Current Electricity - Class 12 Engineering Physics - Extra Questions

For the balanced bridge, the of the two resistances is equal to the ratio of the lengths of the two parts AJ and JB of the wire, i.e.,
$$\dfrac{x}{6\Omega }=\dfrac{40 cm}{60cm}$$ or $$X=4\Omega$$

When the sliding contact is in middle, a resistance of $$R_o/2$$ is connected in series with a parallel combination of $$R$$ and $$R_o/2$$ .

$$R=\cfrac{\rho l}{A}$$
$$\Rightarrow A=\cfrac{\rho l}{R}$$
$$\Rightarrow A=\cfrac{1.84\times 10^{-6}\times 2}{46}$$
$$\Rightarrow A=8\times 10^{-8}$$ $$m^2$$

Krichof's second law is also known as Krichhoff Voltage Law which deals with conservation of 

No value of resistance depend on resistivity  , length of wire and area of 

when deflection is zero $$\Rightarrow$$ voltage of secondary cell=voltage drop on the balancing length of the wire AB
voltage drop on the balancing length of the wire $$\Rightarrow$$ i $${\rho}$$ X
   where i= currnt flowing in primary circuit
      $$\rho$$ =resistance per unit length of the wire
      X=balancing length of wire AB
when deflection in galvanometer is zero $$\Rightarrow {\frac {9}{2}}=(\frac {10}{10})(\frac {9}{12})x$$
$$x=6m.$$

(A)Conductivity depends on temperature. And definitely it will depend on nature of material.
$$A\rightarrow 2,3$$
(B)Conductance depends on conductivity as well as shape and size of conductor means dimensions of conductor.
$$B \rightarrow 1,2\ \&\ 3$$
(C)Since current density depends on conductivity and electric field strength by relation. $$J=\sigma E$$ but temperature and conductor are given therefore,conductivity is fixed.
$$C \rightarrow 4$$
(D)Current depends on resistance and resistance is same as conductance therefore,
$$D \rightarrow 1,2\ \&\ 3$$

The net potential difference applied across the resistor= $$200V-10V=190V$$

Slope of a V-I graph gives value of $$1/R$$ , as we can see that the slope of graph 1 is greater than graph 2 ,so

Green $$\rightarrow 5$$

(a) Working principle of potentiometer
When a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.
Application of Potentiometer for comparing emf's of two cells:
The following figure shows an application of the potentiometer to compare the emf of two cell of emf of two cells of emf $$E_{1}and\;E_{2}$$  
$$E_{1},E_{2}$$ are the emf of the two cells 
1,2,3 form a two way key.
When 1 and 3 are connected $$E_{1}$$ is connected to the galvanometer (G)
Jokey is moved to $$N_{1}$$ which is at a distance li from A to find the balancing length

$$R=AB\times 10^C \pm \  x\% = 62\times 10^3 \pm\  5\%$$

The standard cell (S) of known emf and the cell (X) of unknown emf are connected in parallel ( through a double switch, galvanometer and movable jocky) to a conducting wire AB of uniform cross section. By using double switch, first the standard cell S is placed in circuit. Jocky is then moved along wire AB and its position is adjusted so that no current flows through galvanometer.
Let AD be the length of the conducting wire for this position of jocky.
$$\displaystyle E_{S} \propto l(AD)$$
Now, the cell (X) of unknown emf  is placed in circuit. Jocky is then moved along wire AB and its position is adjusted so that no current flows through galvanometer. Let AD' be the length of the conducting wire for this position of jocky.
$$\displaystyle E_{X} \propto l(AD')$$
$$\displaystyle \dfrac {   E_{S}  }{  E_{X}  }=\dfrac { l(AD)   }{   l(AD') }$$
From the above equation, the emf of cell (X) can be calculated.

Resistance of wire is given by   $$R = \dfrac{\rho L}{A}$$
where $$L$$ is the length of wire and $$A$$ is the cross-section of wire.
So resistance of wire depends directly on the length of wire and inversely on cross-section of wire.

At any junction, the sum of the currents entering the Junction is equal to the sum of currents leaving the Junction.
The algebraic sum of changes in potential around any closed loop involving resistors and cells in the loop is zero.

The direction of induced current will be in forward direction i.e., left to right to increase the flux passing through surface $$S$$ which was decreasing due to voltage end.

Let at x distance it will show no deflection-
then current on $$AB=\dfrac{\varepsilon}{10r}$$
In case of no deflection-
$$\dfrac{9r}{L}X\times \dfrac{\varepsilon}{10r}=\dfrac{\varepsilon}{10r}$$
$$So, x=\dfrac{5L}{9}$$

Wire $$C$$ has maximum resistance because it has maximum length, least thickness and highest resistivity.

Conductivity of a material depends  on two factors
Factor 1: no,of electron in conduction band
Factor 2: Collision of  conduction electron with core atom. Factor 1 increases conductivity and Factor 2 decreases.
 

