so as to get a potential gradient of $$10^{-3}$$ V/cm on the wire?
A potential difference of 100$$\mathrm { V }$$ is applied across a resistor of resistance $$50 \Omega$$ for $$6$$ minutes and $$58$$ seconds. Find the heat produced in 1) joule 2) calorie
What should be the resistance of the shunt used? The potentiometer wire AB shown in figure (32-E26) is 40 cm long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?
Derive the condition of balance for Wheatstone bridge.
Look at the diagram above This is a wheatstone bridge.
You can notice four resistances P,Q,R and S connected on the bridge.
You can also notice a galvanometer connected between points B and D.
The four resistances are adjusted in such a way that the current through the galvanometer becomes zero.
This shows that B and D are at the same electric potential.
$$\therefore V_B=V_D$$
The potential difference across B will be $$V_A-V_B$$
Similarly across Q will be $$V_A-V_D$$
Similarly across R will be $$V_B-V_C$$
Similarly across S will be $$V_D-V_C$$
By Ohm's law we know that $$V=IR$$ where $$I$$ is the current.
$$\therefore V_A - V_B= IP$$ $$, V_A - V_D= IQ$$ $$, V_B - V_C= IR$$ and$$ V_D- V_C= IS$$
$$\because V_B=V_D$$ we get
$$IP=IQ$$ and $$IR=IS$$
By dividing the 1st by the 2nd, we get
$$\cfrac{IP}{IR}=\cfrac{IQ}{IS}$$
$$\therefore \cfrac PQ= \cfrac RS$$ this equation is known as the balancing condition in a wheatstone bridge
In the figure shown below the maximum possible unknown resistance (X) that can be measured by the post office box are $$\displaystyle X_{max}$$ is given by $$\displaystyle R\times 10^{5}\Omega$$ then R is (given that in this experiment we can take out only one plug from arm AB and only one plug from arm BC but from arm AD we can take out many plugs)
A block in the shape of a rectangular solid has a cross-sectional area of $$3.50 \,cm^2$$ across its width, a front-to-rear length of $$15.8$$ cm, and a resistance of $$935 \Omega$$. The block’s material contains $$5.33 \times 10^{22}$$ conduction electrons/$$m^3$$ . A potential difference of $$35.8$$ V is maintained between its front and rear faces. (a) What is the current in the block? (b) If the current density is uniform, what is its magnitude? What are (c) the drift velocity of the conduction electrons and (d) the magnitude of the electric field in the block?
Why are fairy decorative lights always connected in parallel ?
What do you understand by electric energy?
Write the condition of the balanced state of the Wheatstone bridge.
When the Wheatstone bridge is balanced, then the ratio of the resistances of ratio arms are equal.
$$\dfrac {P}{Q} = \dfrac {R}{S}$$.
$$\dfrac {P}{Q} = \dfrac {R}{S}$$.
Describe the principle and working of a meterbridge
Give reasons:
(i) The wires in the coil of resistance box are turned twice.
(ii) In Wheatstone bridge, initially a cell key and later galvanometer is pressed.
Why should the current be not passed through potentiometer wire for a long time?
A steady current flows in a metallic conductor of the non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed?
What will be the effect on the position of zero deflection if only the current flowing through the potentiometer wire is increased?
On what factors does the potential gradient of the wire depend?
Why should not the jockey be slided along the potentiometer wire?
Define or describe a Potentiometer.
In Wheatstone's meter-bridge experiment, the null point is obtained in the middle one-third portion of the wire. Why is it recommended?
Are Kirchhoff's laws applicable to both AC and DC currents?
Is Ohms law universally applicable for all conducting elements? If not, give examples of elements that do not obey Ohms law.
In a house two $$60\ W$$ electric bulbs are lighted for $$4$$ hours, and three $$100\ W$$ bulbs for $$5$$ hours everyday. Calculate the electric energy consumed in $$30$$ days.
Give statements of Kirchhoff's junction and loop laws. Find out the balance condition of Wheatstone's bridge with the help of these laws.
"In an electric circuit, the algebraic sum of current at any junction point must be zero."
i.e., $$\sum I = 0$$
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kircchoff's First Law or Junction Law.
It is based on the law of Conservation of Charge. When current in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.
In figure (6.1) at junction $$O$$,
$$I_{1} + I_{2} - I_{3} - I_{4} + I_{5} = 0$$
or $$I_{1} + I_{2} + I_{5} = I_{3} + I_{4} .... (1)$$
In figure $$(6.1)$$ the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.
Kircchoff's Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., $$\sum V = 0 .... (1)$$
The algebraic sum of the emf's in any loop of a circuit is equal to the sum of the voltage across the resistance.
$$\sum E = \sum IR .... (2)$$
Sign convention for applying Loop Rule
1. The product of current and resistance $$(IR)$$ is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance $$(IR)$$ is taken as negative if the resistor is traversed in the opposite direction of assumed current.
In loop we can move from $$A$$ to $$B$$ then the product of $$IR$$ take as positive $$V_{AB} = +IR$$. [In the direction of current]
In a loop we can move from $$B$$ to $$A$$ then the product of $$IR$$ take as negative.
$$\therefore V_{BA} = -IR$$ [Opposite direction of current]
2. When we move from $$B$$ to $$A$$ then $$E$$ is taken positive $$(+)$$. When we move from $$A$$ to $$B$$, then $$E$$ is taken negative $$(-)$$.
Loop law can be understood by the figure 6.4.
In the given figure for point $$d$$ by the Junction Law, current flowing through the resistance $$R\%$$ will be
$$I_{3} = I_{1} + I_{2} ..... (3)$$
In the loop $$adcba$$ by the Kircchoff's Loop Law.
$$I_{1}R_{1} - I_{2}R_{2} = \epsilon_{1} - \epsilon_{2}$$
or $$I_{2}R_{2} = I_{2}R_{2} = \epsilon_{2} - \epsilon_{1} .... (4)$$
In the loop $$defcd$$ by the Loop Law
$$I_{3}R_{3} + I_{2}R_{2} = \epsilon_{0} .... (5)$$
By solving equation $$(3), (4)$$ and $$(5)$$ we can find out electric currents in different branches as well as we can find potential differences across the resistance.
Construction : The construction of Wheatstone's bridge is shown in following diagram (figure 6.5) in which a quadrilateral $$ABCD$$ is prepared by connecting four resistance $$P, Q, R$$ and $$S$$. A cell is connected between the points $$A$$ and $$C$$ through key $$K_{1}$$ which is known as battery key and a galvanometer between the point $$B$$ and $$D$$ through another key $$K_{2}$$ which is known as galvanometer key. If on pressing the key $$K_{1}$$ and $$K_{2}$$, there is no deflection in galvanometer, then the bridge is said to be balanced condition.
In this situation the relation between all the resistance is given by:
$$\dfrac {P}{Q} = \dfrac {R}{S}$$
Principle of Wheatstone Bridge and Condition of Balance
When battery key $$K_{1}$$ is pressed, then main current $$I$$ starts flowing in the circuit. At junction $$A$$ this current splits in two parts $$I_{1}$$ and $$I_{2}$$ as shown in figure 6.6. These current $$I_{2}$$ and $$I_{2}$$ again obtain two paths at junctions $$B$$ and $$D$$ respectively. Now following three situations are possible:
(i) When $$V_{B} > V_{D}$$. then current $$I_{2}$$ passes as such through resistance $$S$$ but current $$I_{1}$$ splits in two parts at junction $$B$$, one part $$I_{g}$$ passes through galvanometer showing deflection in one direction while rest current $$(I_{1} - I_{g})$$ passes through $$Q$$.
(ii) When $$V_{B} < V_{D}$$, then opposite situation to previous situation (i) is observed i.e. current $$I_{2}$$ splits in two parts at $$D$$. Now the current $$I_{g}$$ produces deflection in galvanometer in opposite direction to previous deflection.
(iii) When $$V_{B} = V_{D}$$, then no current passes through galvanometer i.e. galvanometer shows no deflection because now $$I_{g} = 0$$.
The situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called 'Bridge circuit' Thus in balanced condition.
$$V_{B} = V_{D}$$
Therefore $$V_{A} - V_{B} = V_{A} - V_{D}$$
or $$I_{1}P = I_{2}R$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {R}{P} .... (1)$$
and $$V_{B} - V_{C} = V_{D} - V_{C}$$
or $$I_{1} \times Q = I_{2} \times S$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {S}{Q} .... (2)$$
Now from equation $$(1)$$ and $$(2)$$
$$\dfrac {R}{P} = \dfrac {S}{Q}$$ or $$\dfrac {R}{S} = \dfrac {P}{Q}$$
or $$\dfrac {P}{Q} = \dfrac {R}{S} .... (3)$$
Therefore, it is clear that in balanced situation of the bridge, the ratio of resistances of any two corresponding arms of quadrilateral $$ABCD$$ is equal to ratio of two remaining corresponding arms".
From equation $$(3)$$
$$S = \dfrac {Q}{P}\times R$$
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below:
(i) $$P$$ and $$Q$$ are called ratio arms.
(ii) $$R$$ is called variable resistance arm.
(iii) $$S$$ is called unknown resistance arm.
i.e., $$\sum I = 0$$
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kircchoff's First Law or Junction Law.
It is based on the law of Conservation of Charge. When current in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.
In figure (6.1) at junction $$O$$,
$$I_{1} + I_{2} - I_{3} - I_{4} + I_{5} = 0$$
or $$I_{1} + I_{2} + I_{5} = I_{3} + I_{4} .... (1)$$
In figure $$(6.1)$$ the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.
Kircchoff's Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., $$\sum V = 0 .... (1)$$
The algebraic sum of the emf's in any loop of a circuit is equal to the sum of the voltage across the resistance.
$$\sum E = \sum IR .... (2)$$
Sign convention for applying Loop Rule
1. The product of current and resistance $$(IR)$$ is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance $$(IR)$$ is taken as negative if the resistor is traversed in the opposite direction of assumed current.
In loop we can move from $$A$$ to $$B$$ then the product of $$IR$$ take as positive $$V_{AB} = +IR$$. [In the direction of current]
In a loop we can move from $$B$$ to $$A$$ then the product of $$IR$$ take as negative.
$$\therefore V_{BA} = -IR$$ [Opposite direction of current]
2. When we move from $$B$$ to $$A$$ then $$E$$ is taken positive $$(+)$$. When we move from $$A$$ to $$B$$, then $$E$$ is taken negative $$(-)$$.
Loop law can be understood by the figure 6.4.
In the given figure for point $$d$$ by the Junction Law, current flowing through the resistance $$R\%$$ will be
$$I_{3} = I_{1} + I_{2} ..... (3)$$
In the loop $$adcba$$ by the Kircchoff's Loop Law.
$$I_{1}R_{1} - I_{2}R_{2} = \epsilon_{1} - \epsilon_{2}$$
or $$I_{2}R_{2} = I_{2}R_{2} = \epsilon_{2} - \epsilon_{1} .... (4)$$
In the loop $$defcd$$ by the Loop Law
$$I_{3}R_{3} + I_{2}R_{2} = \epsilon_{0} .... (5)$$
By solving equation $$(3), (4)$$ and $$(5)$$ we can find out electric currents in different branches as well as we can find potential differences across the resistance.
Construction : The construction of Wheatstone's bridge is shown in following diagram (figure 6.5) in which a quadrilateral $$ABCD$$ is prepared by connecting four resistance $$P, Q, R$$ and $$S$$. A cell is connected between the points $$A$$ and $$C$$ through key $$K_{1}$$ which is known as battery key and a galvanometer between the point $$B$$ and $$D$$ through another key $$K_{2}$$ which is known as galvanometer key. If on pressing the key $$K_{1}$$ and $$K_{2}$$, there is no deflection in galvanometer, then the bridge is said to be balanced condition.
In this situation the relation between all the resistance is given by:
$$\dfrac {P}{Q} = \dfrac {R}{S}$$
Principle of Wheatstone Bridge and Condition of Balance
When battery key $$K_{1}$$ is pressed, then main current $$I$$ starts flowing in the circuit. At junction $$A$$ this current splits in two parts $$I_{1}$$ and $$I_{2}$$ as shown in figure 6.6. These current $$I_{2}$$ and $$I_{2}$$ again obtain two paths at junctions $$B$$ and $$D$$ respectively. Now following three situations are possible:
(i) When $$V_{B} > V_{D}$$. then current $$I_{2}$$ passes as such through resistance $$S$$ but current $$I_{1}$$ splits in two parts at junction $$B$$, one part $$I_{g}$$ passes through galvanometer showing deflection in one direction while rest current $$(I_{1} - I_{g})$$ passes through $$Q$$.
(ii) When $$V_{B} < V_{D}$$, then opposite situation to previous situation (i) is observed i.e. current $$I_{2}$$ splits in two parts at $$D$$. Now the current $$I_{g}$$ produces deflection in galvanometer in opposite direction to previous deflection.
(iii) When $$V_{B} = V_{D}$$, then no current passes through galvanometer i.e. galvanometer shows no deflection because now $$I_{g} = 0$$.
The situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called 'Bridge circuit' Thus in balanced condition.
$$V_{B} = V_{D}$$
Therefore $$V_{A} - V_{B} = V_{A} - V_{D}$$
or $$I_{1}P = I_{2}R$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {R}{P} .... (1)$$
and $$V_{B} - V_{C} = V_{D} - V_{C}$$
or $$I_{1} \times Q = I_{2} \times S$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {S}{Q} .... (2)$$
Now from equation $$(1)$$ and $$(2)$$
$$\dfrac {R}{P} = \dfrac {S}{Q}$$ or $$\dfrac {R}{S} = \dfrac {P}{Q}$$
or $$\dfrac {P}{Q} = \dfrac {R}{S} .... (3)$$
Therefore, it is clear that in balanced situation of the bridge, the ratio of resistances of any two corresponding arms of quadrilateral $$ABCD$$ is equal to ratio of two remaining corresponding arms".
From equation $$(3)$$
$$S = \dfrac {Q}{P}\times R$$
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below:
(i) $$P$$ and $$Q$$ are called ratio arms.
(ii) $$R$$ is called variable resistance arm.
(iii) $$S$$ is called unknown resistance arm.
Conceptual Questions
How does the resistance for copper and for silicon change with temperature? Why are the behaviors of these two materials different?
What do you mean by a potential difference of $$1$$ volt?
Compare emf and potential difference.
Write two differences between emf and potential difference (pd).
Kirchhoff's rules are very useful for analysis of electrical circuits. State Kirchhoff's junction rules.
Kirchhoff's first rule applies to junction points in DC circuits. The junction rule is:
The sum of all the currents entering any junction point is equal to the sum of all the currents leaving that junction point.
How much energy will be spent in $$15$$ minutes in a $$5$$ kilowatt electric heater?
What will be the effect of change in the length and thickness of the electrical wire in the experiment of electric current, potential difference and electric resistance?
Find the value of current I in given circuit.
$$E_{eff}=1.7-0.7=1.0V$$
$$I=\dfrac{V}{R_{eff}}=\dfrac{1}{2+2}=0.25A$$
Figure shows four cells fixed on a board. Draw lines to indicate how you will connect their terminals with wires to make a battery of four cells.
