# JEE Main

Explore popular questions from Current Electricity for JEE Main. This collection covers Current Electricity previous year JEE Main questions hand picked by experienced teachers.

## Mathematics

Current Electricity

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Q 1. If resistance of each wire in the network shown is {tex} r , {/tex} the equivalent resistance between {tex} A {/tex} and {tex} C {/tex} is equal to

A

{tex} r {/tex}

B

{tex} \frac { r } { 2 } {/tex}

{tex} \frac { 2 r } { 3 } {/tex}

D

{tex} \frac { 3 r } { 2 } {/tex}

##### Explanation

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Q 2. In the circuit shown, the potential difference between points {tex} C {/tex} and {tex} B {/tex} will be

A

{tex} ( 8 / 9 ) \mathrm { V } {/tex}

{tex} ( 4 / 3 ) \mathrm { V } {/tex}

C

{tex} ( 2 / 3 ) \mathrm { V } {/tex}

D

4{tex} \mathrm { V } {/tex}

##### Explanation

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Q 3. The current through 2{tex} \Omega {/tex} resistor is

Zero

B

1{tex} \mathrm { A } {/tex}

C

2{tex} \mathrm { A } {/tex}

D

4{tex} \mathrm { A } {/tex}

##### Explanation

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Q 4. The equivalent resistance between points {tex} A {/tex} and {tex} B {/tex} in the circuit shown is

A

4{tex} \Omega {/tex}

B

6{tex} \Omega {/tex}

10{tex} \Omega {/tex}

D

8{tex} \Omega {/tex}

##### Explanation

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Q 5. In the circuit shown in Fig. 14.11 , the reading of voltmeter will be

A

0.8{tex} \mathrm { V } {/tex}

1.33{tex} \mathrm { V } {/tex}

C

1.6{tex} \mathrm { V } {/tex}

D

2.00{tex} \mathrm { V } {/tex}

##### Explanation

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Q 6. A battery of internal resistance 4{tex} \Omega {/tex} is connected to the network of resistance as shown. In order to give the maximum power to the network, the value of {tex} R {/tex} should be

A

{tex} \frac { 4 } { 9 } \Omega {/tex}

B

{tex} \frac { 8 } { 9 } \Omega {/tex}

2{tex} \Omega {/tex}

D

18{tex} \Omega {/tex}

##### Explanation

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Q 7. A cell of emf {tex} E {/tex} is connected across a resistance {tex} R {/tex} . The potential difference between the terminals of the cell is found to be {tex} V . {/tex} The internal resistance of the cell must be

A

{tex} \frac { 2 ( E - V ) V } { R } {/tex}

B

{tex} \frac { 2 ( E - V ) R } { E } {/tex}

{tex} \frac { ( E - V ) R } { V } {/tex}

D

{tex} ( E - V ) R {/tex}

##### Explanation

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Q 8. The equivalent resistance of the network shown in Fig. 14.13 between the base terminals is

A

3{tex} \Omega {/tex}

B

3{tex} \frac { 1 } { 2 } \Omega {/tex}

2{tex} \frac { 2 } { 3 } \Omega {/tex}

D

2{tex} \Omega {/tex}

##### Explanation

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Q 9. {tex} n {/tex} identical cells, each of emf {tex} \varepsilon {/tex} and internal resistance {tex} r , {/tex} are joined in series to form a closed circuit as shown. The potential difference across any one cell is

Zero

B

{tex} E {/tex}

C

{tex} \frac { \mathcal { E } } { n } {/tex}

D

{tex} \frac { n - 1 } { n } \varepsilon {/tex}

##### Explanation

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Q 10. In the given circuit, it is observed that the current {tex} I {/tex} is independent of the value of the resistance {tex} R _ { 6 } . {/tex} Then the resistance value must satisfy

A

{tex} R _ { 1 } R _ { 2 } R _ { 5 } = R _ { 3 } R _ { 4 } R _ { 6 } {/tex}

B

{tex} \frac { 1 } { R _ { 5 } } + \frac { 1 } { R _ { 6 } } = \frac { 1 } { \left( R _ { 1 } + R _ { 2 } \right) } + \frac { 1 } { \left( R _ { 3 } + R _ { 4 } \right) } {/tex}

{tex} R _ { 1 } R _ { 4 } = R _ { 2 } R _ { 3 } {/tex}

D

{tex} R _ { 1 } R _ { 3 } = R _ { 2 } R _ { 4 } = R _ { 5 } R _ { 6 } {/tex}

##### Explanation

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Q 11. A set of {tex} n {/tex} identical resistors, each of resistance {tex} R \Omega {/tex} , when connected in series, has an effective resistance of {tex} x {/tex} ohm. When the resistors are connected in parallel, the effective resistance is {tex} y {/tex} ohm. What is the relation between {tex} R , x , {/tex} and {tex} y ? {/tex}

A

{tex} R = \frac { x y } { ( x + y ) } {/tex}

B

{tex} R = ( y - x ) {/tex}

{tex} R = \sqrt { x y } {/tex}

D

{tex} R = ( x + y ) {/tex}

##### Explanation

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Q 12. A wire with resistance 12{tex} \Omega {/tex} is bent in the form of a circle. The effective resistance between the two points on any diameter of the circle is

A

12{tex} \Omega {/tex}

B

24{tex} \Omega {/tex}

C

6{tex} \Omega {/tex}

3{tex} \Omega {/tex}

##### Explanation

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Q 13. A wire {tex} l = 8 \mathrm { m } {/tex} long of uniform cross-sectional area {tex} A = 8 \mathrm { mm } ^ { 2 } {/tex} has a conductance of {tex} G = 2.45 \Omega ^ { - 1 } . {/tex} The resistivity of material of the wire will be

A

{tex} 2.1 \times 10 ^ { - 7 } \Omega \mathrm { m } {/tex}

B

{tex} 3.1 \times 10 ^ { - 7 } \Omega \mathrm { m } {/tex}

{tex} 4.1 \times 10 ^ { - 7 } \Omega \mathrm { m } {/tex}

D

{tex} 5.1 \times 10 ^ { - 7 } \Omega \mathrm { m } {/tex}

##### Explanation

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Q 14. The potentiometer wire {tex} A B {/tex} is 600{tex} \mathrm { cm } {/tex} long. At what distance from {tex} A {/tex} should the jockey {tex} J {/tex} touch the wire to get zero deflection in the galvanometer?

