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CBSE Questions for Class 10 Physics Electricity Quiz 9 - MCQExams.com
CBSE
Class 10 Physics
Electricity
Quiz 9
The resistors of resistances $$2\Omega, 4\Omega, 5\Omega$$ are connected in parallel. The total resistance of the combination will be:
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$$\dfrac {29}{10}\Omega$$
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$$\dfrac {19}{20}\Omega$$
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$$\dfrac {10}{20}\Omega$$
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$$\dfrac {20}{19}\Omega$$
Explanation
Given: $$R_1=2\ \Omega$$, $$R_2=4\ \Omega$$, $$R_3=5\ \Omega$$
In parallel combination-
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\Rightarrow\dfrac{1}{R}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{5}$$
$$\Rightarrow R=\dfrac{20}{19}\Omega$$
When resistances are connected in series then the same current flows through each resistance.
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0%
True
0%
False
Explanation
When resistances are connected in series, the same current flows through each resistance. Current will be the same through each resistor.
An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in $$10\ min$$. When the other coil is used the water boils in $$40\ min$$. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be
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$$25\ min$$
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$$15\ min$$
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$$8\ min$$
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$$4\ min$$
Two resistances are connected in series as shown in the diagram. What is the current through $$R$$?
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$$1 A$$
0%
$$2 A$$
0%
$$3 A$$
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$$4 A$$
Explanation
The voltage across $$5\Omega$$ resistor is $$V=10V$$
Current,
$$I=\dfrac{V}{R}$$; according to ohm' law
$$\Rightarrow I=\dfrac{10}{5}$$
$$\Rightarrow I=2A$$
Since resistance $$5\Omega$$ and $$R$$ are in series and hence current through both resistance is the same.
Hence current through $$R$$ is also $$2A$$.
Two resistances are connected in parallel and a current is sent through the combination. The current divides itself:
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In the inverse ratio of resistance
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In the direct ratio of resistance
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Equally in both the resistance
0%
In none of the above manner
Explanation
Current flows in inverse ratio of resistance
$$\dfrac{i_1}{i_2}=\dfrac{R_2}{R_1}$$
What do you add to distilled water for making it to conduct electricity.
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sulphuric acid
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water
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milk
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base
In a circuit two or more cells of the same e.m.f. are connected in parallel in order to
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0%
Increase the P.D across a resistance in the circuit
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Decrease the P.D across a resistance in the circuit
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Facilitate drawing more current from the battery system
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Change the e.m.f. across the system of batteries
Explanation
$$E_{eff}=\dfrac{E_1}{r_1}+\dfrac{E_2}{r_2}$$ where r is internal resistancE
It will increase the current ,$$I=\dfrac{E_{eff}}{R}$$
Resistance of $$2\ \Omega$$ and $$3\ \Omega$$ are connected in series. If the potential difference across the $$2\ \Omega$$ resistor is $$3\ V$$, the potential difference across $$3\Omega$$ is:
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$$4.5\ V$$
0%
$$9\ V$$
0%
$$3\ V$$
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$$2\ V$$
Explanation
Given,
Resistors of resistances, $$R_1=2\ \Omega$$ and $$R_2=3\ \Omega$$ are connected in series.
Potential difference across $$2\ \Omega$$ resistor, $$V_1=3\ V$$
Potential difference across $$2\ \Omega$$ resistor, $$V_2$$
According to Ohm's law,
$$V=IR$$
$$\implies I=\dfrac VR$$
$$\implies I=\dfrac {V_1}{R_1}$$
$$I=\dfrac {3}{2}$$
$$I=1.5\ A$$
We know, current in a series circuit remains constant.
$$V_2=IR_2$$
$$V_2=1.5\times 3$$
$$V_2=4.5\ V$$
Human body is a
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good conductor of electricity
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poor conductor of electricity
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semiconductor
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good insulator
Explanation
Human body is a good conductor of electricity which allow electric current to flow.