Define the term
dielectric constant. Give its S.I.unit.
A copper coil has a resistance of $$2.43\ \Omega$$ at $$50^{o}C$$. Find its resistance at $$60^{o}C$$. The resistance temperature coefficient of copper is $$0.0043$$ per $$^{o}C$$ at $$0^{o}C$$.
A combination of two or more cells is called a _________.
At room temperature, $$27^{\circ}C$$, the resistance of a heating element is $$100\, \Omega $$. What is the temperature(in $$^{\circ}C$$) of the element if the resistance is found to be $$117\, \Omega $$? Given that the temperature coefficient of the material of the resistor is $$1.70\times 10^{-4}C^{-1}$$.
For determination of resistance of a coil, which of two methods is better Ohm's Law method or metre bridge method ?
When is the Wheatstone's bridge said to be most sensitive?
Can we measure a resistance of the order of 0.160$$ \Omega$$ using a Wheatstone's bridge? Support your answer with reasoning?
State the underlying principle of a potentiometer.
A combination of a group of cells joined in series is called a _______.
What do you mean by potential gradient?
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
Use Kirchhoff's rule to obtain conditions for the balance condition in a Wheatstone bridge.
We apply Kirchoff's current law in the shown circuit.
At junction B,
$$i_1=i_g+i_3$$
At junction D,
$$i_2+i_g=i_4$$
If current through the galvanometer is zero,
$$i_g=0$$
thus $$i_1=i_3$$
and $$i_2=i_4$$
Applying Kirchoff's voltage law for loop ABDA,
$$i_1P+i_gG=i_2R$$
Applying Kirchoff's voltage law for loop BCDB,
$$i_3Q=i_4S+i_gG$$
When $$i_g=0$$,
$$i_1P=i_2R$$
and $$i_3Q=i_4S$$
But $$i_1=i_3$$ and $$i_2=i_4$$,
Therefore $$\dfrac{P}{Q}=\dfrac{R}{S}$$
What is a potential meter explain its principal
(a)Explain how it can be used to compare the emf of two cells
(b)Explain how it can be used to determine the internal resistance of a cells.
Potentiometer
It is a three terminal resistor with a sliding or rotating contact that format an adjustable voltage devider
(a) compare Emf of two cells
Ref. image 1
Ref. image 2
$$E_k = i(\rho l_1) \,\, E_u = i (\rho l_2)$$
$$\rho = $$ Resistance of wire ab per unit length
$$\dfrac{E_k}{E_u} = \dfrac{l_1}{l_2}$$
(b) Determine internal resistance
ref. image 3
$$i = \dfrac{E}{R + r} $$ __(I)
$$V = E - ir = iR$$ ___(II)
From eq (I) and (II)
$$r = R \left(\dfrac{E}{V} - 1\right)$$
Fig shown a simple potentiometer circuit for measuring a small e.m.f produced buy a thermocouple. The meter wire $$PQ$$ has a resistance of $$5\ \Omega$$. If a balance point is obtained $$0.6\ m$$ along $$PQ$$ when measuring an e.m.f of $$6\ mV$$., what is the value of resistance $$R$$?
An electric heater and a motor of $$750\ W$$ and $$1kW$$ are used for domestic purpose for $$2\ hours$$ and $$1\ 1/2$$ hours per day in the month of January respectively. Calculate the cost of energy used if unit costs $$5.50$$.
Networks of nine conductors connects six points A, B, C, D, E and F as shown
in figure. The figure denoted resistances in ohms. Find the equivalent resistance between A and D.
$$V_E = V_P \Rightarrow I_{EF} = 0$$
$$V_B = V_C \Rightarrow I_{BC} = 0$$
An electric heater of resistance $$8\Omega$$ draws $$15A$$ from the service mains in 2 hrs.Calculate the rate at which heat is developed in the heater.
The voltmeter shown in figure, reads 18 V across the $$50 \Omega$$ resistor. Find the resistance of the voltmeter.
Primary circuit of potentio meter is as shown in the diagram find current in primary.
Given,
$$V=2v,R_{AB}=10\Omega,L=10m, R_1=1\Omega,R_2=20\Omega$$
We kknow that $$V=IR\Rightarrow I=\dfrac{V}{R}=\dfrac{2}{31}=0.064A$$
In the figure, take the potential at B to be zero.
(a) What are the potentials at the point A and C? (b) At which point D of the wire AB, the potential is equal to the potential at C? (c) If the points C and D are connected by a wire, what will be the current through it? (d) If the 4V battery is replaced by a 7.5 V battery, what would be the answers of parts (a) and (b)?
Find the radius of the wire of length $$25\ m$$ needed to prepare a coil of resistence $$25\ \Omega $$ (Resistivity of material of wire is $$3.142 \times {10^{ - 7}}\Omega m$$)
What is the principle of metre-bridge ?
State KCL and explain its sign conventions
Two bulbs of $$40\,W$$ each are lighted for eight hours daily. Find the cost of electrical energy consumed by them in one week at Rs. 3 per unit.
In the potentiometer circuit suppose $$ \varepsilon $$ =12V, $$\ell$$ =1m
if the galvanometer current is zero when x=075m, (a) find the
potential difference V, (b) if v=emf of the battery, find its
internal resistance.
Given $$\varepsilon=12 V$$
$$l=1 m$$
Potential Gradient $$=k=\dfrac{\varepsilon}{l}=12 \dfrac{V}{m}$$
(a) In this situation $$x=0.75$$ is null point
we know that
$$\varepsilon =kl\\ v=12\times 0.75\\ v={ 12 } \times \dfrac { 3 }{ 4 } \\ v=9V\\ $$
In the figure a long uniform potentiometer wire $$AB$$ is having a constant potential gradient along its length. The null points for the two primary cells of emfs $${ \varepsilon }_{ 1 }$$ and $${ \varepsilon }_{ 2 }$$ connected in the manner shown are obtained at a distance of $$120 cm$$ and $$300 cm$$ from the end $$A$$. Find (i) $${ { \varepsilon }_{ 1 } }/{ { \varepsilon }_{ 2 } }$$ and (ii) position of null point for the cell $${ \varepsilon }_{ 1 }$$.
How is the sensitivity of a potentiometer increased?
State Kirchhoff's rules. Explain briefly how these rules are justified.
A potentiometer wire of length 1.0 m has a resistance of 15 $$\Omega$$. It is connected to a 5V battery in series with a resistance of 5 $$\Omega$$. Determine the emf of the primary cell which gives a balance point at 60 cm.
Write the difference between potential difference and emf.
EMF:
$$\rightarrow$$ Work done by a source in taking a unit charge once around complete circuit.
$$\rightarrow$$ EMF exists even when ckt is open
$$\rightarrow$$ EMF is independent of external resistance
$$\rightarrow$$ EMF transforms energy from non electrical form to electrical form
P.D:
$$\rightarrow$$ Work done in moving a charge from one point to another point.
$$\rightarrow$$ P.D. exists only when ckt is closed
$$\rightarrow$$ P.D depends on resistance b/w $$2$$ points & current flowing through it
$$\rightarrow$$ P.D is the actual output voltage across battery.
(i) E$$=$$EMF
V$$=(p-1)$$
$$(E-1r)=v$$ discharge
$$(E-ir)=iR$$
$$i=\dfrac{E}{(R+r)}$$
$$v=iR=\dfrac{ER}{R+r}$$.
(ii) $$v-R-ir=0$$
$$v=E+ir$$
chargings
(iii) $$v=E$$ no internal loss.
Explain Krichhoff's laws with examples.
Kirchhoff's first rule ( Kirchhoff's Current Law or KCL or Junction Rule) :
It states that the sum of the currents flowing towards a junction is equal to the sum of currents leaving the junction.
This is in accordance with the conservation of charge which is the basis of Kirchhoff's current rule.
In fig(i), $$I_1, I_2, I_3, I_4$$ are currents flowing through the respective wires.
Convention: The current flowing towards the junction is taken positive and the current flowing away from the junction is taken as negative.
Hence, $$I_3+(-I_1)+(-I_2)+(-I_4)=0$$
"In an electric network, the algebraic sum of currents meeting at a junction is always zero", $$\sum I=0$$
Kirchhoff's second rule ( Kirchhoff's Voltage Law or KVL Loop rule ) :
It states that the algebraic sum of all potential drops and emfs along any closed path in a network is zero.
OR
The algebraic sum of the emfs in a loop of a circuit is equal to the algebraic sum of the product of current and resistances in it.
Mathematically, the loop rule may be expressed as :
$$\sum E=\sum IR$$
Kirchhoff's second law expresses the conservation of energy.
From fig(ii),
For closed loop BACB:
$$E_1-E_2=I_1R_1+I_2R_2-I_3R_3$$
For closed loop CADC:
$$E_2=I_3R_3+I_4R_4+I_5R_5$$
State Ohm's law. Suggest an experiment to verify it and explain the procedure.
Amount of current flowing is proportional to potential difference /voltage under fixed physical condition.
$$v\alpha I$$
$$V=IR$$
$$R=\dfrac V{I}$$
Experiment to verify Ohm's Law
Aim : To show that the ratio of $$\dfrac VI$$ is a constant for a conductor.
Materials required: $$5$$ dry cells of $$1.5 V$$ each, conducting wires, an ammeter, a voltmeter, thin iron spoke of length $$10 cm$$, LED and key.
Procedure :
- Connect a circuit as shown in the figure.
- Solder the conducting wires to the ends of the iron spoke and close the key.
- Note the reading of ammeter (current) and voltmeter (potential difference) and tabulate them.
- Now connect two cells in the circuit and note the respective readings of ammeter and voltmeter in the above table.
- Repeat the above procedure using three cells, four cells, and five cells respectively.
- Record the values of potential difference, $$V$$ and current, $$I$$ corresponding to each case in the above table.
- Find $$\dfrac VI$$ for each set of values.
- We notice that $$\dfrac VI$$ is constant.
From this we can conclude that P.D difference between the ends of the iron spoke is directly propotionala to the current passing tr=hrough it. This is Ohm's law.
In a meter bridge, the balance point is found at a distance $$l_1$$ with resistances R and S as shown in the figure.
An unknown resistance X is now connected in parallel to the resistance S and the balance point is found at a distance $$l_2$$. Obtain a formula for X in terms of $$l_1, l_2$$ and S.
Write Kirchhoff's first rule (law of junction). Drawing a circuit diagram of Wheatstone bridge, derive condition for zero deflection in the bridge.
According to Kirchoff's current law, in an electrical circuit, total current entering a node equals the total current exiting the node. It is in accordance with the law of conservation of charge.
Circuit diagram of a Wheatstone bridge is shown in the attached figure.
For zero deflection in the galvanometer, no current should flow through it. This happens when voltage across it is zero.
$$V_{P_1} = V_{P_2} ............(i)$$
Now since no current flows through galvanometer,
$$I_1 = I_3$$
Using ohm's Law,
$$\cfrac{V_{P_1} -V_A}{R_1} = \cfrac{V_{P_1} - V_B}{R_2} ...........(ii)$$
Similarly,
$$I_2 = I_4$$
$$\cfrac{V_{P_2} -V_A}{R_3} = \cfrac{V_{P_2} - V_B}{R_4} ...........(iii)$$
Dividing (ii) from (iii), we get:
$$\displaystyle \cfrac{ \frac { V_{P_2} - V_A} {R_3} } {\frac { V_{P_1} - V_A}{R_1}} = \cfrac{ \frac { V_{P_2} - V_B }{R_4} } { \frac {V_{P_1} - V_B}{R_2} }$$
Using (i) and rearranging terms we get,
$$\cfrac{R_1}{R_3} = \cfrac{R_2}{R_4}$$
The above is a necessary and sufficient condition for the galvanometer to show no deflection.
(a) Define potential gradient.
(b) Write Kirchhoff's junction rule.
In the given diagram write the value of current $$I$$.
Write Kirchhoff's first rule. A battery of $$10\ V$$ and negligible internal resistance is connected to the diagonally opposite corners of a cubical network consisting of $$12$$ resistors each of resistance $$2\Omega$$. Determine the equivalent resistance of the network.
According to Kirchhoff's first Law, at any node in an electrical circuit, total currents entering the node equals the currents exiting the node.
Using symmetry, equal currents flows through resistors $$a, b\ \&\ c$$.
Hence, equal potential drops occur across them and they can be connected in parallel near node $$A$$.
Similarly, equal potential drops occur across resistors $$j, k\ \&\ l$$ them and they can be connected in parallel near node $$B$$.
After this, it can be observed that the ends of remaining resistors are at same value of potentials. Hence, the remaining six resistors can be connected in parallel as shown.
For the equivalent circuit, the equivalent resistance can be found as a simple series-parallel resistance circuit.
$$R_{eq}= (2||2||2) + (2||2||2||2||2||2) + (2||2||2)$$
$$= \cfrac{2}{3} + \cfrac{2}{6} + \cfrac{2}{3}$$
$$=\cfrac{5}{3}\Omega$$
Obtain the condition for bridge balance in Wheatstone's bridge.
Wheatstone's bridge: An important application of Kirchoff's law is the Wheatstone's bridge figure.
Wheatstone's network consists of resistances P, Q, R and S connected to form a closed path. A cell of emf E is connected between points A and C. The current I from the cell is divided into $$I_1, I_2, I_3$$ and $$I_4$$ across the four branches. The current through the galvanometer is $$I_g$$. The resistance of galvanometer is $$G$$.
Applying Kirchoff's current law to junction B,
$$I_1-I_g-I_3=0$$ ............$$(1)$$
Applying Kirhcoff's current law of junction D.
$$I_2+I_g-I_4=0$$ ...............$$(2)$$
Applying Kirchoff's voltage law to closed path ABDA
$$I_1P+I_gG-I_2R=0$$ ..............$$(3)$$
Applying Kirchoff's voltage law to closed path ABCDA
$$I_1P+I_3Q-I_4S-I_2R=0$$ .............$$(4)$$
When the galvanometer shows zero deflection, the points B and D are at same potential and $$I_g=0$$. Substituting $$I_g=0$$ in equation $$(1), (2)$$ and $$(3)$$.
$$I_1=I_3$$ ..............$$(5)$$
$$I_2=I_4$$ ...............$$(6)$$
$$I_1P=I_2R$$ ...............$$(7)$$
Substituting the values of $$(5)$$ and $$(6)$$ in equation $$(4)$$
$$I_1P+I_1Q-I_2S-I_2R=0$$
$$I_1(P+Q)=I_2(R+S)$$ ..............$$(8)$$
Dividing $$(8)$$ by $$(7)$$
$$\displaystyle\frac{I_1(P+Q)}{I_1P}=\frac{I_2(R+S)}{I_2R}$$
$$\therefore \displaystyle\frac{P+Q}{P}=\frac{R+S}{R}$$
$$\Rightarrow 1+\displaystyle\frac{Q}{P}=1+\frac{S}{R}$$
$$\therefore \displaystyle\frac{Q}{P}=\frac{S}{R}$$ or $$\frac{P}{Q}=\dfrac{R}{S}$$.
This is the condition for bridge balance. If P, Q and R are known, the resistance S can be calculated.