320{tex} \mathrm { cm } {/tex}

B

120{tex} \mathrm { cm } {/tex}

C

20{tex} \mathrm { cm } {/tex}

D

450{tex} \mathrm { cm } {/tex}

##### Explanation

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Q 15. In Fig. 14.20, the steady state current in 2{tex} \Omega {/tex} resistance is

A

1.5 {tex}\mathrm A {/tex}

0.9{tex} \mathrm { A } {/tex}

C

0.6{tex} \mathrm { A } {/tex}

D

Zero

##### Explanation

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Q 16. Kirchoff's second law is based on the law of conservation of

A

Momentum

B

Charge

Energy

D

Sum of mass and energy

##### Explanation

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Q 17. In the circuit shown in Fig. 14.27, the heat produced in the 5{tex} \Omega {/tex} resistor due to a current flowing in it is 10 calories per second. The heat produced in the 4{tex} \Omega {/tex} resistor is

A

1 {tex} \mathrm { cal } \mathrm { s } ^ { - 1 } {/tex}

2{tex} \mathrm { cal } \mathrm { s } ^ { - 1 } {/tex}

C

3{tex} \mathrm { cal } \mathrm { s } ^ { - 1 } {/tex}

D

4 {tex} \mathrm { cal } \mathrm { s } ^ { - 1 } {/tex}

##### Explanation

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Q 18. In the circuit shown in Fig. {tex} 14.28 , {/tex} the current through

A

the 3{tex} \Omega {/tex} resistor is 0.50{tex} \mathrm { A } {/tex}

B

the 3{tex} \Omega {/tex} resistor is 0.25{tex} \mathrm { A } {/tex}

C

the 4{tex} \Omega {/tex} resistor is 0.50{tex} \mathrm { A } {/tex}

the 4{tex} \Omega {/tex} resistor is 0.25{tex} \mathrm { A } {/tex}

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Q 19. The meter bridge circuit shown in Fig. is balanced when jockey {tex} J {/tex} divides wire {tex} A B {/tex} in two parts {tex} A J {/tex} and {tex} B J {/tex} in the ratio of {tex} 1 : 2 . {/tex} The unknown resistance {tex} Q {/tex} has value

A

1{tex} \Omega {/tex}

3{tex} \Omega {/tex}

C

4{tex} \Omega {/tex}

D

7{tex} \Omega {/tex}

##### Explanation

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Q 20. Two mole of argon are mixed with one mole of hydrogen, then Cp/Cv for the mixture is nearly

A

1.2

B

1.3

1.4

D

1.5

##### Explanation

Average degree of freedom
fav =
νmix = 1 + = 1 + = = 1.4

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Q 21. Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of gas contained in A is and that in B is . The gas in each cylinder is now allowed to expand isothermally to the same final volume 2V. The change in the pressure in A and B are found to be and 1.5 respectively. Then

A

4 = 9

B

2 = 3

3 = 2

D

9 = 4

##### Explanation

For gas in A,

Putting
We get
Similarly for Gas in B,
From eq. (I) and (II) we get 2 = 3
Hence (C) is the correct.

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Q 22. Figure (a), (b), (c) shows three different arrangements of materials 1, 2 and 3 to form a wall. Thermal conductivities are K1/> K2/> K3. The left side of the wall is 20oC higher than the right side. Temperature difference ΔT across the material 1 has following relation, in three cases:

A

ΔTa>ΔTb>ΔTc

ΔTa = ΔTb = ΔTc

C

ΔTa = ΔTb>ΔTc

D

ΔTa = ΔTb<ΔTc

##### Explanation

All are in series therefore current remains same. Hence temperature difference = (current × thermal resistance) are equal for every case.

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Q 23. The temperature drop through a two layer furnace wall is 9000C. Each layer is of equal area of cross-section. Which of the following actions will result in lowering the temperature θ of the interface ?

By increasing the thermal conductivity of outer layer

B

By increasing the thermal conductivity of inner layer

C

By increasing thickness of outer layer

D

By decreasing thickness of inner layer

##### Explanation

H = rate of heat flow
=
Now 100 - θ =
or θ = 1000 -
= 100 -
Now, we can see that θ can be decreased by increasing thermal conductivity of outer layer (K0) and thickness of inner layer (li).

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Q 24. A cylinder of radius R made of material of thermal conductivity K1 is surrounded by a cylindrical shell of inner radius R and outer radius 3R made of a material of thermal conductivity K2. The two ends of the combined system are maintained at two different temperatures. What is the effective thermal conductivity of the system?

A

K1 + K2

C

D

##### Explanation

= +
= +
Keq =

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Q 25. A rod of length l and cross-section area A has a variable thermal conductivity given by K = α T, where α is a positive constant and T is temperature in Kelvin. Two ends of the rod are maintained at temperatures T1 and T2 (T1> T2). Heat current flowing through the rod will be -

A

B

C

##### Explanation

Heat current i = - KA
idX = - KA dT

⇒ il = - A α
⇒ i = A α