Answer-(A)
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the lengths and radii of the wires are in the ratio of $$4/3$$ and $$2/3$$, then the ratio of the currents passing through the wire will be
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$$3$$
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$$\dfrac{1}{3}$$
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$$\dfrac{8}{9}$$
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$$2$$
Explanation
$$R=\rho\dfrac{l}{A}$$
$$\implies R=\rho\dfrac{l}{\pi r^2}$$
$$\implies R\propto \dfrac{l}{r^2}$$
Hence, $$\dfrac{R'}{R}=\dfrac{l'}{r'^2}\dfrac{r^2}{l}$$
$$\dfrac{R'}{R}=\dfrac{l'}{l}\dfrac{r^2}{(r')^2}$$
$$\implies \dfrac{R'}{R}=\dfrac{4/3}{(2/3)^2}$$
$$\implies \dfrac{R'}{R}=3$$
Since, $$I\propto \dfrac{1}{R}$$
Hence, $$\dfrac{I'}{I}=\dfrac{1}{3}$$
Answer-(B)
A resistor of $$6 k\Omega $$ with tolerance $$10$$% and another of $$4k\Omega $$ with tolerance $$10$$% are connected in series. The tolerance of combination is about
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$$5$$%
0%
$$10$$%
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$$12$$%
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$$15$$%
Explanation
Let the resistances are $$R_1= 6\ k\Omega$$ and $$R_2=4\ k\Omega$$.
Tolerence in the resistances : $$\Delta R_1 = 600\Omega$$
$$\Delta R_2 = 400\Omega$$
Total resistance of the combination $$R_{eq} = R_1+R_2 =6+4 =10k\Omega$$
Total tolerance in the equivalent resistance: $$\Delta R_{eq} =\Delta R_1+\Delta R_2 =600+400 = 1k\Omega $$
Hence percentage tolerance of combination $$ = \dfrac{1k\Omega}{10k\Omega} \times 100 =10$$ %
The resistance of the series combination of two resistances is S. When they are joined in parallel the total resistance is P. If S = nP, then the minimum
possible value of n is
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0%
4
0%
3
0%
2
0%
1
Explanation
Let two resistor be $$R_{1} \ and \ R_{2}$$
In series combination the equivalent resistance is given by $$ S = R_{1} + R_{2}$$
In parallel combination the equivalent resistance is given by $$ P = \dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$$
By relation S = nP
We can write it $$R_{1} + R_{2} = n\dfrac{R_{1} \times R_{2}}{R_{1} + R_{2}}$$
Rearranging the term $$(R_{1} + R_{2})^2 = n R_{1} \times R_{2}$$
$$R_1^2 + R_2^2+ 2R_1R_2=nR_1R_2$$
$$R_1^2 + R_2^2 + (2-n)R_1R_2 = 0$$
Dividing the whole equation with $$R_2^2$$, we will get
$$(\dfrac{R_1}{R_2})^2 + 1 + (2-n) \dfrac{R_1}{R_2} = 0$$
Put $$\dfrac{R_1}{R_2} = k$$
$$k^2 + (2-n)k +1 = 0$$
For real value of k, $$D \geq 0$$
$$(2-n)^2 - 4 \geq 0$$
For minimum value of n,
$$(2-n)^2 - 4 = 0$$
$$4 + n^2 - 4n-4=0$$
$$n^2 - 4n = 0$$
$$n(n-4)=0$$
$$\implies n=0 \ or \ 4$$
Since $$n$$ can't be 0,
So $$ n = 4$$
Four bulbs marked $$40W, \ 250V$$ are connected in series with $$250\ V$$ mains, the total power consumed is
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$$10W$$
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$$40W$$
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$$320W$$
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$$160W$$
Explanation
Power of each bulb,
$$P = 40 W$$
Total power consumed when four bulbs are connected in series
$$\dfrac{1}{P_s} = \dfrac{1}{P_1}+$$
$$\dfrac{1}{P_2} + \dfrac{1}{P_3}+\dfrac{1}{P_4}$$
$$\therefore$$
$$\dfrac{1}{P_s} = \dfrac{1}{P}+$$
$$\dfrac{1}{P} + \dfrac{1}{P}+\dfrac{1}{P}$$
We get
$$P_s =\dfrac{P}{4}$$
$$\Rightarrow P_s = \dfrac{40}{4}$$
$$\Rightarrow P_s =10 W $$
Two resistors $$400\ \Omega $$ and $$800\ \Omega$$ are connected in series with a 6 V battery. the potential difference measured by voltmeter across $$400\ \Omega $$ resistor is:
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$$2\ V$$
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$$1.95\ V$$
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$$3.8\ V$$
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$$4\ V$$
Explanation
Total resistance of the circuit (series combination) $$R_s = R_1+R_2$$
$$\therefore$$ $$R_s = 400+800 =1200\Omega$$
Current, $$I=\dfrac{E}{R_s}=\dfrac{6}{1200}$$
$$I=0.005\ A$$
Potential difference across $$R_1$$, $$V' = IR_1$$
$$\therefore$$ $$V' = 0.005 \times 400$$
$$V' =2\ V$$
What is the minimum number of bulbs, each marked $$60W, \ 40V,$$ that can work safely when connected in series with a $$240V$$ mains supply?