Wheatstone's network consists of resistances P, Q, R and S connected to form a closed path. A cell of emf E is connected between points A and C. The current I from the cell is divided into $$I_1, I_2, I_3$$ and $$I_4$$ across the four branches. The current through the galvanometer is $$I_g$$. The resistance of galvanometer is $$G$$.
Applying Kirchoff's current law to junction B,
$$I_1-I_g-I_3=0$$ ............$$(1)$$
Applying Kirhcoff's current law of junction D.
$$I_2+I_g-I_4=0$$ ...............$$(2)$$
Applying Kirchoff's voltage law to closed path ABDA
$$I_1P+I_gG-I_2R=0$$ ..............$$(3)$$
Applying Kirchoff's voltage law to closed path ABCDA
$$I_1P+I_3Q-I_4S-I_2R=0$$ .............$$(4)$$
When the galvanometer shows zero deflection, the points B and D are at same potential and $$I_g=0$$. Substituting $$I_g=0$$ in equation $$(1), (2)$$ and $$(3)$$.
$$I_1=I_3$$ ..............$$(5)$$
$$I_2=I_4$$ ...............$$(6)$$
$$I_1P=I_2R$$ ...............$$(7)$$
Substituting the values of $$(5)$$ and $$(6)$$ in equation $$(4)$$
$$I_1P+I_1Q-I_2S-I_2R=0$$
$$I_1(P+Q)=I_2(R+S)$$ ..............$$(8)$$
Dividing $$(8)$$ by $$(7)$$
$$\displaystyle\frac{I_1(P+Q)}{I_1P}=\frac{I_2(R+S)}{I_2R}$$
$$\therefore \displaystyle\frac{P+Q}{P}=\frac{R+S}{R}$$
$$\Rightarrow 1+\displaystyle\frac{Q}{P}=1+\frac{S}{R}$$
$$\therefore \displaystyle\frac{Q}{P}=\frac{S}{R}$$ or $$\frac{P}{Q}=\dfrac{R}{S}$$.
This is the condition for bridge balance. If P, Q and R are known, the resistance S can be calculated.
Draw. Wheatstone's bridge circuit and write the condition for its balance.
Wheatstone's bridge is shown in the above figure with resistances $$P,Q,R$$ and $$S$$.
The Wheatstone's bridge is said to be balance if no current flows through the galvanometer $$G$$.
Condition for balanced Wheatstone's bridge : $$\dfrac{P}{Q} = \dfrac{R}{S}$$
State the working principle of potentiometer. Explain with the help of circuit diagram how the emf of two primary cells are compared by using the potentiometer.
Working principle: Potential difference across length of potentiometer wire is proportional to the length of the wire.
$$\epsilon =kL$$ where$$\epsilon$$ is the emf and k is the potential gradient.
In the figure points 1, 2, 3 form a two key way.
Consider first position of key where 1 and 3 are connected so the galvanometer is connected to $$\epsilon_1$$.
Jockey is moved at a point till $$l_1$$ distance from $$A_1$$ to point $$N_1$$.
$$\epsilon =kL$$ where$$\epsilon$$ is the emf and k is the potential gradient.
In the figure points 1, 2, 3 form a two key way.
Consider first position of key where 1 and 3 are connected so the galvanometer is connected to $$\epsilon_1$$.
Jockey is moved at a point till $$l_1$$ distance from $$A_1$$ to point $$N_1$$.
Then $$\epsilon_1=kl_1$$
Similarly, if another emf $$\epsilon_2$$ is balanced against $$l_2 (AN_2)$$
$$\epsilon_2=kl_2$$
Dividing both equations we finally get:
$$\dfrac{\epsilon_1}{\epsilon_2}=\dfrac{l_1}{l_2}$$
Similarly, if another emf $$\epsilon_2$$ is balanced against $$l_2 (AN_2)$$
$$\epsilon_2=kl_2$$
Dividing both equations we finally get:
$$\dfrac{\epsilon_1}{\epsilon_2}=\dfrac{l_1}{l_2}$$
Write the principle of working of a meter bridge.
$$l_{1}+l_{2}=L$$
Meter bridge is constructed on the principle based on balanced wheat stone bridge
Jockey $$J$$ moved along the slide wire $$AB$$ and there will onw position where galvanometer show no current.
Now the meter bridge is balanced
Balancing condition is
$$\dfrac{R}{l_{1}}=\dfrac{S}{l_{2}}$$
Derive the balancing condition of a Wheatstone bridge.
Wheatstone bridge (balanced) - Let $$i$$ be the current from battery $$E$$ .At point A , current $$i_{1}$$ flows through resistance $$P$$ and current $$i-i_{1}$$ flows through $$R$$. In balanced state , no current flows through BD , hence point B and D are at same potential .Therefore current $$i_{1}$$ flows through resistance $$Q$$ also and current $$i-i_{1}$$ flows through $$S$$.
Applying Kirchhoff's loop rule in closed mesh $$ABDA$$ ,
$$i_{1}P-(i-i_{1})R=0$$
or $$i_{1}P=(i-i_{1})R$$ ...................eq1
In closed mesh $$BCDB$$ ,
$$i_{1}Q-(i-i_{1})S=0$$
or $$i_{1}Q=(i-i_{1})S$$ ...................eq2
Dividing eq1 from eq2 , we get
$$P/Q=R/S$$
This is the condition for balance in a Wheatstone's bridge .
Derive the condition for balance of a Wheatstone's bridge using Kirchhoff's rules.
Based on the circuit given in the figure , we have three known
resistance's $$R_1, R_2, R_3$$ and an unknown variable resistor $$R_X
$$, a source of voltage, and a sensitive ammeter.
Kirchhoff's first rule is applied to find the currents in junctions B and D:
$$I_3-I_x+I_g=0$$
$$I_1-I_2-I_g=0$$
Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:
$$I_3.R_3-I_g.R_g-I_1.R_1=0$$
$$I_x.R_X-I_2.R_2+I_g.R_g=0$$
The bridge is balanced and $$I_g$$ = 0, so the second set of equations can be rewritten as:
$$I_3.R_3=I_1.R_1$$
$$I_x.R_X=I_2.R_2$$
From first rule of Kirchoff.
$$I_3=I_x$$
$$I_1=I_2$$
$$\frac{I_3.R_3}{I_x.R_X}=\frac{I_1.R_1}{I_2.R_2}$$
$$\frac{R_1}{R_2}=\frac{R_3}{R_x}$$
Kirchhoff's first rule is applied to find the currents in junctions B and D:
$$I_3-I_x+I_g=0$$
$$I_1-I_2-I_g=0$$
Now Kirchoff's second rule is used to find the voltage in the loops ABD and BCD:
$$I_3.R_3-I_g.R_g-I_1.R_1=0$$
$$I_x.R_X-I_2.R_2+I_g.R_g=0$$
The bridge is balanced and $$I_g$$ = 0, so the second set of equations can be rewritten as:
$$I_3.R_3=I_1.R_1$$
$$I_x.R_X=I_2.R_2$$
From first rule of Kirchoff.
$$I_3=I_x$$
$$I_1=I_2$$
$$\frac{I_3.R_3}{I_x.R_X}=\frac{I_1.R_1}{I_2.R_2}$$
$$\frac{R_1}{R_2}=\frac{R_3}{R_x}$$
State Kirchhoff's law for an electrical network. Using these laws, deduce the condition for balance in a Wheatstone bridge.
Kirchhoff's first law (junction rule) - In an electrical circuit , the algebraic sum of currents meeting at a junction , is always zero .
Kirchhoff's second rule (loop rule) - In any closed mesh of an electrical circuit , the algebraic sum of the e.m.fs. is equal to the algebraic sum of products of resistances and current flowing through them .
Wheatstone bridge (balanced) - Let $$i$$ be the current from battery $$E$$ .At point A , current $$i_{1}$$ flows through resistance $$P$$ and current $$i-i_{1}$$ flows through $$R$$. In balanced state , no current flows through BD , hence point B and D are at same potential .Therefore current $$i_{1}$$ flows through resistance $$Q$$ also and current $$i-i_{1}$$ flows through $$S$$.
Applying Kirchhoff's loop rule in closed mesh $$ABDA$$ ,
$$i_{1}P-(i-i_{1})R=0$$
or $$i_{1}P=(i-i_{1})R$$ ...................eq1
In closed mesh $$BCDB$$ ,
$$i_{1}Q-(i-i_{1})S=0$$
or $$i_{1}Q=(i-i_{1})S$$ ...................eq2
Dividing eq1 from eq2 , we get
$$P/Q=R/S$$
This is the condition for balance in a Wheatstone's bridge .
State and explain Kirchhoff's Laws in the current electricity.
Kirchhoff's laws: (i) The algebraic sum of current meeting at an junction in a circuit is zero. In this law, the current flowing towards the junction are considered as $$+$$ve and those flowing away from the junction as $$-$$ve currents.
As shown in the figure, we have
$${i}_{1} - {i}_{2} - {i}_{3} + {i}_{4} - {i}_{5} = 0$$
or $${i}_{1} + {i}_{4} = {i}_{2} + {i}_{3} + {i}_{5}$$
(ii) In any closed mesh (or loop) of an electrical circuit, be algebraic sum of the product of the currents and resistances is equal to the total e.m.f. of the mesh.
If we go along the direction of conventional current, be potential difference will be taken as negative and opposite to it will be positive.
Inside the cell, if we move from low to high potential along the direction of conventional current the e.m.f. will be positive.
For loop 1,
$${E}_{2} - {i}_{2}{R}_{2} - \left({i}_{1} + {i}_{2}\right) {R}_{3} = 0$$
or $${E}_{2} = {i}_{2}{R}_{2} + \left({i}_{1} + {i}_{2}\right) {R}_{3}$$
For loop 2,
$${i}_{2}{R}_{2} - {E}_{2} - {i}_{1}{R}_{1} + {E}_{1} = 0$$
or $${E}_{1} - {E}_{2} = {i}_{1}{R}_{1} - {i}_{2}{R}_{2}$$
The figure is a part of a closed circuit. Find the currents $$i_1, i_2$$ and $$i_3$$.
A resistance of $$R$$ $$\Omega$$ draws current from a potentiometer. The potentiometer wire, $$AB$$, has a total resistance of $$R_{0}$$ $$\Omega$$. A voltage $$V$$ is supplied to the potentiometer. Derive an expression for the voltage across $$R$$ when the sliding contact is in the middle of the potentiometer wire.
Let the length of coir AB be 'l'
mid point of AB be M
Voltage across AB is V
$$\Rightarrow$$ Potential gradient of AB$$=\dfrac{V}{l}$$
Since, sliding contact is at M
$$\Rightarrow$$ Voltage across resistance R will be equal to potential across AM
$$\Rightarrow$$ Voltage across $$R=($$potential gradient of AB$$)\times ($$ length of AM$$)$$
$$=\dfrac{V}{l}\times \dfrac{
l}{2}$$
l}{2}$$
$$=\dfrac{V}{2}$$
$$\Rightarrow$$ Voltage across$$R=\dfrac{V}{2}$$.
Draw a labelled diagram of experiment explaining Ohm's Law.
Experimental verification of Ohm's law
Procedure :
- Draw the circuit diagram as shown above.
- Arrange the apparatus as per the circuit diagram.
- Clean the ends of the connecting wires with sand paper and make them shiny.
- Make the connections as per circuit diagram. All connections must be neat and tight. Take care to connect the ammeter and voltmeter with their correct polarity. (+ve to +ve and -ve to -ve).
- Determine the zero error and least count of the ammeter and voltmeter and record them.
- Adjust the rheostat to pass a low current.
- Insert the key K and slide the rheostat contact to see whether the ammeter and voltmeter are showing deflections properly.
- Adjust the rheostat to get a small deflection in ammeter and voltmeter.
- Record the readings of the ammeter and voltmeter.
- Take atleast six sets of readings by adjusting the rheostat gradually.
- Plot a graph with V along x-axis and I along y-axis.
- The graph will be a straight line which verifies Ohm's law.
- Determine the slope of the V-I graph. The reciprocal of the slope gives resistance of the wire.
Veena's car radio will run from a 12 V car battery that produces a current of 0.20 A even when the car engine is turned off. The car battery will not longer operate when it has lost $$1.2 \times 10^6 J$$ of energy. If Veena gets out of the car, leaving the radio on by mistake, how long will it take for the car battery to go completely dead, i.e., lose all energy?
How will the position of balance point in the potentiometer will be affected if, (I) current in the potentiometer be increased, (ii) length of the potentiometer wire be doubles?
On which conservation principle is Kirchoff's Second Law of electrical networks based?
The resistance of a conductor at $$20^o$$C is $$3.15$$ ohm and at $$100$$ $$^oC$$ is $$3.75$$ohm. Determine the temperature co-efficient of resistance of the conductor. What will be the resistance of the conductor at $$0$$ $$^oC$$?
Write down the principle of potentiometer. Why is it called an ideal voltmeter?
The potentiometer wire AB shown in the figure is $$40cm$$ long. Where should the free end of the galvanometer be connected on AB so that the galvanometer may show zero deflection?
A resistance of $$R$$ draws the current form of the potentiometer. The potentiometer wire, $$AB$$, has a total resistance of $$R_{0}$$. A voltage $$V$$ is supplied to the perpendicular. Derive an expression for the voltage across $$R$$ when the sliding contact is in the middle of potentiometer wire.
State Kirchoff's current law and also give the sign convention for current.
Kirchhoff's current law states that at any node(junction) in an electric circuit, the sum currents flowing in to that node is equal to the sum of current flowing out of that node.
Sign convention:- current entering at any junction is considered as positive.
and current leaving the junction is considered as negative.
In given circuit,
By Kirchhoff's current law,
$$i_2+i_3=i_1+i_4$$
In a potentiometer experiment, balancing length is found to be $$120$$cm for a cell $$E_1$$ of emf $$2V$$. What will be the balancing length for another cell $$E_2$$ of emf $$1.5$$V? (No other changes are made in the experiment.)
Describe the construction and working of a potentiometer. Explain, drawing circuit diagram, how emfs of two cells can be compared with its help?
The potentiometer is a device used to compare the e.m.f of two cells. It works on the principle that when a constant current flows through a wire of uniform cross-sectional area, a potential difference between its two points, is directly proportional to the length of the wire between the two points.
Construction: It consists of a long uniform cross-sectional area and of $$10m$$ in length. The material of the wire should have a high resistivity and low-temperature coefficient. The wires are joined in series by using thick copper strips. A matter scale is also attached to the wooden board.
Comparing E.m.f of two cells:
Using a potentiometer, we can determine the e.m.f of a cell by obtaining the balancing length $$l$$ of the potentiometer wire is equal to the e.m.f of the cell, as no current is being drawn from the cell, Then,
$$E\alpha l$$
or $$E=kl\quad $$ {$$k:Potential gradient$$}
For comparing e.m.f of two cells,
$$\dfrac{E_{1}}{E_2}=\dfrac{kl_1}{kl_2}=\dfrac{l_1}{l_2}$$
Derive the relation between the resistances of the arms of a Wheatstone bridge in its balance condition. Write its two applications.
Look at the diagram above This is a wheatstone bridge.
You can notice four resistances P,Q,R and S connected on the bridge.
You can also notice a galvanometer connected between points B and D.
The four resistances are adjusted in such a way that the current through the galvanometer becomes zero.