Kirchhoff's voltage law is in accordance with the conservation of energy and Kirchhoff's current law is in accordance with the conservation of charge.

$$R=\cfrac{\rho l}{A}$$
$$A=\cfrac{64\times 10^{-6}\times 198\times 10^{-2}}{7}$$
$$\Rightarrow A=1.810\times 10^(-5)$$ $$m^2$$
And $$A=\pi r^2=1.810\times10^{-5}$$
$$\Rightarrow r=2.4$$ $$mm$$

$$i=V/R=\cfrac{20+10}{32+1}=\cfrac{30}{33}Amp$$

$$4-7K\Omega\pm 10\%$$

$$\overline {FE}  =  - \overline {Fm}$$

Since three identical cells are in series.

Given that ,

The unit of electrical resistance is ' $$\Omega$$ '

$$T=T_0+\dfrac{1}{\alpha}(\dfrac{R}{R_0}-1)=20^\circ C+\Bigg(\dfrac{1}{4.5\times 10^{-3}/K}\Bigg)\Bigg(\dfrac{9.67\, \Omega}{1.1\,\Omega}-1\Bigg)=1.8\times 10^3\, ^\circ C$$

Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the
value of $$α$$ used in this calculation is not inconsistent with the other units involved. The table has been used. 

(a) In Eq. , we let $$ρ = 2ρ_0$$ where $$ρ_0$$ is the resistivity at $$T_0 = 20°C$$ :
                               
                                                $$ρ-ρ_0=2ρ_0-ρ_0 =ρ_0\alpha(T-T_0),$$  

and solve for the temperature T: $$T=T_0+\dfrac{1}{\alpha}=20^{\circ}+\dfrac{1}{4.3\times10{-3}/K}=250^{\circ}C$$

(b) Since a change in Celsius is equivalent to a change on the Kelvin temperature scale,
the value of $$α$$ used in this calculation is not inconsistent with the other units involved. It
is worth noting that this agrees well with Fig. 

i) Let $$]varepsilon$$ be emf and r the internal resistance of each cell.The equation of terminal potential difference.
$$V=\varepsilon_{eff}-i r_{int}$$ becomes
$$V=3\varepsilon -i r_{int}$$ ........(i)
Where $$r_{int}$$ is effective (total) internal resistance.
From fig., where $$i=0,V=6.0 V$$
$$\therefore$$ From (i), $$6=3\varepsilon-0\Rightarrow \varepsilon=\dfrac{6}{3}=2V$$
i.e., emf of each cell, $$\varepsilon=2V$$
Thus,emf of each cell, $$\varepsilon=2V$$

i) By increasing R the current through AB decreases, so potential gradient decreases.

For parallel combination,
Net emf, $$\varepsilon=\dfrac{\varepsilon_1r_2+\varepsilon_2r_2}{r_1+r_2}$$
Net internal resistance , $$r_{int}=\dfrac{r_1r_2}{r_1+r_2}$$
For series combination,
Net emf, $$\varepsilon=\varepsilon_1+\varepsilon_2$$
Net internal resistance = $$r_{int}=r1+r_2$$
Given $$\varepsilon_1=1 V,\varepsilon_2=2V$$ and $$r_1=2\Omega,r_2=1\Omega,R_{ext}=R$$
$$\therefore$$ Current $$I_1=\dfrac{\varepsilon_1+\varepsilon_2}{r_1+r_2+R}=\dfrac{1+2}{2+1+r}=\dfrac{3}{3+R}A$$ .......(i)
Current , $$I_2=\dfrac{(\varepsilon_1r_2+\varepsilon_2r_1)(r_1+r_2)}{R+{(r_1r_2/r_1+r_2)}}$$ ...................(ii)
$$=\dfrac{(1\times 1+2\times 2)/(2+1)}{R+(2\times 1)/(2+1)}=\dfrac{5/3}{R+\dfrac{2}{3}}=\dfrac{5}{3R+2}$$
Given $$I_1=I_2$$
$$\therefore \dfrac{3}{3+R}=\dfrac{5}{3R+2}$$ or $$9R+6=15+5R$$
$$4R=9 \Rightarrow R=\dfrac{9}{4}=2.25 \Omega$$

i) Unknown emf $$\varepsilon_2$$ is given by
$$\dfrac{\varepsilon_2}{\varepsilon_1}=\dfrac{l_2}{l_1}\Rightarrow\varepsilon_2=\dfrac{l_2}{l_1}\varepsilon_1$$
Given $$\varepsilon_1=1.5V,I_1=60 cm,I_2=80 cm$$
$$\therefore \varepsilon_2=\dfrac{80}{60}\times 1.5 V=2.0 V$$
ii)The circuit will not work if emf of driver cell is 1 V (less than that of cell in secondary circuit, because total voltage across wire AB is 1 V whcih cannotbalance the voltage V.
iii) No, since a t balance point no current flows through galvanomater G i.e., cell remains in open cicuit.

