JEE Advanced Previous Year Questions (2018 - 2024): Current Electricity | Physics for JEE Main & Advanced PDF Download

2024

No Question in JEE Advanced Exam from Chapter Current Electricity in Year 2023 !

2023

No Question in JEE Advanced Exam from Chapter Current Electricity in Year 2023 !

2022

Q1: Two resistances R₁ = XΩ and R₂ = 1Ω are connected to a wire AB of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from   0.2 mm at A to   1 mm at B. A galvanometer (G) connected to the center of the wire,   50 cm from each end along its axis, shows zero deflection when A and B are connected to a battery. The value of X is ____________. [JEE Advanced 2022 Paper 2]

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Ans: 5
Solution: According to the given condition wire is having different cross-section at two junction. (Frustum shaped conductor)

For the shown conductor in the diagram.

Hence, Resistance of left  50 cm wire

Resistance of Right  50 cm wire

For wheatstone balanced condition,

Q2:  In the following circuit . The charge stored in C₃ is ____________ μC.          [JEE Advanced 2022 Paper 1]

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Ans:

Q3: The figure shows a circuit having eight resistances of 1Ω each, labelled R1 to R8, and two ideal batteries with voltages  ε₁=12 V and  ε₂=6 V.  

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Ans:(a), (b), (c) & (d)

Solution:

Point A and B are at same potential so they can be merged/folded.

For loop 1

For loop 2

= 1.2 A

Current in R₃ = 24/5 = 4.8A

Current in R₄ = 12/5 = 2.4A

2021

Q1: In order to measure the internal resistance r₁ of a cell of emf E, a meter bridge of wire resistance R₀ = 50Ω, a resistance R₀/2, another cell of emf E/2 (internal resistance r) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r₁ = __________ Ω. [JEE Advanced 2021 Paper 2]

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Ans: 3

Solution: 

Resistance of potential wire, R₀ = 50Ω

Resistance of 100 m wire = 50Ω

So, resistance of 72 cm wire = 50 / 100 × 72 = 36Ω

Current,

2020

Q1: In the balanced condition, the values of the resistances of the four arms of a Wheatstone bridge are shown in the figure below. The resistance R₃ has temperature coefficient 0.0004∘C^-1. If the temperature of R₃ is increased by 100∘C, the voltage developed between S and T will be ________ volt.               [JEE Advanced 2020 Paper 2]

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Ans: 0.27 

Solution:

Now,

Q2: Shown in the figure is a semicircular metallic strip that has thickness t and resistivity ρ. Its inner radius is R₁ and outer radius is R₂. If a voltage V₀ is applied between its two ends, a current I flows in it. In addition, it is observed that a transverse voltage ΔV develops between its inner and outer surfaces due to purely kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is schematic and not drawn to scale)    [JEE Advanced 2020 Paper 1]

(a)
(b) the outer surface is at a higher voltage than the inner surface
(c) the outer surface is at a lower voltage than the inner surface
(d)

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Ans: (a), (c) & (d)

Solution: 

All the elements are in parallel.

So, Option (a) is correct.

Electrons are revolving in a circular path so there is a centripetal force on the electron, which provide a transverse electric field.
The direction of transverse electric field is radially out side hence inner surface is at higher potential.
So, option (C) is correct.

2019

Q1: A parallel plate capacitor of capacitance C has spacing d between two plates having area A. The region between the plates is filled with N dielectric layers, parallel to its plates, each with thickness, . The dielectric constant of the m^th layer is . For a very large N(>10³), the capacitance C is .

The value of α will be______
[∈₀ is the permittivity of free space.] [JEE Advanced 2019 Paper 1]

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Ans: 1
Solution:Parallel plate capacitor,

Integration on both sides, we get

Therefore, α = 1.

Q2: In the circuit shown, initially there is no charge on the capacitors and keys S₁ and S₂ are open. The values of the capacitors are C₁ = 10μF, C₂ = 30μF and C₃ = C₄ = 80μF. 

View Answer  

Ans: (b) & (c)
Solution: Just after closing of switch charge on any capacitor is zero.
∴  Replace all capacitors by conducting wires.

Current flow in the circuit,

At steady state for S₁ closed is

At steady state potential difference between 'P' and 'Q' is 4 volt.

Now when switch S₂ is closed (Then equivalent circuit)