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$$2$$
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$$4$$
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$$6$$
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$$8$$
Explanation
Resistance of each bulb
$$R = \dfrac{V^2}{P}$$
$$\Rightarrow R = \dfrac{(40)^2}{60}$$
$$\Rightarrow R = 26.67\Omega$$
Maximum current that can flow through the bulb without damaging it
$$I_{max}= \dfrac{P}{V}$$
$$\Rightarrow I_{max} = \dfrac{60}{40}$$
$$\Rightarrow I_{max} = 1.5\ A$$
Thus minimum resistance required in the circuit
$$R_{min} = \dfrac{V_s}{I_{max}}$$
$$\Rightarrow R_{min} = \dfrac{240}{1.5}$$
$$\Rightarrow R_{min} = 160 \Omega$$
Let $$n$$ number of bulbs are connected in series in the circuit.
Equivalent resistance of $$n$$ bulbs
$$R_{eq} = nR = 26.67 n$$
But,
$$26.67 n \leq 160$$
$$\Rightarrow n \leq 6$$
Thus minimum number of bulbs required is $$6$$.
Two wire of same metal have same length but their cross-sections are in the ratio $$3:1$$. They are joined in series. The resistance of thick wire is $$10$$. The total resistance of combination will be
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$$10 \Omega$$
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$$20 \Omega$$
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$$40 \Omega$$
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$$100 \Omega$$
Explanation
Resistance of the wire $$R = \dfrac{\rho L}{A}$$
where $$\rho$$ is the resistivity and $$L$$ and $$A$$ is the length and cross-section area of the wire, respectively.
$$\implies$$ $$R\propto \dfrac{1}{A}$$ (for same $$L$$ and $$\rho$$)
Ratio of cross-sectional area of thick wire to thin wire $$\dfrac{A_1}{A_2} = \dfrac{3}{1}$$
Thus ratio of resistance of thin wire to thick wire $$\dfrac{R_2}{R_1} = \dfrac{A_1}{A_2} = \dfrac{3}{1}$$
Or
$$\dfrac{R_2}{10}= \dfrac{3}{1}$$
$$\implies$$ $$R_2 = 30\Omega$$
Total resistance of the (series) combination $$R_{eq} = R_1 + R_2$$
$$\implies$$ $$R_{eq} = 10 + 30 = 40 \Omega$$
A $$100W, 200V$$ bulb is connected to a $$160V$$ supply. The actual power consumption would be
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$$185W$$
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$$100W$$
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$$54W$$
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$$64W$$
Explanation
Resistance of the bulb $$R = \dfrac{V^2}{P}$$
where $$P = 100 W$$ and $$V = 200$$ V.
$$\therefore$$ $$R = \dfrac{(200)^2}{100} = 400\Omega$$
Actual power consumption $$P' = \dfrac{V_1^2}{R}$$
where $$V_1 = 160$$ V
$$\therefore$$ $$P' = \dfrac{(160)^2}{400} = 64 W$$
$$1$$ kilowatt hr $$=$$ __________ joules.
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$$10.6 \times {10}^{6}$$
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$$3.6 \times {10}^{6}$$
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$$30.6 \times {10}^{6}$$
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$$3.6 \times {10}^{5}$$
Explanation
$$kWh$$ is the commercial unit of energy.
$$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$$ joules
You are given several identical resistance each of value $$R=10$$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of $$5$$ which can carry a current of $$4$$ ampere. The minimum number of resistance's of the type $$R$$ that will be required for this job is
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$$4$$
0%
$$10$$
0%
$$8$$
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$$20$$
Explanation
Resistance of each resistor $$R = 10$$
Equivalent resistance of each segment $$R' = R+R = 2R$$
$$\therefore$$ $$R' = 2\times 10 = 20$$
Such $$4$$ segments are connected in parallel to each other.