This shows that B and D are at the same electric potential.
$$\therefore V_B=V_D$$
The potential difference across B will be $$V_A-V_B$$
Similarly across Q will be $$V_A-V_D$$
Similarly across R will be $$V_B-V_C$$
Similarly across S will be $$V_D-V_C$$
By Ohm's law we know that $$V=IR$$ where $$I$$ is the current.
$$\therefore V_A - V_B= IP$$ $$, V_A - V_D= IQ$$ $$, V_B - V_C= IR$$ and $$ V_D- V_C= IS$$
$$\because V_B=V_D$$ we get
$$IP=IQ$$ and $$IR=IS$$
By dividing the 1st by the 2nd, we get
$$\cfrac{IP}{IR}=\cfrac{IQ}{IS}$$
$$\therefore \cfrac PQ= \cfrac RS$$ this equation is known as the balancing condition in a wheatstone bridge
The length of potentiometer wire is $$10$$m and is connected in series with an accumulator. The e.m.f. of a cell balances against $$250$$ cm length of wire. If the length of potentiometer wire is increased by $$1$$m, calculate the new balancing length of wire.
Why is Wheatstone bridge (Metre Bridge) method not suitable for measurement of very low and very high resistance?
A battery of emf $$E_0$$ = 12V is connected across a 4m long uniform wire having resistance $$\frac{4\Omega}{m}$$. The cells of small emfs $$\varepsilon_1$$ = 2V and $$\varepsilon_2$$ = 4V having internal resistance $$2\Omega$$ and $$6\Omega$$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point N, find the distance of point N from the point A.
Given Resistance of AB wire$$=\dfrac{4\Omega}{m}$$ (as this is per meter)
but AB $$=4m$$ So,
Resistance of AB $$=16\Omega$$
Potential gradient of AB $$=\dfrac{V}{L}$$
$$\varepsilon_{0}=i(8+16)$$
$$=\dfrac{12}{24}=i$$
$$i=0.5A$$
$$\varepsilon_{0}-4=V_{AB}$$
$$12-4=V_{AB}$$
$$V_{AB}=8 volt$$
Potential Gradient$$=\dfrac{V_{AB}}{L}=\dfrac{8}{4}=2\dfrac{V}{m}$$
Potential gradient$$=2\dfrac{V}{m}$$
Net EMF$$=\varepsilon_{eq}=\dfrac{\dfrac{\varepsilon_{1}}{r_{1}}-\dfrac{\varepsilon_{2}}{r_{2}}}{\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}}$$
$$=\dfrac{1-\dfrac{2}{3}}{\dfrac{1}{2}+\dfrac{1}{6}}$$
$$=\dfrac{\dfrac{1}{3}}{\dfrac{4}{6}}$$
$$=\dfrac{1}{3}\dfrac{6}{4}$$
$$=0.5V$$
If at N, there is no deflection
So we know that
$$=\varepsilon=kl$$
k=potential gradient
l=length at null point occurs
$$0.5=2\times l$$
$$l=25cm$$
The length of potentiometer wire is 10 m and is connected in series with an accumulator. The e.m.f. of a cell balances against 250 cm length of wire. If the length of potentiometer wire is increased by 1m, calculate the new balancing length of wire.
Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges ?
State the working principle of potentiometer explain with the help of circuit diagram. How the emf of two primary cells are compared by using the potentiometer.
The principle of potentiometer:
The potential drop along any length of the wire is directly proportional to that length. When a constant current flows through a wire of uniform cross section and composition then, $$v\alpha 1$$
When the key $$K$$ is closed, a constant current flows through the potentiometer wire. By closing key $$K_1$$, the null point can be obtained by cell $$E_1$$. The jockey is moved along the wire and adjusted still galvanometer shoes no deflection.
Suppose, $$AJ_1=l_1$$ is the balancing length for cell $$E_1$$.
Then $$E_1=kl_2$$ where $$k\rightarrow$$ Potential gradient
Now, let the null point of cell $$E_2$$ be obtained by closing key $$K_2$$
Let $$AJ_2=l_2$$
Then, $$E_2=Kl_2$$
$$\therefore \dfrac{E_2}{E_1}=\dfrac{I_2}{I_1}$$
Hence, by increasing a length of a potentiometer's wire, the sensitivity of a potentiometer can be increased.
Find the internal resistance in the given circuit.
State Kirchoff's rules. Use these rules to find the values of current $${I}_{1},{I}_{2},{I}_{3}$$ in the circuit diagram given
Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of $$4\Omega$$ in series with a combination of two resistors ( $$8\Omega$$ each ) in parallel and a voltmeter across parallel combination. Each of them dissipate maximum energy and canwithstand a maximum power of $$16W$$ without melting. Find the maximum current that can flow through the three resistors.
Maximum current through $$4\Omega$$ resister $$=\sqrt{\dfrac{P}{R}}$$
$$=\sqrt{\dfrac{16}{4}}=2\ A$$
MAximum current through each $$8\Omega$$ resister $$=\dfrac{1}{2}\times2=1\ A$$
As shown in following circuit $$A, B$$ and $$C$$ are three ammeter. $$0.5\ A$$ current is shown by ammeter $$B$$.
(a) Find the current passing through Ammeter $$A$$ and $$C$$.
(b) Find total resistance of circuit.
(c) Find emf of the battery (Take resistance of each ammeter as zero.)
Resistance of all the ammeter $$= 0$$
Current through $$6 \Omega$$ resistor $$= 0.5 A$$
Potential drop across $$6 \Omega$$ and $$3 \Omega$$ is same ($$\because$$ they are in parallel)
$$6 \times 0.5 = 3 \times x = 0 \, x = 1A $$ __(I)
Total current $$= x + 0.5 = 1.5 A$$ __(II)
Equivalent Resistance of the circuit
$$R_{eq} = \dfrac{6 \times 3}{6 + 3} + 2 = 4 \Omega$$ __(III)
$$V = i R_{eq} = 0$$ $$V = 1.5 \times 4 = 6 V$$ __(IV)
a) Current through $$A = 1.5 A$$ (from eq. (II))
current through $$C = 1A$$ (from eq (I))
b) Total resistance $$= 4 \Omega$$ (from eq. (III))
c) Emf of Battery $$= 6V $$ (from eq. (IV))
Two diametrically opposite points of a metal ring are connected to terminals of the left gap of meter bridge. In the right gap, resistance of $$15 \Omega$$ is introduced. If the null point is obtained at a distance of $$40\ cm$$ from the left end, find resistance of the wire that is bent in the shape of the ring.
What's the difference between the electrostatic potential of the positive and negative end of an electric cell?
In the adjoint figure, Calculate $$V_{A}\ -\ V_{B}$$ ?
A charge $$q$$ is placed at the centre of the line joining two equal charges $$Q$$. The system of the three charges will be in equilibrium if $$q$$ is equal to?
What is potentimeter? Write its principle.
A coil and a magnet moves with their constant speed $$5$$ m/sec. and $$3$$ m/sec. respectively, towards each other, then induced emf in coil is $$16$$ mV. If both are moves in same direction, find induced emf in coil?
State any two possible sources of errors in meter bridge experiment.
An electric fan runs from the 220 V mains. The current flowing through it is 0.5 A. At what rate is the electrical energy transformed by the fan? How much energy is transformed in 2 min?
What is the advantages of using thick metallic strips of join wires in a potentiometer?
State the two Kirchhoff's laws. Explain briefly, how these rules are justified?
Potentiometer wire length is $$10\, m$$, having a total resistance of $$10\Omega$$. If a battery of emf $$2$$ volts (of negligible internal resistance) and a rheostat are connected to it them the potential gradient is $$20\, mV/m$$; find the resistance imparted through the rheostat
A battery having a steady voltage is connected to the ends of 10 m long potentiometer wire.The null point for a Leclaire cell is obtained at 750 cm.If the length of potentiometer wire is increased by 1 m,the new balancing length in metres will be:-
Find the current in each branch ?
Ref. image
By star method
$$AC = \dfrac{5 \times 5 \times 5 + 5 \times 5}{5} \Rightarrow \dfrac{75}{5} = 15$$
$$AB = 15$$
$$AC= 15$$
Ref. image
$$\begin{array}{l} { R_{ net } }=6.13 \\ i=\dfrac { v }{ r } =\dfrac { { 540 } }{ { 6.13 } } \\ \therefore i=8.1459\, A \end{array}$$
A battery of e.m.f. 15 V and internal resistance 2 $$\Omega$$ is connected to two resistors of resistances 4 $$\Omega$$ and 6 $$\Omega$$joined in series. Find the electrical energy spent per minute in 6 $$\Omega$$ resistor.
What is the current of $$I_2$$?
$$ \begin{array}{l} resistance=12k\Omega \\ { I_{ net } }=\dfrac { { 12 } }{ { 12 } } =1mA \\ v=IR \\ I\propto \dfrac { 1 }{ R } \\ Hence, \\ { I_{ 2 } }=0.5mA \end{array}$$
in a meter bridge$$,$$ the balance point is found to be $$40cm$$ from one end $$($$say end A$$).$$ if a resistance of $$10\Omega $$ is connected in series with $$R,$$ a balance point is obtained $$60 cm$$ from end $$A.$$ What are the values of $$R$$ and $$S$$ $$($$S is other resistance in series with $$R)?$$
If $$R=1 \Omega$$, the current in the branch x is :
$$\begin{array}{l} \dfrac { { x-v } }{ 1 } +\dfrac { { x-0 } }{ 1 } +\dfrac { { x-0 } }{ 1 } =0 \\ \Rightarrow 3x=v \\ \Rightarrow x=\dfrac { v }{ 3 } \end{array}$$
Now,
$$\begin{array}{l} { R_{ eq } }=\dfrac { { 3R } }{ 5 } \\ \therefore V=\dfrac { { 3R } }{ 5 } \times 1=\dfrac { { 3R } }{ 5 } \end{array}$$
Now,
$$\begin{array}{l} { i_{ 1 } }=\dfrac { V }{ R } =\dfrac { 3 }{ 5 } \\ { i_{ 1 } }-{ i_{ 2 } }=\dfrac { { -x } }{ R } =\dfrac { { -1 } }{ 5 } \\ \Rightarrow { i_{ 2 } }={ i_{ 1 } }+\dfrac { 1 }{ 5 } \\ \Rightarrow { i_{ 2 } }=\dfrac { 4 }{ 5 } A \end{array}$$
Which material is used for metere bridge wire and why ?
Electric power =$$\dfrac { { V }^{ 2 } }{ .............. } $$
State Kirchoff's law for an electrical network. Using these laws deduce the condition for balance in a wheatstone bridge.
Three resistors $$2\Omega,4\Omega$$ and $$5\Omega$$ are combined in parallel. What is the total resistance of the combination?
Kirchoff's law,
1.Loop Law- The algebraic sum of voltage drop across each component is equal to the supplied voltage by the source.
2.Junction Law- The net (incoming + outgoing) current at each junction should be zero.
In case of Wheatstone bridge we know that the current through the galvanometer should be zero i.e $$I_g=0$$
Applying Loop law to closed loop ADBA
$$-I_1R_1 + 0 +I_2R_2=0$$ ..... (1) because no source in this arm and the current in arm BD is zero.
For loop CBDC
$$-I_2R_4 +0 +I_1R_3=0$$ ...... (2)
From eqaution-1 $$\dfrac{I_1}{I_2}=\dfrac{R_1}{R_2}$$
From eqaution-1 $$\dfrac{I_1}{I_2}=\dfrac{R_4}{R_3}$$
so $$\dfrac{R_1}{R_2}=\dfrac{R_4}{R_3}$$
Suppose the resulting resistance is $$R$$ then for parallel combination $$\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}$$ so $$R=1.05\Omega$$
An electric fan runs from the 230$$\mathrm { V }$$ mains . The current flowing through it is $$0.4 \mathrm { A }$$ . At what rate electrical energy transferred by the fan?
A wire with an area of cross-section as $$10 \,mm^2$$ has a resistance of $$5 \,\Omega$$, when a potential difference across its ends is $$25 \,V$$. Calculate the drift velocity of electrons. Given the number density of electrons as $$5 \times 10^{20}$$ electrons per cubic meter $$(e/m^{-3})$$.
Write the statement of Kirchhoff's Junction law and loop law.
What do you understand by electricity at rest?
(a) Find the current in the $$20\Omega $$ resistor shown in the figure.
(b) If a capacitor of capacitance $$4\mu F$$ is joined between the points $$A$$ and $$B$$, what would be the electrostatic energy stored in it in steady-state ?
Taking circuit, $$ABCDA$$,
$$10i + 20(i -i_1)-5=0$$
$$\Rightarrow 10i + 20i - 20i_1 - 5 = 0$$
$$\Rightarrow 30i - 20i_1 - 5 = 0$$ ...(1)
Taking circuit $$ABFEA$$,
$$20(i - i_1) - 5 - 10i_i = 0$$
$$\Rightarrow 10i - 20i_1 - 10i_1 - 5 =0$$
$$\Rightarrow 20i - 30i_1 - 5 = 0$$ ...(2)
From (1) and (2)
$$(90 - 40)i_1 = 0$$
$$\Rightarrow i_1 = 0$$
$$30i - 5 = 0$$
$$\Rightarrow i = 5/30 = 0.16A$$.
Current through $$20\Omega$$ is $$0.16A$$.
Two metallic wires $$P_{1}$$ and $$P_{2}$$ of the same material and same length but different cross - sectional areas $$A_{1}$$ and $$A_{2}$$ are joined together and then connected to a source of emf. find the ratio of the drift. Velocities of free electrons in the wires $$P_{1}$$ and $$P_{2}$$ if the wires are connected (i) in series, and (ii) in parallel.
We know that,
$$I=neA\,v_d \Rightarrow v_d=\dfrac{1}{neA}$$
Let $$R_1$$ and $$R_2$$ be resistance of $$P_1$$ and $$P_2$$ and $$A_1$$ and $$A_2$$ are their cross sectional areas respectively.
$$\therefore R_1=p\dfrac{1}{A_1}$$ and $$R_2=p\dfrac{1}{A_2}$$
When connected in seried
$$\therefore \dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{c}{(\dfrac{p^l}{A_1}+\dfrac{p^l}{A_2})naA_1}}{\dfrac{c}{(\dfrac{pl}{A_1}+\dfrac{p^l}{A_2})neA_2}}=\dfrac{A_2}{A_1}$$
When,connected in parallel,
$$\dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{\dfrac{\varepsilon}{pl}.\dfrac{1}{neA_1}}{A_1}}{\dfrac{\varepsilon}{\dfrac{pl}{A_2}}.\dfrac{1}{neA_2}}=1$$
$$I=neA\,v_d \Rightarrow v_d=\dfrac{1}{neA}$$
Let $$R_1$$ and $$R_2$$ be resistance of $$P_1$$ and $$P_2$$ and $$A_1$$ and $$A_2$$ are their cross sectional areas respectively.
$$\therefore R_1=p\dfrac{1}{A_1}$$ and $$R_2=p\dfrac{1}{A_2}$$
When connected in seried
$$\therefore \dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{c}{(\dfrac{p^l}{A_1}+\dfrac{p^l}{A_2})naA_1}}{\dfrac{c}{(\dfrac{pl}{A_1}+\dfrac{p^l}{A_2})neA_2}}=\dfrac{A_2}{A_1}$$
When,connected in parallel,
$$\dfrac{v_{d_1}}{v_{d_2}}=\dfrac{\dfrac{\dfrac{\varepsilon}{pl}.\dfrac{1}{neA_1}}{A_1}}{\dfrac{\varepsilon}{\dfrac{pl}{A_2}}.\dfrac{1}{neA_2}}=1$$
Draw the circuit diagram of a meter bridge to explain how it is based on Wheatstone bridge.
