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

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Q 1. A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if

A

both the length and the radius of the wire are halved.

both the length and the radius of the wire are doubled.

C

the radius of the wire is doubled.

D

the length of the wire is doubled.

Explanation

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Q 2. A piece of copper and another of germanium are cooled from room temperature to {tex} 80 ^ { \circ } \mathrm { K } . {/tex} The resistance of

A

each of them increases

B

each of them decreases

C

copper increases and germanium decreases

copper decreases and germanium increases

Explanation

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Q 3. In the circuit {tex} P \neq R , {/tex} the reading of the galvanometer is same with switch {tex} S {/tex} open or closed. Then

{tex} I _ { R } { = } I _ { G } {/tex}

B

{tex} I _ { P } = I _ { G } {/tex}

C

{tex} I _ { Q } = I _ { G } {/tex}

D

{tex} I _ { Q } = I _ { R } {/tex}

Explanation

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Q 4. The three resistance of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation.

{tex} \mathrm { III } < \mathrm { II } < \mathrm { IV } < \mathrm { I } {/tex}

B

{tex} \mathrm { II } < \mathrm { III } < \mathrm { IV } < \mathrm { I } {/tex}

C

{tex} \mathrm { I } < \mathrm { IV } < \mathrm { III } < \mathrm { II } {/tex}

D

{tex} \mathrm { I } < \mathrm { III } < \mathrm { II } < \mathrm { IV } {/tex}

Explanation


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Q 5. Shown in figure is a Post Office box. In order to calculate the value of external resistance, it should be connected between

A

{tex} B ^ { \prime } {/tex} and {tex} C^{\prime} {/tex}

{tex} A {/tex} and {tex} D {/tex}

C

{tex} C {/tex} and {tex} D {/tex}

D

{tex} B {/tex} and {tex} D {/tex}

Explanation

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Q 6. Find out the value of current through {tex} 2 \Omega {/tex} resistance for the given circuit.

zero

B

{tex} 2 A {/tex}

C

{tex} 5 A {/tex}

D

{tex} 4 A {/tex}

Explanation

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Q 7. A {tex} 4 \mu F {/tex} capacitor, a resistance of {tex} 2.5 M \Omega {/tex} is in series with {tex} 12 \mathrm { V } {/tex} battery. Find the time after which the potential difference across the capacitor is {tex}3{/tex} times the potential difference across the resistor. {tex} [ \text { Given } \ln ( 2 ) = 0.693 ] {/tex}

{tex} 13.86 s {/tex}

B

{tex} 6.93 \mathrm { s } {/tex}

C

{tex} 7 \mathrm { s } {/tex}

D

{tex} 14 \mathrm { s } {/tex}

Explanation


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Q 8. A resistance of {tex} 2 \Omega {/tex} is connected across one gap of a metre- bridge (the length of the wire is {tex} 100 \mathrm { cm } {/tex} ) and an unknown resistance, greater than {tex} 2 \Omega , {/tex} is connected across the other gap. When these resistances are interchanged, the balance point shifts by {tex} 20 \mathrm { cm } . {/tex} Neglecting any corrections, the unknown resistance is

{tex} 3 \Omega {/tex}

B

{tex} 4 \Omega {/tex}

C

{tex} 5 \Omega {/tex}

D

{tex} 6 \Omega {/tex}

Explanation


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Q 9. Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, {tex} 100 \mathrm { W } , 60 \mathrm { W } {/tex} and {tex} 40 \mathrm { W } {/tex} bulbs have filament resistances {tex} R _ { 100 } , R _ { 60 } {/tex} and {tex} R _ { 40 } {/tex} respectively, the relation between these resistances is

A

{tex} \frac { 1 } { R _ { 100 } } = \frac { 1 } { R _ { 40 } } + \frac { 1 } { R _ { 60 } } {/tex}

B

{tex} R _ { 100 } = R _ { 40 } + R _ { 60 } {/tex}

C

{tex} R _ { 100 } > R _ { 60 } > R _ { 40 } {/tex}

{tex} \frac { 1 } { R _ { 100 } } > \frac { 1 } { R _ { 60 } } > \frac { 1 } { R _ { 40 } } {/tex}

Explanation