Thus equivalent resistance of the circuit $$R_p = \dfrac{R'}{4}$$
$$\implies$$ $$R_p = \dfrac{20}{4} = 5$$
Maximum current flowing through each resistor is one ampere i.e. $$i =1 A$$
Thus total current flowing through the circuit $$I = i+i+i+i$$
$$\therefore$$ $$I = 4i = 4 A$$
Thus minimum $$8$$ resistors are required to satisfy the above conditions.
A $$30\ V,\ 90\ W$$ lamp is to be operated on a $$120\ V\ DC$$ line. For proper glow, a resistor of ..... should be connected in series with the lamp.
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$$10$$
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$$20$$
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$$30$$
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$$40$$
Explanation
Resistance of the lamp $$R = \dfrac{V^2}{P}$$
$$\therefore$$ $$R = \dfrac{30^2}{90}$$
$$\implies$$ $$R = 10 \Omega$$
Current flowing through the lamp $$I = \dfrac{P}{V}$$
$$\therefore$$ $$I = \dfrac{90}{30} =3 A $$
DC line voltage $$V_d = 120$$ V
Resistance of the line must be $$R' = \dfrac{V_d}{I}$$
We get $$R' = \dfrac{120}{3} = 40 \Omega$$
Let $$R_{1}$$ be the external resistance to be connected in series with the lamp for proper glow.
$$\therefore$$ $$R + R_1 = R'$$
OR $$10 + R_1 = 40$$
$$\implies$$ $$R_1 = 30 \Omega$$
$$1 Wh$$ (Watt hour) is equal to :
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$$36\times { 10 }^{ 5 }J$$
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$$36\times { 10 }^{ 4 }J$$
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$$3600 J$$
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$$3600 { Js }^{ -1 }$$
Explanation
Watt is a unit of power such that $$1W =1\dfrac{J}{s}$$
Also we know $$1h = 3600$$ $$s$$
$$\therefore$$ $$1Wh = 1\dfrac{J}{s} \times 3600 $$ $$s = 3600$$ $$J$$
Aluminium is preferred in overhead power cables because:
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It is a good conductor
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It acts as an insulator, in case of lightening
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It prevents accidents
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It is lightweight
Explanation
Aluminum is preferred in overhead power cables because it is a good conductor, so it can transport current with minimal losses while also being lightweight enough to not need the large support structure.
A 30V-90W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is
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$$10 \Omega$$
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$$30 \Omega$$
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$$20 \Omega$$
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$$40 \Omega$$
Explanation
Resistance of lamp
$$R_0 = \dfrac{V^2}{P} = \dfrac{(30)^2}{90} = 10 \Omega$$
Current in the lamp
$$I = \dfrac{V}{R_0} = \dfrac{30}{10} = 3A$$
As the lamp is operated on 120V DC, then resistance becomes
$$R' = \dfrac{V'}{i} = \dfrac{120}{3} = 40 \Omega$$
For proper glow, a resistance R is joined in series with the bul
$$R' = R + R_0$$
$$\Rightarrow R^{\alpha} = R' - R_0 = 40 - 10 = 30 \Omega$$
In the following circuit, the 1 $$\Omega$$ resistor dissipates power P. If the resistor is replaced by 9 $$\Omega$$, the power dissipated in it is
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$$P$$
0%
$$3P$$
0%
$$9P$$
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$$P/3$$
Explanation
Equivalent resistance in the circuit $$= (3+1)=4\Omega$$
By Ohm's law,
$$10 \ V=(4 \ \Omega)(i \ Ampere)$$
$$i=2.5 \ A$$
Power through a resistance is given by $$P=I^2R$$
Power through $$1 \Omega,$$
$$P_{1 \Omega}=i^2\times 1=6.25 \ W$$
Given, $$P_{1\Omega}=P$$
When $$1\Omega$$ is replaced by $$9\Omega$$, equivalent resistance $$=(3+9)=12\Omega$$
By Ohm's law,
$$10 \ V=(12 \ \Omega)(i' \ Ampere)$$
Current $$i'=\dfrac{10}{12} \ A$$
Power dissipated through replaced resistor, $$P_{9\Omega}=(\dfrac{10}{12})^2\times 9=6.25=P_{1\Omega}=P$$
Option A is correct.