Meter Bridge:-
An electrical instrument based on the wheat stone bridge which is used for measuring the unknown resistance of any material is called as meter bridge.
It consists of a wire whose length is one meter and has uniform cross-sectional area.
Meter bridge works on the principle of Wheatstone bridge. Here, we need to find the unknown resistance '$$S$$'.
Here, length of meter bridge wire $$= l = 1m = 100cm$$ (given)
When the pointer $$C$$ is moved along $$AB$$ such that there is so current in Galvanometer (G). This Wheatstone bridge arrangement which is down to the right side of the diagram
According to Wheatstone bridge principle :-
$$\dfrac{R_{AD}}{R_{AC}} = \dfrac{R_{BD}}{R_{BC}}$$
$$\Rightarrow \dfrac{R}{R_{AC}} = \dfrac{S}{R_{BC}}$$ ....(1)
$$R_{AC} = \dfrac{\rho l_{AC} }{A} = \dfrac{R_o}{l} \times x = \left(\dfrac{R_ox}{l}\right)$$
For $$R_{BC} ;- l_{BC} = (100 - x)$$
$$R_{BC} = \left(\dfrac{\varrho l_{BC}}{A}\right) = \dfrac{R_o}{l} \times (100 - x)$$
$$R_{BC} = \dfrac{R_o(100-x)}{l}$$
Put value of $$R_{AC}$$ and $$R_{BC}$$ in equation (1)
$$\Rightarrow \dfrac{R}{R_{AC}} = \dfrac{5}{R_{BC}}$$
$$\Rightarrow \dfrac{R}{\left(\dfrac{R_ox}{l}\right)} = \dfrac{S}{\dfrac{R_o(100-x)}{l}}$$
$$\Rightarrow \dfrac{R}{x} = \dfrac{S}{100-x}$$
$$\Rightarrow S = \dfrac{R}{x} \times (100- x)$$
$$S = \dfrac{R(100-x)}{x}$$
Here, $$S$$ is unknown
$$R$$ is known
This is the required formula to find unknow resistance value $$S$$.
The potential difference applied across a given resistance is altered so that heat produced per second increases by a factor of 16.By what factor does the applied pd change?
A current of $$1.0 A$$ exists in a copper wire of cross-section $$1.0 mm^2$$. Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is $$9000 kg m^{-3}$$.
Name a device which helps to maintain potential difference across a conductor (say a , bulb)
If a potential difference of $$10 \,V$$ causes a current of $$2 \,A$$ to flow for 1 minute , how much energy is transferred ?
(a) Name a device that helps to measure the potential difference across a conductor.
(b) How much energy is transferred by a $$12 \,V$$ power supply to each coulomb of charge which it moves around a circuit ?
(a) How does the resistance of a pure metal change if its temperature decreases ?
(b) How does the presence of impurities in a metal affect its resistance ?
An electric lamp is marked as $$26\ W, 220\ V$$. It is used for $$10$$ hours daily. Calculate energy consumed in unit of $$kWh$$ per day.
The current through the battery and resistors 1 and 2 in
Figure(a) is $$2.00$$ A. Energy is transferred from the current to
thermal energy $$E_{th}$$ in both resistors. Curves 1 and 2 in Figure(b) give that thermal energy $$E_{th}$$ for resistors 1 and 2, respectively, as
a function of time t. The vertical scale is set by $$E_{th,s}=40.0 \,mJ$$,
and the horizontal scale are set by $$t_s=5.00\, s$$. What is the power of
the battery?
A 5.0 A current is set up in a circuit for 6.0 min by a rechargeable battery with a 6.0 V emf. By how much is the chemical energy of the battery reduced?
How much electrical energy is transferred to thermal energy
in $$2.00$$ h by the electrical resistance of $$400 \,\Omega$$ when the potential
applied across it is $$90.0$$ V?
Fill in the following blanks with suitable words :
Resistance is measured in...........................The resistance of a wire increases as the length.......................; as the
temparature................; and as the cross - sectional area...............
Draw a $$V-I$$ graph for a conductor at two different temperature. What conclusion do you draw from your graph for the variation of resistance of conductor with temperature?
In the graph, $$T_1>T_2$$. The straight line $$A$$ is steeper than the line $$B$$, which leads us to conclude that the resistance of conductor is more at high temperature $$T_1$$, than at low temperature $$T_2$$. Thus, we can say that resistance of a conductor increases with the increase in temperature.
In Figure, $$R_1=6.00\Omega , R_2=18.0\Omega $$, and the ideal battery has emf $$\mathscr{E}=12.0V$$ . What are the (a) size and (b) direction (left or right) of current $$i_1$$? (c) How much energy is dissipated by all four resistors in 1.00 min?
Figure shows a battery connected across a uniform resistor $$R_o$$. A sliding contact can move across the resistor from $$x=0cm$$ at the left to $$x=10cm$$ at the right. Moving the contact changes how much resistance is to the left of the contact and how much is to the right. Find the rate at which energy is dissipated in resistor R as a function of x. Plot the function for $$\mathscr{E}=50V,R=2000\Omega$$, and $$R_o=100\Omega $$
Answer the following:
i) State the principle of working a potentiometer.
ii) Draw a circuit diagram to compare the emf of two primary cells.Write the formula used .How can the sensitivity of a potentiometer be increased?
iii) Write two possible causes for one sided deflection in the potentiometer experiment.
i) Principle of Potentiometer : When a constant current flows through a wire of uniform area of cross-section, the potential drop across any length of the wire is directly proportional to the length.
Circuit Diagram: It consists of a long resistance wire AB of uniform cross-section.Its one end A is connected to the positive terminal of battery B1 whose negative terminal is connected to the other end B of the wire through key k and a rheostat (Rh).The battery B1 connected in circuit is called the driver battery and this circuit is called the primary circuit.By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A a=to B.The fall of potential per unit length of wire is called potential gradient.It is denoted by 'k'.A cell is connected such that its positive terminal is connected to end A and the negative terminal to a jockey through the galvanometer G.This circuit is called the secondary circuit.
In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when k=jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.
The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection.Such a point P is called null deflection point.
VAB is the potential between points A and B and L metre be the length of wire, then the potential gradient
k=VPABLk=VPABL
In this way the emf of a cell may be determined by a potentiometer.
Circuit Diagram: It consists of a long resistance wire AB of uniform cross-section.Its one end A is connected to the positive terminal of battery B1 whose negative terminal is connected to the other end B of the wire through key k and a rheostat (Rh).The battery B1 connected in circuit is called the driver battery and this circuit is called the primary circuit.By the help of this circuit a definite potential difference is applied across the wire AB; the potential falls continuously along the wire from A a=to B.The fall of potential per unit length of wire is called potential gradient.It is denoted by 'k'.A cell is connected such that its positive terminal is connected to end A and the negative terminal to a jockey through the galvanometer G.This circuit is called the secondary circuit.
In primary circuit the rheostat (Rh) is so adjusted that the deflection in galvanometer is on one side when k=jockey is touched on wire at point A and on the other side when jockey is touched on wire at point B.
The jockey is moved and touched to the potentiometer wire and the position is found where galvanometer gives no deflection.Such a point P is called null deflection point.
VAB is the potential between points A and B and L metre be the length of wire, then the potential gradient
k=VPABLk=VPABL
In this way the emf of a cell may be determined by a potentiometer.
An electric bulb of power $$60 \,W$$ is used for $$6 \,h$$ per day. Calculate the 'units' of energy consumed in one day by the bulb.
Answer the following:
Why are the connections between the resistors in a meter bridge made of thick copper strips?
Figure shows a section of a circuit. The resistances are $$R_1=2.0\Omega R_2=4.0\Omega$$, and $$R_3=6.0\Omega$$, and the indicated current is $$i=6.0A$$. The electric potential difference between points A and B that connect the section to the rest of the circuit is $$V_A-V_B=78V$$. (a) Is the device represented by Box absorbing or providing energy to the circuit, and (b) at what rate ?
When is a Wheatstone's bridge most sensitive?
What do you understand by the term current ? State and define its S.I. unit.
Answer the following questions:
1) i) State the principle on which a potentiometer works.How can a given potentiometer be made more sensitive?
ii) In th egraph shown below for two potentiometers,state with reason which of the two potentiometer,A or B, is more sensitive.
2) Two metallic wires, P1 and P2 of the same material and same length but different cross-sectional area.A1 and A2 are joined together and connected to a source of emf.Find the ratio of the drift velocities of free electrons in the two wires when they are connected (i) in series and (ii) in parallel.
Draw the circuit diagram of potentiometer which can be used to determine the internal resistance (E) of a given cell of emf.Describe a method to find the internal resistance of a primary cell.
Determination of Internal Resistance of Potentiometer.
Circuit: A battery $$B_1$$ a rhecostat (Rh) and a key K is connected across the ends A and B of the potentiometer wire such that positive terminal of battery is connected to point A.This completes the primary circuit.
Now the given cell C is connected such that its positive terminal is connected to A and negative terminal is jockey K through a galvanometer. A resistance (R) and a key $$K_1$$ are connected across the cell.This completes the secondary circuit.
Method:
Initially key K is closed and a potential difference is applied across the wire AB.Now rheostat Rh is so adjusted that on touching the jockey K at ends A and B of potentiometer wire, the deflection in the galvanometer is on both sides.Suppose that in this position the potential gradient on the wire is k.
Now Key $$k_1$$ is kept open and the position of null deflection is obtained by sliding and pressing the jockey on the wire,Let this position be $$P_1$$ and $$AP_1=l_1$$
In this situation the cell is in open circuit, therefore the terminal potential difference will be equal to the emf of cell , i.e.,
$$emf \, \varepsilon=kl_1$$.....(i)
Now s suitable resistance R is taken in the resistance box and key $$K-1$$ is closed.Again, the position of null point is obtained on the wire by using jockey J.Let this position on wire be $$P_2$$ and $$AP_2=l_2$$.
In this situation the cell is in closed circuit,therefore the terminal potential difference (V) of cell will be equal to the potential difference across external resistance R,i.e.,
$$V=kl_2$$.........(ii)
Dividing (i) by (ii),we get $$\dfrac{\varepsilon}{V}=\dfrac{l_1}{l_2}$$
$$\therefore$$ Internal resistance of cell, $$r=(\dfrac{\varepsilon}{V}-1)R=(\dfrac{l_1}{l_2}-1)R$$
From this formula r may be calculated.
Circuit: A battery $$B_1$$ a rhecostat (Rh) and a key K is connected across the ends A and B of the potentiometer wire such that positive terminal of battery is connected to point A.This completes the primary circuit.
Now the given cell C is connected such that its positive terminal is connected to A and negative terminal is jockey K through a galvanometer. A resistance (R) and a key $$K_1$$ are connected across the cell.This completes the secondary circuit.
Method:
Initially key K is closed and a potential difference is applied across the wire AB.Now rheostat Rh is so adjusted that on touching the jockey K at ends A and B of potentiometer wire, the deflection in the galvanometer is on both sides.Suppose that in this position the potential gradient on the wire is k.
Now Key $$k_1$$ is kept open and the position of null deflection is obtained by sliding and pressing the jockey on the wire,Let this position be $$P_1$$ and $$AP_1=l_1$$
In this situation the cell is in open circuit, therefore the terminal potential difference will be equal to the emf of cell , i.e.,
$$emf \, \varepsilon=kl_1$$.....(i)
Now s suitable resistance R is taken in the resistance box and key $$K-1$$ is closed.Again, the position of null point is obtained on the wire by using jockey J.Let this position on wire be $$P_2$$ and $$AP_2=l_2$$.
In this situation the cell is in closed circuit,therefore the terminal potential difference (V) of cell will be equal to the potential difference across external resistance R,i.e.,
$$V=kl_2$$.........(ii)
Dividing (i) by (ii),we get $$\dfrac{\varepsilon}{V}=\dfrac{l_1}{l_2}$$
$$\therefore$$ Internal resistance of cell, $$r=(\dfrac{\varepsilon}{V}-1)R=(\dfrac{l_1}{l_2}-1)R$$
From this formula r may be calculated.
Explain with the help of circuit diagram how the value of unknown resistance can be determined using a Wheatstone Bridge. Find the formula used.
Determination of unknown resistance by Wheatstone Bridge.The circuit diagram is completed as shown in fig.P and Q are each $$10\Omega$$ resistance,RB is a resistance box and X is unknown resistance to be measured.B is battery with key $$K_1$$ (in series,G is galvanometer with key $$K_2$$ in series.)
The battery key $$K_1$$ is pressed first and smallest resistance in RB is introduced by pressing galvanometer by $$K_2$$,the defection in galvanometer is noted.Now resistance in RB is introduced by pressing galvanometer key the deflection should be on other side.This is the main precaution before staring the experiment.
Now suitable value of resistancein RB is chosen so that on pressing the galvanometer key,there is no deflection in galvanometer.This resistance R is noted.Now formula used is
$$\dfrac{P}{Q}=\dfrac{R}{X}$$
$$X=\dfrac{Q}{P}R$$ can be calculated.
The battery key $$K_1$$ is pressed first and smallest resistance in RB is introduced by pressing galvanometer by $$K_2$$,the defection in galvanometer is noted.Now resistance in RB is introduced by pressing galvanometer key the deflection should be on other side.This is the main precaution before staring the experiment.
Now suitable value of resistancein RB is chosen so that on pressing the galvanometer key,there is no deflection in galvanometer.This resistance R is noted.Now formula used is
$$\dfrac{P}{Q}=\dfrac{R}{X}$$
$$X=\dfrac{Q}{P}R$$ can be calculated.
When the chemicals in the electric cell are used up, the electric cell stops producing electricity. The electric cell is then replaced with a new one. In case of rechargeable batteries (such as the type used in mobile phones. camera and inverters), they are used again and again. How?
If an electric iron of $$1200\ W$$ is used for $$30$$ minutes every day, find electric energy consumed in the month of April.
How does one understand the temperature dependence of resistivity of a semiconductor ?
If an appliance of power $$ P $$ watt is used for time $$ t $$ hour. How much electrical energy is consumed in $$ kWh $$ ?
What is the advantage of using thick metallic strips to join wires in a potentiometer?
What are the advantages of the null-point method in a Wheatstone bridge? What additional measurements would be required to calculate R (unknown) by any other method?
Explain the meaning of the term e.m.f. of a cell.
Fill in the blanks:
The long line represents .......... and short line represents ........... terminal of electric battery's symbol.
Write the statement of Junction and Loop Law of Kirchhoff's.
"In an electric circuit, the algebraic sum of current at any junction point must be zero."
i.e., $$\sum I = 0$$
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kircchoff's First Law or Junction Law.
It is based on the law of Conservation of Charge. When current in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.