Given $$R_{27}=R(say),R_T=R+\dfrac{20}{100}R=1.2R,T_1=27+273=300k$$
From relation.
$$R_T=R_{27}[1+\alpha(T_2-300)]$$

Graph to show current voltage relationship of an ohmic resistor.

Expression :

$$\rho$$ - resistivity
$$R$$ - resistance
$$l$$ - length of conductor
$$A$$ - area of cross-section

$$V =4V$$

$$I=0.4A$$

Total resistance $$R_{eq}=?$$

$$R_{eq} = V/I = 0.4/4 = 10\Omega$$

From Ohm's law, we have 
$$R= \dfrac{V}{I}$$

Given: 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 as shown are obtained at a distance of $$120 cm$$ and $$300 cm$$ from the end

$$R=\dfrac{v}{1}=\dfrac{1}{0.1}=10$$
p.d across R in series with x= $$0.025\times 100$$
$$=2.5v$$
P.d across $$\times =$$ (p.d across both-pd across R)
P.d across $$\times =(2.5-1.0)=1.5$$
current $$\times =0.1A$$
So $$x=\dfrac{v}{1}=\dfrac{1.5}{0.1}=15$$

i) $$\dfrac{R}{X}=\dfrac{r|}{r(100-l)}$$
$$\Rightarrow \dfrac{R}{X}=\dfrac{1}{100-l}$$ ..........(i)
When bothe R and X are doubled and then interchanged, the new balance length becomes l given by 
$$\dfrac{2X}{2R}=\dfrac{l'}{(100-l')}$$ ...........(ii)
$$\Rightarrow \dfrac{X}{R}=\dfrac{l'}{100-l'}$$
From (i) and (ii)
$$\dfrac{100-l}{l}=\dfrac{l^2}{100-l^2}\Rightarrow I=(100-l)$$
ii) If galvanometer and battery are interchanged , there is no effect on the balance point.

a) Kirchhoff's Law
i) First law (or junction law): The algebraic sum of currents meeting at any junction is zero,i.e.,
$$\sum^{l-0}$$
This law is based on conservation of charge.
ii) Second law (or loop law): The algebric sum of potential differences of different circuit elements of a closed circuit (or mesh) is zero.i.e.
$$\sum ^{V=0}$$
This law is based on conservation of energy.
b) Let l1 and l2 be the currents leaving the positive, terminals of the cells, and at the point B
$$l=l1+l2$$ .....................(i)
Let V be the potential difference between points A and B of the combination of the cells,
so
$$v=E_1-l1r1$$ ..............(ii) across the cells 
and $$V=E_2-l1r2$$ ...........(iii)
From equation (i),(ii) and (iii),we get
$$I=\dfrac{(E_1-V)}{r_1}+\dfrac{(E_2-V)}{r_2}$$
$$=(\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2})-V(\dfrac{1}{r_1}+\dfrac{1}{r_2})$$ ..................(iv)
Fib (b) shows the equivalent cell,so for the same potential difference
$$V=E_{eq}-I_{rq}$$
or $$I=\dfrac{E_{eq}}{r_q}-\dfrac{V}{r_q}$$ ........(v)
On comparing Eq.(iv) and (v), we get
$$\dfrac{E_{eq}}{r_q}=\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}$$
and $$\dfrac{1}{r_q}=\dfrac{1}{r_1}+\dfrac{1}{r_2}\Rightarrow r_q=\dfrac{r_1r_2}{r_1+r_2}$$
On further solving,we have 
$$E{eq}=\dfrac{E_{1r_{2}}+E_{2r_1}}{r_1+r_2}$$

(a) An instrument that measures electric current in a circuit is called "ammeter". The unit of electric current is ampere $$(A)$$ . $$1$$ ampere is constituted by the flow of $$1$$ coulomb of charge through any point in an electric circuit in $$1$$ second.
(ii) Following graph was plotted between $$V$$ and $$I$$ values.
At potential difference $$0.8\ V$$ ,
$$\dfrac {V}{I} = \dfrac {0.8}{0.3} = \dfrac {8}{3} .... (1)$$
At potential difference $$1.2\ V$$ ,
$$\dfrac {V}{I} = \dfrac {1.2}{0.45} = \dfrac {8}{3} .... (2)$$
At potential difference $$1.6\ V$$ ,
$$\dfrac {V}{I} = \dfrac {1.6}{0.6} = \dfrac {8}{3} .... (3)$$
Conclusion: If $$I$$ be the current through $$XY$$ resistor and $$V$$ be the potential difference across it, then the ratio $$\dfrac {V}{1}$$ constant.
$$\Rightarrow V \infty I$$ and $$Ohm's$$
Law is obeyed.

(a) The ratio of potential difference(V) and current(I) is known as resistance(R).