The equivalent resistance between A and B in the following figure is
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$$120\Omega $$
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$$40\Omega $$
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$$30\Omega $$
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$$22.5\Omega $$
Two electric bulbs have ratings respectively of $$25 \ W, 220 \ V$$ and $$100 \ W, 220 \ V$$. If the bulbs are connected in series with a supply of $$440\ V$$, which bulb will fuse?
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$$25\ W$$ bulb
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$$100\ W$$ bulb
0%
Both of these
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None of these
Explanation
Power is given by $$ P = \dfrac{V^2}{R}$$, where $$P$$ is Power, $$V$$ is voltage across bulb and $$R$$ is resistance of bulb.
Hence, $$R = \dfrac{V^2}{P}$$
Resistance of $$25\ W\ bulb = \dfrac {220\times 220}{25} = 1936\ \Omega$$
Using Ohm's law, $$V = iR$$
Value of current that can pass through it $$i= \dfrac{V}{R} = \dfrac {220}{1936} = 0.11\ A$$
Resistance of $$100\ W\ bulb = \dfrac {220\times 220}{100} = 484\ \Omega$$
Value of current that can pass through it $$i = \dfrac{V}{R} = \dfrac {220}{484} = 0.45\ A$$
When connected in series to $$440 \ V$$ supply, the equivalent resistance is given by,
$$R_{equi} = R_1+R_2 = 1936+484=2420 \ \Omega$$
Calculating the value of the current $$i = \dfrac {440}{2420} = 0.18\ A$$
Thus $$0.18 \ A >0.11 \ A$$
The value of current flowing in the final circuit is greater than that allowed by the $$25W \ bulb$$, so it will fuse.
The total electrical resistance between the points $$A$$ and $$B$$ of the circuit shown is:
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$$9.23\Omega $$
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$$15\Omega $$
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$$30\Omega $$
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$$100\Omega $$
Explanation
$$\cfrac { 1 }{ Req. } =\cfrac { 1 }{ 30\Omega } +\cfrac { 1 }{ 20\Omega } +\cfrac { 1 }{ 40\Omega } $$
$$Req.=\cfrac { 120 }{ 13 } \Omega =9.23\Omega $$
Which of the following are the properties of fuse wire?
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Made of alloy of tin
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Has a low melting point
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Connected in series with main supply
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All of the above
Explanation
All the given options hold true for fuse wire. A fuse wire should have these characteristics:
low melting point,
high conductivity and
least deterioration due to oxidation
It is connected in series with main supply.
When the electric current through the fuse exceeds a certain limit, the fuse wire melts and breaks.
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0%
True
0%
False
Explanation
(True)
The statement is true. The electric fuse works on the principle of heating effect of electric current. The amount of heating caused depends on the amount of current flowing through the wire. If current exceeds a certain limit, due to excessive heating the fuse wire will automatically melt.
The resistivity of the material of potentiometer wire is $$5\times 10^{-5}\Omega m$$ and its area of cross-section is $$5\times 10^{-6} m^{2}$$. If $$0.2\ A$$ current is flowing through the wire, then the potential drop per metre length of the wire is
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$$0.1\ Vm^{-1}$$
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$$0.5\ Vm^{-1}$$
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$$0.25\ Vm^{-1}$$
0%
$$0.2\ Vm^{-1}$$
0%
$$0.001\ Vm^{-1}$$
Explanation
Given, $$i = 0.2\ A$$
Resistivity, $$\rho = 5\times 10^{-6}\Omega - m$$
and $$A = 5\times 10^{-6} m^{2}, l = 1\ m$$
We know that, $$V = iR$$
$$\Rightarrow V = \dfrac {i\rho L}{A}\left [\because R = \dfrac {\rho l}{A}\right ]$$
$$\Rightarrow V = \dfrac {0.2\times 5\times 10^{-6}\times 1}{5\times 10^{-6}}$$
$$\Rightarrow V = 0.2\ Vm^{-1}$$.