In figure (6.1) at junction $$O$$,
$$I_{1} + I_{2} - I_{3} - I_{4} + I_{5} = 0$$
or $$I_{1} + I_{2} + I_{5} = I_{3} + I_{4} .... (1)$$
In figure $$(6.1)$$ the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.
Kircchoff's Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., $$\sum V = 0 .... (1)$$
The algebraic sum of the emf's in any loop of a circuit is equal to the sum of the voltage across the resistance.
$$\sum E = \sum IR .... (2)$$
Sign convention for applying Loop Rule
1. The product of current and resistance $$(IR)$$ is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance $$(IR)$$ is taken as negative if the resistor is traversed in the opposite direction of assumed current.
In loop we can move from $$A$$ to $$B$$ then the product of $$IR$$ take as positive $$V_{AB} = +IR$$. [In the direction of current]
In a loop we can move from $$B$$ to $$A$$ then the product of $$IR$$ take as negative.
$$\therefore V_{BA} = -IR$$ [Opposite direction of current]
2. When we move from $$B$$ to $$A$$ then $$E$$ is taken positive $$(+)$$. When we move from $$A$$ to $$B$$, then $$E$$ is taken negative $$(-)$$.
Loop law can be understood by the figure 6.4.
In the given figure for point $$d$$ by the Junction Law, current flowing through the resistance $$R\%$$ will be
$$I_{3} = I_{1} + I_{2} ..... (3)$$
In the loop $$adcba$$ by the Kircchoff's Loop Law.
$$I_{1}R_{1} - I_{2}R_{2} = \epsilon_{1} - \epsilon_{2}$$
or $$I_{2}R_{2} = I_{2}R_{2} = \epsilon_{2} - \epsilon_{1} .... (4)$$
In the loop $$defcd$$ by the Loop Law
$$I_{3}R_{3} + I_{2}R_{2} = \epsilon_{0} .... (5)$$
By solving equation $$(3), (4)$$ and $$(5)$$ we can find out electric currents in different branches as well as we can find potential differences across the resistance.
i.e., $$\sum I = 0$$
The sum of currents entering a junction is equal to the sum of currents leaving out at the junction. This law is called Kircchoff's First Law or Junction Law.
It is based on the law of Conservation of Charge. When current in a circuit are steady, charges can not accumulate at any junction. So whatever charge flows towards the junction in any time interval, an equal charge must flow away from that junction in the same time interval.
In figure (6.1) at junction $$O$$,
$$I_{1} + I_{2} - I_{3} - I_{4} + I_{5} = 0$$
or $$I_{1} + I_{2} + I_{5} = I_{3} + I_{4} .... (1)$$
In figure $$(6.1)$$ the currents flowing towards the junction are taken as positive and current flowing away from the junction are taken as negative.
Kircchoff's Second Law or Loop Law
This law is applicable for closed electrical circuit. The algebraic sum of the potential in a closed loop must be zero.
i.e., $$\sum V = 0 .... (1)$$
The algebraic sum of the emf's in any loop of a circuit is equal to the sum of the voltage across the resistance.
$$\sum E = \sum IR .... (2)$$
Sign convention for applying Loop Rule
1. The product of current and resistance $$(IR)$$ is taken as positive if the resistor is traversed in the same direction of assumed current. The product of current and resistance $$(IR)$$ is taken as negative if the resistor is traversed in the opposite direction of assumed current.
In loop we can move from $$A$$ to $$B$$ then the product of $$IR$$ take as positive $$V_{AB} = +IR$$. [In the direction of current]
In a loop we can move from $$B$$ to $$A$$ then the product of $$IR$$ take as negative.
$$\therefore V_{BA} = -IR$$ [Opposite direction of current]
2. When we move from $$B$$ to $$A$$ then $$E$$ is taken positive $$(+)$$. When we move from $$A$$ to $$B$$, then $$E$$ is taken negative $$(-)$$.
Loop law can be understood by the figure 6.4.
In the given figure for point $$d$$ by the Junction Law, current flowing through the resistance $$R\%$$ will be
$$I_{3} = I_{1} + I_{2} ..... (3)$$
In the loop $$adcba$$ by the Kircchoff's Loop Law.
$$I_{1}R_{1} - I_{2}R_{2} = \epsilon_{1} - \epsilon_{2}$$
or $$I_{2}R_{2} = I_{2}R_{2} = \epsilon_{2} - \epsilon_{1} .... (4)$$
In the loop $$defcd$$ by the Loop Law
$$I_{3}R_{3} + I_{2}R_{2} = \epsilon_{0} .... (5)$$
By solving equation $$(3), (4)$$ and $$(5)$$ we can find out electric currents in different branches as well as we can find potential differences across the resistance.
Write the mathematical form of Kirchhoff's junction law.
How can the sensitivity of the potentiometer be increased?
Why the potential gradient of the wire of potentiometer is based on the temperature?
Why copper wire can not be used in potentiometer?
Why the cross-sectional area of the wire should be uniform in a potentiometer?
In a wall, there is a metallic framed window $$(120 cm \times 50cm) $$. The total resistance of a frame is $$ 0.01 \Omega $$. Determine the charges flowing on opening the frame from $$ 90^o $$. [If $$ H=0.36 G] $$
What is standardisation of potentiometer? Write the process of it and draw the necessary diagram.
Standardisation of Potentiometer
For the potentiometer process of finding practical value of potential gradient is called standardisation of potentiometer.
For this process, a standard cell of known emf is required which is connected in secondary circuit as shown in figure.
Standard cell is a type of cell whose emf remains constant for long time. Cadmium cell or Daniel cell is used as standard cell. Its emf is $$1.0186\ V$$ and $$1.08\ V$$ at $$20^{\circ}C$$ respectively.
For finding potential gradient from standard cell, we slide jockey on wire and find out the balancing length where zero deflection is obtained in galvanometer. If the emf of standard cell is es then from the principle of potentiometer.
$$\epsilon_{s} = kxl_{0}$$
$$k = \dfrac {\epsilon_{s}}{l_{0}} ....(1)$$.
For the potentiometer process of finding practical value of potential gradient is called standardisation of potentiometer.
For this process, a standard cell of known emf is required which is connected in secondary circuit as shown in figure.
Standard cell is a type of cell whose emf remains constant for long time. Cadmium cell or Daniel cell is used as standard cell. Its emf is $$1.0186\ V$$ and $$1.08\ V$$ at $$20^{\circ}C$$ respectively.
For finding potential gradient from standard cell, we slide jockey on wire and find out the balancing length where zero deflection is obtained in galvanometer. If the emf of standard cell is es then from the principle of potentiometer.
$$\epsilon_{s} = kxl_{0}$$
$$k = \dfrac {\epsilon_{s}}{l_{0}} ....(1)$$.
What is the sensitivity of potentiometer, how can it be increased?
Sensitivity of Potentiometer
A potentiometer is sensitive if:
1. It is capable of measuring very small potential difference.
2. It shows a significant change in balancing length for a small change in the potential difference being measured. The sensitivity of a potentiometer depends upon the potential gradient along the wire.
Smaller the potential gradient, greater will be the sensitivity of the Potentiometer.
The sensitivity of potentiometer can be increased by reducing the potential gradient. This can be done in two ways.
1. For a given potential difference, the sensivity can be increased by the increasing the length of the potentiometer wire.
2. For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of a rheostat. When in primary circuit current decreases then potential difference across the wire decreases so what without changing in primary circuit for reducing potential gradient the total length of the wire increases.
Graph between $$V$$ and $$L$$
$$\because$$ The slope of graphical line
$$= \tan \theta = \dfrac {Y - axis}{X - axis}$$
or $$Slope = \dfrac {V}{L} =$$ potential gradient $$= k$$.
A potentiometer is sensitive if:
1. It is capable of measuring very small potential difference.
2. It shows a significant change in balancing length for a small change in the potential difference being measured. The sensitivity of a potentiometer depends upon the potential gradient along the wire.
Smaller the potential gradient, greater will be the sensitivity of the Potentiometer.
The sensitivity of potentiometer can be increased by reducing the potential gradient. This can be done in two ways.
1. For a given potential difference, the sensivity can be increased by the increasing the length of the potentiometer wire.
2. For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of a rheostat. When in primary circuit current decreases then potential difference across the wire decreases so what without changing in primary circuit for reducing potential gradient the total length of the wire increases.
Graph between $$V$$ and $$L$$
$$\because$$ The slope of graphical line
$$= \tan \theta = \dfrac {Y - axis}{X - axis}$$
or $$Slope = \dfrac {V}{L} =$$ potential gradient $$= k$$.
Write down the energy changes in the following electrical devices with respect to their use.
| Device | Use | Energy change |
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What may be factors affecting the induced emf ?
What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Energy is precious, especially electrical energy. Society must be convinced of the necessity of reducing the consumption of electrical energy. Prepare and propogate posters for this purpose.
What is battery? How do you arrange the cells in a battery?
Combination of two or more cells is called battery. The cells should be arranged as the positive terminal of one cell is connected to the negative terminal of the next cell It is shown in the below figure.
Fill in the blanks:
The combination of two or more cells is called a ...............
A bulb of $$60\,W$$ in a classroom remains switched on for a long time from $$9$$ am to $$5$$ pm due to the negligence of students in the class.
Calculate in joule the electric energy consumed by the bulb ?
A bulb of $$60\,W$$ in a classroom remains switched on for a long time from $$9$$ am to $$5$$ pm due to the negligence of students in the class.
If similar negligence had happened in $$3$$ classrooms, how many units of electric energy would have been wasted ?
Did the resistance increase or decrease when the temperature was increased?
Two lamps, one rated $$60\ W$$ at $$220\ V$$ and the other $$40\ W$$ at $$220\ V$$, are connected in parallel to the electric supply at $$220\ V$$.
(a) Draw a circuit diagram to show the connections
(b) Calculate the current drawn from the electric supply.
(c) Calculate the total energy consumed by the two lamps together when they operate for one hour.
(c) $$E = P \times t = (40\ W + 60\ W) \times 1\ h $$
$$ = 100\ Wh $$ or $$0.1\ kWh $$.
$$ = 100\ Wh $$ or $$0.1\ kWh $$.
Current of $$4\ A$$ flows through a $$12\ V$$ car headlight bulb for $$10$$ minutes. How much energy transfer occurs during this time?
Figure shows a circuit containing a battery of electromotive force (e.m.f.) $$12\,V$$ and a heater of resistance $$6.0\Omega.$$
State what is meant by electromotive force (e.m.f.).
An electric bulb has marking 110 V, 100 W on it.
a. How much energy is used per second by the circuit?
b. What is the resistance of electric bulb?
How is the direction of electric current related to the direction of flow of electrons in a wire?
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor $$R=10.0\Omega$$ is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf $$\epsilon$$?
Describe the experiment to find internal resistance of a cell by Potentiometer under the following points.
(i) Labelled electric circuit diagram.
(ii) Derivation of formula
(iii) Observation table
(iv) Two precautions.
(i) Labelled electric circuit diagram: Potentiometer wire, Rh$$\rightarrow$$ variable resistance, C- accumulator, D-experimental cell, G-galvanometer, R.B- resistance box, $$K_1, K_2$$-keys, J-jockey.
Note: Direction of current is shown by arrows.
(ii) Derivation of formula: Let the e.m.f. of the experimental cell is E and its internal resistance is r. If the resistance R from the resistance box is connected with this cell then the potential difference at the ends of the cell becomes V.
$$E=I(R+r)$$ and $$V=IR$$
then $$r=\left(\displaystyle\frac{E}{V}-1\right)R$$ ..........(1)
Let the potential gradient of the potentiometer be v.
When R is not connected, then let the balance point is found at distance $$l_1$$ from A, then
$$E=vl_1$$ .........(2)
Now when R is connected with the cell, then let the balance point is obtained at a distance $$l_2$$ from A. The potential difference at the ends of the cell
$$V=vl_2$$ ..........(3)
Substituting eqns. (2) and (3) in (1),
$$r=\left(\displaystyle\frac{vl_1}{vl_2}-1\right)R=\left(\displaystyle\frac{l_1}{l_2}-1\right)R$$
(iii) Observation table:
(iv) Two precautions: (1) the diameter of potentiometer wire should be unifrom throughout, otherwise potential gradient will not remain same everywhere.
(2) Shunt should be connected with the galvanometer before null position but near the null point the shunt should be removed.
Note: Direction of current is shown by arrows.
(ii) Derivation of formula: Let the e.m.f. of the experimental cell is E and its internal resistance is r. If the resistance R from the resistance box is connected with this cell then the potential difference at the ends of the cell becomes V.
$$E=I(R+r)$$ and $$V=IR$$
then $$r=\left(\displaystyle\frac{E}{V}-1\right)R$$ ..........(1)
Let the potential gradient of the potentiometer be v.
When R is not connected, then let the balance point is found at distance $$l_1$$ from A, then
$$E=vl_1$$ .........(2)
Now when R is connected with the cell, then let the balance point is obtained at a distance $$l_2$$ from A. The potential difference at the ends of the cell
$$V=vl_2$$ ..........(3)
Substituting eqns. (2) and (3) in (1),
$$r=\left(\displaystyle\frac{vl_1}{vl_2}-1\right)R=\left(\displaystyle\frac{l_1}{l_2}-1\right)R$$
(iii) Observation table:
| S.No. | When cell is in open circuit balancing length $$l_1$$(cm) | When cell is in closed circuit balancing length $$l_2$$(cm) | Resistance used from resistance box R(ohm) | Internal resistance $$r=R\left(\displaystyle\frac{l_1}{l_2}-1\right)$$ohm |
(2) Shunt should be connected with the galvanometer before null position but near the null point the shunt should be removed.
$$'n'$$ cells, each of emf $$'e'$$ and internal resistance $$'r'$$ are joined in series to form a row. $$'m'$$ such rows are connected in parallel to form a battery of $$N = mn$$ cells. This battery is connected to an external resistance $$'R'$$
Show that current $$'I'$$ flowing through the external resistance $$'R'$$ is given by:
$$I = \dfrac {Ne}{mR + nr}$$
$$I = \dfrac {ne}{\left (\dfrac {nr}{m} + R\right )} = \dfrac {ne\times m}{nr + mR} = \dfrac {nme}{nr + mR}$$
but $$nm = N$$
So, $$I = \dfrac {Ne}{nr + mR}$$
but $$nm = N$$
So, $$I = \dfrac {Ne}{nr + mR}$$
Deduce the principle of potentiometer with the help of necessary circuit diagram.
Derive the expression to find the unknown resistance in the balanced condition of wheastone bridge.
$$P=$$ Resistance of $$AB=kl$$
$$G=$$ Resistance of $$BC=k(100-l)$$
Applying wheatstone bridge condition
$$\dfrac{R}{X}=\dfrac{P}{G}=\dfrac{kl}{k(100-l)}$$
or $$\boxed{X=\dfrac{(100-l)}{l})R}\leftarrow $$Unknown resistance
What is the effect of temperature on the resistance of a metal? The resistance of a platinum resistances thermometer at $$0^o$$C temperature is $$3.0\Omega$$ and at $$100^o$$C it is $$3.75\Omega$$. Its resistance at an unknown temperature is $$3.15\Omega$$. Find the value of the unknown temperature.
In the circuit shown in figure, $${ E }_{ 1 }=3V,{ E }_{ 2 }=2V,{ E }_{ 3 }=1V$$ and $${ r }_{ 1 }={ r }_{ 2 }={ r }_{ 3 }=1ohm$$
(a) Find the potential difference between the points $$A$$ and $$B$$ and the currents through each branch.
(b) If $${r}_{2}$$ is short circuited and the point $$A$$ is connected to point $$B$$ through a resistance $$R$$, find the currents through $${E}_{1},{E}_{2},{E}_{3}$$ and the resistor $$R$$.
Define emf and internal resistance of a cell
Consider the potentiometer circuit arranged as in figure. The potentiometer wire is $$600$$cm long. At what distance from the point A should the jockey touch the wire to get zero deflection in the galvanometer?
Explain the principle of potentiometer.
The resistance of an iron wire and a copper wire at $$20^{o}C$$ are $$3.9\Omega$$ and $$4.1\Omega$$ respectively. At what temperature will the resistance be equal ? Temperature coefficient of resistivity for iron is $$5.0\times 10^{-3}K^{-1}$$ and for copper it is $$4.0\times 10^{-3}K^{-1}$$. Neglect any thermal expansion.
On what principle is Kirchhoff's current law based?
State the working principle of a potentiometer with the help of the circuit diagram, explain how a potentiometer is used to compare the emf's of the two primary cell obtain the required expression used for comparing the emf's.
Calculate the steady current through the $$2\Omega$$ resistor in the circuit shown in the figure.
The resistance of the rheostat shown in figure is $$30 \Omega.$$ Neglecting the meter resistance, find the minimum and maximum currents through the ammeter as the rheostat is varied.
Define e.m.f. of a cell. How can you compare the emf of two cells using potentiometer?
The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, it is measured in volts (V)
Two lamps, one rated $$40 W \,at \,220 V$$ and the other $$60 W \,at\, 220 V$$, are connected in parallel to the electric supply at $$220 V$$.
(a) Draw a circuit diagram to show the connections.
(b) Calculate the current drawn from the electric supply.
(c) Calculate the total energy consumed by the two lamps together when they operate for one hour.
What is Wheatstone bridge? Deduce the balanced condition of it with the help of Kirchhoff's law.
Construction : The construction of Wheatstone's bridge is shown in following diagram (figure 6.5) in which a quadrilateral $$ABCD$$ is prepared by connecting four resistance $$P, Q, R$$ and $$S$$. A cell is connected between the points $$A$$ and $$C$$ through key $$K_{1}$$ which is known as battery key and a galvanometer between the point $$B$$ and $$D$$ through another key $$K_{2}$$ which is known as galvanometer key. If on pressing the key $$K_{1}$$ and $$K_{2}$$, there is no deflection in galvanometer, then the bridge is said to be balanced condition. In this situation the relation between all the resistance is given by:
$$\dfrac {P}{Q} = \dfrac {R}{S}$$
Principle of Wheatstone Bridge and Condition of Balance
When battery key $$K_{1}$$ is pressed, then main current $$I$$ starts flowing in the circuit. At junction $$A$$ this current splits in two parts $$I_{1}$$ and $$I_{2}$$ as shown in figure 6.6. These current $$I_{2}$$ and $$I_{2}$$ again obtain two paths at junctions $$B$$ and $$D$$ respectively. Now following three situations are possible:
(i) When $$V_{B} > V_{D}$$. then current $$I_{2}$$ passes as such through resistance $$S$$ but current $$I_{1}$$ splits in two parts at junction $$B$$, one part $$I_{g}$$ passes through galvanometer showing deflection in one direction while rest current $$(I_{1} - I_{g})$$ passes through $$Q$$.
(ii) When $$V_{B} < V_{D}$$, then opposite situation to previous situation (i) is observed i.e. current $$I_{2}$$ splits in two parts at $$D$$. Now the current $$I_{g}$$ produces deflection in galvanometer in opposite direction to previous deflection.
(iii) When $$V_{B} = V_{D}$$, then no current passes through galvanometer i.e. galvanometer shows no deflection because now $$I_{g} = 0$$. The situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called 'Bridge circuit' Thus in balanced condition.
$$V_{B} = V_{D}$$
Therefore $$V_{A} - V_{B} = V_{A} - V_{D}$$
or $$I_{1}P = I_{2}R$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {R}{P} .... (1)$$
and $$V_{B} - V_{C} = V_{D} - V_{C}$$
or $$I_{1} \times Q = I_{2} \times S$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {S}{Q} .... (2)$$
Now from equation $$(1)$$ and $$(2)$$
$$\dfrac {R}{P} = \dfrac {S}{Q}$$ or $$\dfrac {R}{S} = \dfrac {P}{Q}$$
or $$\dfrac {P}{Q} = \dfrac {R}{S} .... (3)$$
Therefore, it is clear that in balanced situation of the bridge, the ratio of resistances of any two corresponding arms of quadrilateral $$ABCD$$ is equal to ratio of two remaining corresponding arms".
From equation $$(3)$$
$$S = \dfrac {Q}{P}\times R$$
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below:
(i) $$P$$ and $$Q$$ are called ratio arms.
(ii) $$R$$ is called variable resistance arm.
(iii) $$S$$ is called unknown resistance arm.
$$\dfrac {P}{Q} = \dfrac {R}{S}$$
Principle of Wheatstone Bridge and Condition of Balance
When battery key $$K_{1}$$ is pressed, then main current $$I$$ starts flowing in the circuit. At junction $$A$$ this current splits in two parts $$I_{1}$$ and $$I_{2}$$ as shown in figure 6.6. These current $$I_{2}$$ and $$I_{2}$$ again obtain two paths at junctions $$B$$ and $$D$$ respectively. Now following three situations are possible:
(i) When $$V_{B} > V_{D}$$. then current $$I_{2}$$ passes as such through resistance $$S$$ but current $$I_{1}$$ splits in two parts at junction $$B$$, one part $$I_{g}$$ passes through galvanometer showing deflection in one direction while rest current $$(I_{1} - I_{g})$$ passes through $$Q$$.
(ii) When $$V_{B} < V_{D}$$, then opposite situation to previous situation (i) is observed i.e. current $$I_{2}$$ splits in two parts at $$D$$. Now the current $$I_{g}$$ produces deflection in galvanometer in opposite direction to previous deflection.
(iii) When $$V_{B} = V_{D}$$, then no current passes through galvanometer i.e. galvanometer shows no deflection because now $$I_{g} = 0$$. The situation is said to be balance condition of Wheatstone bridge. It is clear that current is flowing in the circuit but there is no effect of current on galvanometer arm which resembles with the bridge on a river. This is why this circuit is called 'Bridge circuit' Thus in balanced condition.
$$V_{B} = V_{D}$$
Therefore $$V_{A} - V_{B} = V_{A} - V_{D}$$
or $$I_{1}P = I_{2}R$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {R}{P} .... (1)$$
and $$V_{B} - V_{C} = V_{D} - V_{C}$$
or $$I_{1} \times Q = I_{2} \times S$$
or $$\dfrac {I_{1}}{I_{2}} = \dfrac {S}{Q} .... (2)$$
Now from equation $$(1)$$ and $$(2)$$
$$\dfrac {R}{P} = \dfrac {S}{Q}$$ or $$\dfrac {R}{S} = \dfrac {P}{Q}$$
or $$\dfrac {P}{Q} = \dfrac {R}{S} .... (3)$$
Therefore, it is clear that in balanced situation of the bridge, the ratio of resistances of any two corresponding arms of quadrilateral $$ABCD$$ is equal to ratio of two remaining corresponding arms".
From equation $$(3)$$
$$S = \dfrac {Q}{P}\times R$$
Thus, by determining the balanced situation, unknown resistance can be determined.
The arms of Wheatstone bridge are known as particulars names given below:
(i) $$P$$ and $$Q$$ are called ratio arms.
(ii) $$R$$ is called variable resistance arm.
(iii) $$S$$ is called unknown resistance arm.
Draw the necessary diagram for measuring small resistance by potentiometer?
Measurement of Small Resistance
In figure necessary circuit is shown.
Primary circuit is as usual. In secondary circuit, unknown small resistance $$r$$, known resistance $$R$$, rheostat $$Rh$$, key $$K_{2}$$ and emf $$\epsilon'$$ all are connected in series. Higher potential end of $$R$$, connects at point $$A$$ which has higher potential, while lower potential end of $$R$$ and $$r$$ connects with the ends $$1$$ and $$3$$ of two way key. Terminal $$2$$ of two way key connects with galvanometer end. Galvanometer also connects with jockey.
Working : Key $$K_{1}$$ is closed, key $$K_{2}$$ is also closed, now plug is inserted between terminal $$1$$ and $$2$$ of two way key. Now find out potential difference across resistance $$R$$ by potentiometer. If $$I$$ current flowing through secondary circuit, potential difference $$V$$ across the ends of resistance $$R$$ and the balancing length is $$l_{1}$$ for $$V$$.
By the principle of potentiometer
$$V = kl_{1} .... (1)$$
but $$V = IR$$ (by Ohm's law)
$$IR = kl_{1} .... (2)$$
Eliminate the plug between $$1$$ and $$2$$ and insert it between $$2$$ and $$3$$. Now $$R$$ and $$r$$ connect in series. Current $$I$$ remains same in this circuit also now the potential difference across the ends of the $$(R + r)$$ is $$V_{1}$$ and balancing length for it is $$l_{2}$$
$$V_{1} = kl_{2}$$
but $$V_{1} = I(R + r)$$
Hence $$I(R + r) = kl_{2} .... (3)$$
From equations (2) and (3)
$$\dfrac {I(R + r)}{IR} = \dfrac {kl_{2}}{kl_{1}}$$
$$1 + \dfrac {r}{R} = \dfrac {l_{2}}{l_{1}}$$ or $$\dfrac {r}{R} = \dfrac {l_{2}}{l_{1}} - 1$$
$$r = \left (\dfrac {l_{2} - l_{1}}{l_{1}}\right )R ....(4)$$
Putting the values $$l_{1}, l_{2}$$ and $$R$$, then internal resistance of primary cell $$(r)$$ can be determined.
In figure necessary circuit is shown.
Primary circuit is as usual. In secondary circuit, unknown small resistance $$r$$, known resistance $$R$$, rheostat $$Rh$$, key $$K_{2}$$ and emf $$\epsilon'$$ all are connected in series. Higher potential end of $$R$$, connects at point $$A$$ which has higher potential, while lower potential end of $$R$$ and $$r$$ connects with the ends $$1$$ and $$3$$ of two way key. Terminal $$2$$ of two way key connects with galvanometer end. Galvanometer also connects with jockey.
Working : Key $$K_{1}$$ is closed, key $$K_{2}$$ is also closed, now plug is inserted between terminal $$1$$ and $$2$$ of two way key. Now find out potential difference across resistance $$R$$ by potentiometer. If $$I$$ current flowing through secondary circuit, potential difference $$V$$ across the ends of resistance $$R$$ and the balancing length is $$l_{1}$$ for $$V$$.
By the principle of potentiometer
$$V = kl_{1} .... (1)$$
but $$V = IR$$ (by Ohm's law)
$$IR = kl_{1} .... (2)$$
Eliminate the plug between $$1$$ and $$2$$ and insert it between $$2$$ and $$3$$. Now $$R$$ and $$r$$ connect in series. Current $$I$$ remains same in this circuit also now the potential difference across the ends of the $$(R + r)$$ is $$V_{1}$$ and balancing length for it is $$l_{2}$$
$$V_{1} = kl_{2}$$
but $$V_{1} = I(R + r)$$
Hence $$I(R + r) = kl_{2} .... (3)$$
From equations (2) and (3)
$$\dfrac {I(R + r)}{IR} = \dfrac {kl_{2}}{kl_{1}}$$
$$1 + \dfrac {r}{R} = \dfrac {l_{2}}{l_{1}}$$ or $$\dfrac {r}{R} = \dfrac {l_{2}}{l_{1}} - 1$$
$$r = \left (\dfrac {l_{2} - l_{1}}{l_{1}}\right )R ....(4)$$
Putting the values $$l_{1}, l_{2}$$ and $$R$$, then internal resistance of primary cell $$(r)$$ can be determined.
Answer the following questions:
i) Using Kirchhoff's rules,calculate the current in the arm AC of the given circuit.
ii) On what principle does the meter bridge work? Why are the metal strips used in the bridge?
How much energy is given to each coulomb of charge passing through a $$6 V$$ battery?
What is the meter bridge? On which principle does it work? Explaining its construction obtain the expression for determining unknown resistance with the help of meter bridge. Given necessary diagram also.
It is the simplest practical application of the Wheatstone bridge that is used to measure an unknown resistance.
Construction : It consists of usually one meter long Manganin or constant on wire of uniform cross-sectional area, stretched along a meter scale fixed over a wooden board and with its two ends soldered to two $$L-shaped$$ thick copper strips $$M$$ and $$N$$. Between these two copper strips, another copper strip $$I$$ is fixed so as to provide two gaps $$ab$$ and $$a_{1}b_{1}$$. A resistance box $$R.B.$$ is connected in the gap $$ab$$ and the $$R.B.G$$ unknown resistance $$S$$ is connected to the gap $$ab$$. and the unknown resistance $$S$$ is connected to the gap $$a_{1}b_{1}$$. A source of emf $$e$$ is connected across $$AC$$. A movable jockey and a galvanometer are connected at $$BD$$ as shown in the figure.
Working : After taking out a suitable resistance $$R$$ from the resistance box, we touch the jockey at point $$A$$ and $$C$$ and observe opposite deflection in galvanometer. If galvanometer shows deflection in one direction then we change the value of known resistance until galvanometer shows opposite deflection.
Now the jockey is moved along the wire $$AC$$ till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge. If $$P$$ and $$Q$$ are the resistance of the parts $$AB$$ and $$BC$$ of the wire, then for the balanced condition of the bridge we have
$$\dfrac {P}{Q} = \dfrac {R}{S} .... (1)$$
If resistance of unit length of the wire is $$\sigma \dfrac {ohm}{cm}$$ and the balance point occurs at $$l\ cm$$ from $$A$$ then the length of part $$BC$$ will be $$(100 - l) cm$$.
$$P =$$ Resistance of wire of $$AB$$ part $$= \sigma l .... (2)$$
$$Q =$$ Resistance of wire of $$BC$$ part $$= \sigma (100 - l) .... (3)$$
In equation (1) putting the value $$P$$ and $$Q$$
$$\dfrac {R}{S} = \dfrac {\sigma l}{\sigma (100 - l)}$$
$$S = \dfrac {(100 - l)}{l}R ..... (4)$$
Knowing $$I$$ and $$R$$, unknown resistance can be determined.
The formula of determination of specific resistance of unknown resistance $$(S)$$
We know that
Unknown resistance $$(S) = \rho \dfrac {P}{A}$$
Where $$l' =$$ Length of unknown resistance
$$A =$$ Area of cross section $$= \pi r^{2}$$
From equation (4)
$$\dfrac {\rho l'}{\pi r^{2}} = \dfrac {(100 - l)}{l} R$$
$$\therefore \rho \dfrac {(100 - l)\times \pi r^{2}}{l \times l'} ohm\times cm$$
Limitations of Meter Bridge:
1. Copper strips have their own resistance but in the calculation of meter bridge formula we neglect this resistance so that our results are errorful. For minimising this error, at time of performing experiment we should calculate unknown resistance by interchanging the position of unknown resistance and known resistance. In this process we take average value of observations by which error can be minimised.