The ratio of resistance of two copper wire of the same length and of same cross-sectional area, when connected in series to that when connected in parallel, is
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$$1 : 1$$
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$$1 : 2$$
0%
$$2 : 1$$
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$$4 : 1$$
0%
$$1 : 4$$
Explanation
Let $$R$$ be the resistance of copper wire. In the first condition,
$$R_{series} = R + R = 2R .... (i)$$
In the second condition,
$$\dfrac {1}{R_{parallel}} = \dfrac {1}{R} + \dfrac {1}{R} = \dfrac {1}{R_{parallel}} = \dfrac {1 + 1}{R}$$
$$R_{parallel} =\dfrac {R}{2} ..... (ii)$$
On dividing Eq. (i) by Eq. (ii), we get
$$\dfrac {R_{series}}{R_{parallel}} = \dfrac {2R}{\dfrac {R}{2}} = \dfrac {4}{1}$$.
Two bulbs of $$250 \ V$$ and $$100 \ W$$ are first connected in series and then in parallel with a supply of $$250 \ V$$. Total power in each of the case will be respectively
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$$100 \ W, 50 \ W$$
0%
$$50\ W, 100 \ W$$
0%
$$200 \ W, 150 \ W$$
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$$50 \ W, 200 \ W$$
A wire of resistance 'R' is cut into 'n' equal parts. These parts are then connected in parallel with each other. The equivalent resistance of the combination is?
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nR
0%
$$R/n$$
0%
$$n/R^2$$
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$$R/n^2$$
A student's $$9.0V, 7.5W$$ portable radio was left on from $$9:00\ P.M.$$ until $$3:00 A.M.$$ How much charge passed through the wires?
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$$18000C$$
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$$24000C$$
0%
$$6000C$$
0%
$$12000C$$
Explanation
As we know,
Power, $$P = V\cdot I$$
$$I = \dfrac {P}{V}$$
$$\Rightarrow I = \dfrac {Q}{t} = \dfrac {P}{V}$$
$$\Rightarrow Q = \dfrac {P}{V} \times t = \dfrac {7.5}{9}\times 3600\times 6$$
$$= 18000\ C$$
$$[\because t = 9\ PM$$ to $$3\ AM = 6\ hr = 6\times 3600s]$$
The diagram shows identical lamps X and connected in series with a battery. The lamps light with normal brightness.If a third lamp Z is connected in parallel with lamp X, then what will happens to the brightness of the lamp Y?
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Brighter than normal
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Normal
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Dimmer than normal
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Very dim (cannot be seen)
Explanation
The third lamp Z is connected parallel to X, so the current gets divided among X and Z. So both bulbs Xand Z light dimmer and the bulb Y lights brighter than earlier because current through it increases.
Observe the given figure and answer the following questions. The bulb will glow when a ______ is placed in between the probes.
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0%
Crayon
0%
Comb
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Key
0%
Piece of stone
Explanation
Materials that allow electric current to pass through them are called conductors.
Key is a metallic object. When it is placed in between probes then the current can easily flow in the circuit.
Fuse is the most important safety device, used for protecting the circuits due to
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0%
Short circuiting and overloading
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Overloading and earthing
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Earthing only
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Short circuiting, overloading and earthing.
$$4$$ bulbs rated $$100$$W each, operate for $$6$$ hours per day. What is the cost of the energy consumed in $$30$$ days at the rate of Rs. $$5$$/kWh?
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Rs. $$360$$
0%
Rs. $$90$$
0%
Rs. $$120$$
0%
Rs. $$400$$
Explanation
$$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$$kWh$$=2.4$$kWh
Consumed in $$30$$ days$$=30\times 2.4=72$$
Total cost $$=72\times 5=360$$Rs.
Consider four circuits shown in the figure below. In which circuit power dissipated is highest?
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0%
0%
0%
0%
Explanation
We know that $$\rho =\dfrac { { \varepsilon }^{ 2 } }{ R } \\ \rho \propto \dfrac { 1 }{ R } (\varepsilon = constant)$$ (1)
so we calculate $${ R }_{ eq }$$ in all options as shown in fig
we know that $$\rho \propto \dfrac { 1 }{ R } $$
In (a) option resistance is minimum, so power dissipated will be highest
Equivalent resistance between A and B is.
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3 $$\Omega $$
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6 $$\Omega $$
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1.5 $$\Omega $$
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4.5 $$\Omega $$
Ravi connected three bulbs with the cells and a switch as shown. When switch is moved to ON position.
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The bulb X will glow first.
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The bulb Y will glow first.
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The bulbs Z and X will glow first.
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All the bulbs will glow simultaneously.