2. The resistance of end points affect the sensitivity of meter bridge. To nulify this effect Carey-Foster's bridge is used.
3. Current should not flow for long time through the wire otherwise wire will heat up and the resistance of wire will be changed.
4. Jockey should not slide tightly on resistance wire otherwise it's cross-section area will change. That is why the resistance of the wire will be changed.
Construction : It consists of usually one meter long Manganin or constant on wire of uniform cross-sectional area, stretched along a meter scale fixed over a wooden board and with its two ends soldered to two $$L-shaped$$ thick copper strips $$M$$ and $$N$$. Between these two copper strips, another copper strip $$I$$ is fixed so as to provide two gaps $$ab$$ and $$a_{1}b_{1}$$. A resistance box $$R.B.$$ is connected in the gap $$ab$$ and the $$R.B.G$$ unknown resistance $$S$$ is connected to the gap $$ab$$. and the unknown resistance $$S$$ is connected to the gap $$a_{1}b_{1}$$. A source of emf $$e$$ is connected across $$AC$$. A movable jockey and a galvanometer are connected at $$BD$$ as shown in the figure.
Working : After taking out a suitable resistance $$R$$ from the resistance box, we touch the jockey at point $$A$$ and $$C$$ and observe opposite deflection in galvanometer. If galvanometer shows deflection in one direction then we change the value of known resistance until galvanometer shows opposite deflection.
Now the jockey is moved along the wire $$AC$$ till there is no deflection in the galvanometer. This is the balanced condition of the Wheatstone bridge. If $$P$$ and $$Q$$ are the resistance of the parts $$AB$$ and $$BC$$ of the wire, then for the balanced condition of the bridge we have
$$\dfrac {P}{Q} = \dfrac {R}{S} .... (1)$$
If resistance of unit length of the wire is $$\sigma \dfrac {ohm}{cm}$$ and the balance point occurs at $$l\ cm$$ from $$A$$ then the length of part $$BC$$ will be $$(100 - l) cm$$.
$$P =$$ Resistance of wire of $$AB$$ part $$= \sigma l .... (2)$$
$$Q =$$ Resistance of wire of $$BC$$ part $$= \sigma (100 - l) .... (3)$$
In equation (1) putting the value $$P$$ and $$Q$$
$$\dfrac {R}{S} = \dfrac {\sigma l}{\sigma (100 - l)}$$
$$S = \dfrac {(100 - l)}{l}R ..... (4)$$
Knowing $$I$$ and $$R$$, unknown resistance can be determined.
The formula of determination of specific resistance of unknown resistance $$(S)$$
We know that
Unknown resistance $$(S) = \rho \dfrac {P}{A}$$
Where $$l' =$$ Length of unknown resistance
$$A =$$ Area of cross section $$= \pi r^{2}$$
From equation (4)
$$\dfrac {\rho l'}{\pi r^{2}} = \dfrac {(100 - l)}{l} R$$
$$\therefore \rho \dfrac {(100 - l)\times \pi r^{2}}{l \times l'} ohm\times cm$$
Limitations of Meter Bridge:
1. Copper strips have their own resistance but in the calculation of meter bridge formula we neglect this resistance so that our results are errorful. For minimising this error, at time of performing experiment we should calculate unknown resistance by interchanging the position of unknown resistance and known resistance. In this process we take average value of observations by which error can be minimised.
2. The resistance of end points affect the sensitivity of meter bridge. To nulify this effect Carey-Foster's bridge is used.
3. Current should not flow for long time through the wire otherwise wire will heat up and the resistance of wire will be changed.
4. Jockey should not slide tightly on resistance wire otherwise it's cross-section area will change. That is why the resistance of the wire will be changed.
Conceptual Questions
Use the atomic theory of matter to explain why the resistance of a material should increase as its temperature increases.
Observe and record the meter reading in your house for $$10$$ consecutive days. Based on this, find out the average consumption per day. Find out methods to reduce consumption and record them. Present your findings in the Energy Club.
Explain electrical energy and derive its formula.
With the help of a labelled diagram, show that the balancing condition of a Wheatstone bridge is $$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$ where the terms have their usual meaning.
It consists of four resistances $$R_1, R_2, R_3$$ and $$R_4$$ which are connected to form a quadrilateral ABCD as shown in the figure.
Let I be the current supplied by the cell. At the junction 'A', the current I is divided into two parts; $$I_1$$ through.
$$R_1$$ and $$I_1 + I_2$$ (Kirchhoff's law).
When the network is balanced $$V_B = V_D$$ ie $$I_g =0$$ So the current through resistance $$R_2$$ along BC is $$I_1$$ and the current through the resistance $$R_4$$ is $$I_2$$ along DC.
By applying Kirchhoff's 2nd law to closed loop 'ABGDA', we get
$$-I_1R_1 + 0 \times G + I_2R_3 = 0$$
$$I_1R_1 = I_2R_3$$ ...(1)
Similarly, by applying Kirchhoff's 2nd law to the closed loop 'BCDGB', we get
$$-I_1R_2 + I_2R_4 + 0 \times G = 0$$
$$\therefore I_1R_2 = I_2R_4 $$...(2)
Dividing equation (1) by equation (2), we get .
$$\frac{R_1}{R_2} = \frac{R_3}{R_4}$$ or $$\frac{R_1}{R_3} = \frac{R_2}{R_4}$$ .....(3)
This is the balancing condition of Wheatstone's network
What is potentiometer? Explain its principle. Describe the method of determining a small resistance by potentiometer. Make necessary diagram also.
Principle of
Potentiometer
Potentiometer : Potentiometer is a apparatus
with help of which the e.m.f. of a cell and potential difference can be
measured accurately.
Principle : Suppose $$AB$$ is a resistance of
uniform area of cross section and length $$LAB$$. If a standard cell of emf
$$E$$ and negligible internal resistance is connected across the ends of wire
$$A$$ and $$B$$ with the help of connection wires of negligible resistance,
then the potential difference across the wire will be $$E$$. Therefore
potential gradient.
$$k = \dfrac {E}{L_{AB}} .... (1)$$
If we consider a point $$C$$ at a distance $$I$$
from $$A$$, then
$$V_{AC} = kI .... (2)$$
Now if a cell of unknown emf $$E'$$ is connected
between $$A$$ and $$C$$ as shown in the figure, then the deflection in
galvanometer will depend on the fact that whether $$V_{AC} > E'$$ or $$E'
> V_{AC}$$. When $$V_{AC} = E'$$, then galvanometer will show no deflection.
This fact can be understood well with the help of following equivalent circuit
between $$A$$ and $$C$$. This is null position galvanometer.
$$E' = V_{AC}$$
or $$E' = kI ... (3)$$
This is the principle of potentiometer.
If $$E' > E$$, then galvanometer will show
the deflection in one direction only. In this situation the measurement of
$$E'$$ is not possible.
If the resistance of potentiometer wire is
$$RAB$$ and the current is $$I$$ then from the ohm's law
$$E = IR_{AB} .... (4)$$
From the equation (1)
$$k = \dfrac {IR_{AB}}{L_{AB}}$$
Let $$\dfrac {R_{AB}}{L_{AB}} =$$ Density of
wire is represented by the $$\rho$$
$$\therefore k = \rho I .... (5)$$
The potential gradient
$$(k) = \dfrac {\text {The potential drop across
the}R_{AB}}{\text {Length of potentiometer wire}(L_{AB})}$$
$$\therefore k = \dfrac {V_{AB}}{L_{AB}} =
\dfrac {I_{AB} \times R_{AB}}{L_{AB}} .... (6)$$
By the circuit diagrma 6.11 (b) the current flow
$$I = I_{AB} = \dfrac {E}{\text {Resistance of
circuit}} .... (7)$$
The resistance of the circuit $$= R_{AB} + R_{h}
+ r$$
$$\therefore I = I_{AB} = \dfrac {E}{R_{AB} +
R_{h} + r}$$
Put in equation (6)
$$\therefore k = \dfrac {E}{(R_{AB} + R_{h} +
r)} \times \dfrac {R_{AB}}{L_{AB}} .... (8)$$
Value of potential gradient under different
conditions:
Case - 1 : When only the battery of negligible
internal resistance is connected to potentiometer.
$$\therefore r = 0$$ and $$R_{h} = 0$$
From the equation (8)
$$k = \dfrac {E}{(R_{AB} + 0 + 0)} \times \dfrac
{R_{AB}}{L_{AB}}$$
$$\therefore k = \dfrac {E}{R_{AB}} \times
\dfrac {R_{AB}}{L_{AB}} = \dfrac {E}{L_{AB}} .... (9)$$
Case - 2 : When the battery of internal
resistance $$(r)$$ is connected to the potentiometer
$$\therefore R_{h} = 0$$
From equation (8)
$$k = \dfrac {E}{(R_{AB} + r + 0)} \times \dfrac
{R_{AB}}{L_{AB}}$$
$$\therefore k = \left (\dfrac {E}{R_{AB} +
r}\right ) \times \dfrac {R_{AB}}{L_{AB}} .... (10)$$
Case - 3 : When a some other resistance or
rheostat of resistance $$(Rh)$$ is connected in series with the battery in the
circuit of potentiometer. Then the potential gradient is calculated by equation
(8).
$$k = \left (\dfrac {E}{R_{AB}+ r + R_{h}}\right
) \times \dfrac {R_{AB}}{L_{AB}}$$
Factors effecting the value of potential
gradient:
From the equation (5)
$$k = \rho I$$
Put the value of $$(\rho) = \dfrac
{R_{AB}}{L_{AB}}$$
$$\therefore k = \dfrac {R_{AB}I}{L_{AB}} ....
(11)$$
We know that
$$R_{AB} = \dfrac {\rho_{AB}\times
L_{AB}}{\alpha_{AB}}$$
but $$\alpha_{AB} = \pi r^{2}$$
[where $$r =$$ radius of potentiometer]
$$\rho_{AB} =$$ Specific resistance
From the equation (11)
$$k = \dfrac {I}{L_{AB}} \times \dfrac
{\rho_{AB}}{\pi r^{2}} L_{AB}$$
$$k = \dfrac {\rho_{AB} I}{\pi r^{2}} ....
(12)$$
$$\therefore$$ The potential gradient depends
upon the following factors
(i) $$k\propto I .... (13)$$
(ii) $$k\propto \rho_{AB} .... (14)$$
(iii) $$k\propto \dfrac {1}{r^{2}} .... (15)$$.
Measurement of Small Resistance
In figure necessary circuit is shown.
Primary circuit is as usual. In secondary
circuit, unknown small resistance $$r$$, known resistance $$R$$, rheostat
$$Rh$$, key $$K_{2}$$ and emf $$\epsilon'$$ all are connected in series. Higher
potential end of $$R$$, connects at point $$A$$ which has higher potential,
while lower potential end of $$R$$ and $$r$$ connects with the ends $$1$$ and $$3$$
of two way key. Terminal $$2$$ of two way key connects with galvanometer end.
Galvanometer also connects with jockey.
Working : Key $$K_{1}$$ is closed, key $$K_{2}$$
is also closed, now plug is inserted between terminal $$1$$ and $$2$$ of two
way key. Now find out potential difference across resistance $$R$$ by
potentiometer. If $$I$$ current flowing through secondary circuit, potential
difference $$V$$ across the ends of resistance $$R$$ and the balancing length
is $$l_{1}$$ for $$V$$.
By the principle of potentiometer
$$V = kl_{1} .... (1)$$
but $$V = IR$$ (by Ohm's law)
$$IR = kl_{1} .... (2)$$
Eliminate the plug between $$1$$ and $$2$$ and
insert it between $$2$$ and $$3$$. Now $$R$$ and $$r$$ connect in series.
Current $$I$$ remains same in this circuit also now the potential difference
across the ends of the $$(R + r)$$ is $$V_{1}$$ and balancing length for it is
$$l_{2}$$
$$V_{1} = kl_{2}$$
but $$V_{1} = I(R + r)$$
Hence $$I(R + r) = kl_{2} .... (3)$$
From equations (2) and (3)
$$\dfrac {I(R + r)}{IR} = \dfrac
{kl_{2}}{kl_{1}}$$
$$1 + \dfrac {r}{R} = \dfrac {l_{2}}{l_{1}}$$ or
$$\dfrac {r}{R} = \dfrac {l_{2}}{l_{1}} - 1$$
$$r = \left (\dfrac {l_{2} - l_{1}}{l_{1}}\right
)R ....(4)$$
Putting the values $$l_{1}, l_{2}$$ and $$R$$,
then internal resistance of primary cell $$(r)$$ can be determined.
Conceptual Questions
A ski resort consists of a few chairlifts and several interconnected downhill runs on the side of a mountain, with a lodge at the bottom. The chairlifts are analogous to batteries, and the runs are analogous to resistors. Describe how two runs can be in series.Describe how three runs can be in parallel. Sketch a junction between one chairlift and two runs. State Kirchhoffs junction rule for ski resorts. One of the skiers happens to be carrying a skydivers altimeter.She never takes the same set of chairlifts and runs twice, but keeps passing you at the fixed location where you are working. State Kirchhoffs loop rule for ski resorts.
Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Explanation:
Given that the lifts are analogous to batteries, and the runs are analogous to resistors.
So from all the figures. The resistors represent the runs while the lift represents the battery.
Kirchhoff’s junction rule states that at any junction, the sum of the altimeter attained moving into and out of that junction are equal.
While
Kirchhoff’s loop rule states that the algebraic sum of the number of lifts used in any closed loop is equal to zero
Additional Problems(69)
A young man owns a canister vacuum cleaner marked "535 \,W [at] 120 \,V" and a Volkswagen Beetle, which he wishes to clean. He parks the car in his apartment parking lot and uses an inexpensive extension cord $$15.0 \,m$$ long to plug in the vacuum cleaner. You may assume the cleaner has constant resistance.
(a) If the resistance of each of the two conductors in the extension cord is $$0.900 \,\Omega$$, what is the actual power delivered to the cleaner?
(b) If instead the power is to beat least $$525 \,W$$, what must be the diameter of each of two identical copper conductors in the cord he buys?
(c) Repeat part (b) assuming the power is to be at least $$532 \,W$$.
Class 12 Medical Physics Extra Questions
- Alternating Current Extra Questions
- Atoms Extra Questions
- Current Electricity Extra Questions
- Dual Nature Of Radiation And Matter Extra Questions
- Electric Charges And Fields Extra Questions
- Electromagnetic Induction Extra Questions
- Electromagnetic Waves Extra Questions
- Electrostatic Potential And Capacitance Extra Questions
- Magnetism And Matter Extra Questions
- Moving Charges And Magnetism Extra Questions
- Nuclei Extra Questions
- Ray Optics And Optical Instruments Extra Questions
- Semiconductor Electronics: Materials, Devices And Simple Circuits Extra Questions
- Wave Optics Extra Questions