Explanation
All bulbs will glow simultaneously and instantly without any delay. It is because the current is setup in the circuit the moment the switch is closed to provide it a closed path to flow. The three bulbs are in series, so equal current flows through all of them.
Two wires of same material have lengths $$L$$ and $$2L$$ cross-sectional areas $$4A$$ and $$A$$ respectively. The ratio of their resistances would be
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0%
$$1:1$$
0%
$$1:8$$
0%
$$8:1$$
0%
$$1:2$$
Explanation
As $$R=\cfrac { \rho l }{ A } $$
$$R\propto \cfrac { l }{ A } $$
$$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { { l }_{ 1 } }{ { A }_{ 1 } } \times \cfrac { { A }_{ 2 } }{ { l }_{ 2 } } $$
$$\cfrac { { R }_{ 1 } }{ { R }_{ 2 } } =\cfrac { L }{ 4A } \times \cfrac { A }{ 2L } =1:8$$
$$\begin{bmatrix} { l }_{ 1 }=L \\ { l }_{ 2 }=2L \\ { A }_{ 1 }=4A \\ { A }_{ 2 }=A \end{bmatrix}$$
$$n$$ resistors each of resistance $$R$$ first combine to give maximum effective resistance and then combine to given minimum effective resistance. The ratio of the maximum to minimum resistance is:
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$$n$$
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$$n^2$$
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$$n^2-1$$
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$$n^3$$
Explanation
Maximum effective resistance will be when all the resistors are in series combination. So,
$$R_{eff_{max}}=n\ R$$
Minimum effective resistance will be when all the resistors are in parallel combination. So,
$$R_{eff_{min}}=R/n$$
$$\therefore \ \dfrac {R_{eff_{max}}}{R_{eff_{min}}}=\dfrac {n\ R}{R/ n}=n^2$$
Which of the following I-V graph represents ohmic conductors?
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0%
0%
0%
Explanation
Ohm's law $$I=\dfrac{V}{R}$$ is an equation of straight line. Hence I-V characteristics for ohmic conductors is also straight line and its slope gives, $$m=\dfrac{1}{resistance}$$ of the conductor.
Option A
A wire of resistance $$12\: ohm/meter\: $$is bent to form a complete circle of radius $$10cm$$. The resistance between its two diametrically opposite points, A and B as shown in the figure is?
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$$3\Omega$$
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$$6\pi \Omega$$
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$$6\Omega$$
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$$0.6\pi \Omega$$
Explanation
Total length of wire, $$l= 2\pi\times 0.1=0.2\pi\ m$$
Resistance per unit length, $$r=12\Omega/m$$
Let us assume two semicircular part of wire connected in parallel between $$A$$ and $$B$$
Resistance of each part $$,R= r \times l = 12\times \cfrac{0.2\pi}{2}= 1.2 \pi\ \Omega$$
Two equal resistors (R) are connected in parallel between A and B so equivalent resistance is equal to $$R/2$$.
$$\therefore$$ Resistance between $$A$$ and $$B = \dfrac R2 = \dfrac{1.2 \pi}2= 0.6\pi \ \Omega$$
Hence, option $$(D)$$ is correct.
An infinite ladder network of resistances is constructed with $$1\Omega$$ and $$2\Omega$$ resistance as shown in figure. The $$6$$V battery between A and B has negligible internal resistance. The equivalent resistance between A and B is?
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$$1\Omega$$
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$$2\Omega$$
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$$3\Omega$$
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$$4\Omega$$
Explanation
$$ \textbf {Step 1: Find Smallest Repeating unit [Refer Fig. 1]} $$
$$ \textbf{Step 2: Replace with equivalent resistance R. [Refer Fig. 2]} $$
After the first smallest repeating unit, the remaining circuit is also same as the initial circuit.
Therefore, if we assume the equivalent resistance between A & B as $$R$$, then the resistance of the remaining circuit will also be
$$R$$
as shown in the figure.
$$ \textbf {Step 3: Find equivalent resistance between A and B and equate with R. [Refer Fig. 3]} $$
From figure, the equivalent resistance between A & B will be equal to the equivalent resistance of the remaining circuit. So,
$$ R_{AB} = 1 + \dfrac {2R}{2+R} $$
$$\Rightarrow\ \ R = 1 + \dfrac{2R}{2+R} $$
$$\Rightarrow\ \ 3R + 2 = R^2 + 2R $$
$$\Rightarrow\ \ R^2 - R - 2 = 0 $$
$$\therefore\ \ \ \ \boxed{R = 2 \Omega}$$ (Here, Negative value can is rejected)
Hence Equivalent resistance between points A and B is 2$$\Omega$$.
Two coils have a combined resistance of $$12 \Omega$$, when combined series and $$5/3 \Omega$$, when connected in parallel. Their respective resistances are:
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$$6$$ and $$6$$ ohms
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$$10$$ and $$2$$ ohms
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$$5$$ and $$7$$ ohms
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$$4$$ and $$8$$ ohms
Explanation
Let us assume the two resistances to be $$R_1$$ and $$R_2$$.
Let resistances in series and parallel be $$R_s$$ and $$R_p$$ respectively.
Given:
$$R_{s}=12 \Omega=R_1+R_2$$
So,
$$R_1+R_2=12$$ . . . . (i)
$$R_2=12-R_1$$ . . . . (ii)
And,
$$R_{p}=\dfrac{5}{3}\Omega \Rightarrow \dfrac{1}{R_{p}}=\dfrac{3}{5}$$
So,
$$\dfrac{3}{5}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$$
$$\Rightarrow \dfrac{3}{5}=\dfrac{R_1+R_2}{R_1R_2}$$ . . . (iii)
Using $$eq. (i)$$ and $$eq. (ii)$$ in $$eq. (iii)$$
$$\dfrac{3}{5}=\dfrac{12}{R_1(12-R_1)}$$
$$\Rightarrow 12R_1-R_1^2=20$$
$$\Rightarrow R_1^2-12R_1+20=0$$
This is a quadratic equation in $$R_1$$. Let's solve it by factorisation.
$$R_1^2-10R_1-2R_1+20=0$$
$$R_1(R_1-10)-2(R_1-10)=0$$
$$(R_1-10)(R_1-2)=0$$
$$\Rightarrow R_1-10=0$$ or $$R_1-2=0$$
$$\Rightarrow R_1=10$$ or $$R_1=2$$
Putting $$R_1=10$$ in $$eq.(i)$$, $$R_2=2$$
or
Putting $$R_1=2$$ in $$eq.(i)$$, $$R_2=10$$
So option B is correct.
You have been provided with four $$400$$$$\ \Omega$$ resistors each. The number of ways in which these can be combined to have different equivalent resistances is
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Seven different combinations and seven different equivalent resistances
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Eight different combinations and seven different equivalent resistances
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Nine different combinations and eight different equivalent resistances
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Ten different combinations and nine different equivalent resistances.
A copper cylindrical tube has inner radius a and outer radius b. The resistivity is $$\rho$$. The resistance of the cylinder between the two rends is?
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$$\dfrac{\rho l}{b^2-a^2}$$
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$$\dfrac{\rho l}{2\pi (b-a)}$$
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$$\dfrac{\rho l}{\pi (b^2-a^2)}$$
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$$\dfrac{\pi(b^2-a^2)}{\rho l}$$
Explanation
$$R=\cfrac{\rho l}{A}\\=\cfrac{\rho l}{\pi(b^2-a^2)}\\A=\pi b^2-\pi a^2\\=\pi (b^2-a^2)$$
Find the equivalent resistance between points A and B in the given circuit
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$$6\Omega$$
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$$12\Omega$$
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$$4\Omega$$
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$$18\Omega$$
Explanation
using wheatstone bridge,
$$\dfrac{6}{3}=\dfrac{12}{6}\Rightarrow$$ It is a balanced wheatstone bridge, hence no current in $$12\Omega$$ i.e $$PQ$$ branch or equivalent resistance b/w $$A\times B$$.
$$6$$ & $$12\Omega$$ are in series
$$=6\Omega +12\Omega$$
$$=18\Omega$$
similarly, $$3\Omega$$ & $$6\Omega$$, equi $$=3+6=9\Omega$$
$$18\Omega$$ & $$9\Omega$$ are in parallel
$$\dfrac{1}{R_e}=\dfrac{1}{18}+\dfrac{1}{9}=\dfrac{3}{18}$$
$$\boxed{R_e=6\Omega}$$
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Practice Class 10 Physics Quiz Questions and Answers
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