Current Electricity JEE Advanced Previous Year Questions With Solutions

The branch of physics which deals with the dynamics of charges is called current electricity. The charges, in general, can be positive ions, negative ions, electrons or holes. A current is said to exist in the wire only when there is a net flow of charge through any section of the wire. The conventional direction of the current is the direction in which positive charges flow, namely from the point of higher potential to the point of lower potential.

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Question 1) Two cells of emf 2E and E with internal resistance r₁ and r₂respectively are connected in series to an external resistor R (see figure). The value of R, at which the potential difference across the terminals of the first cell becomes zero is

(A) r₁ – r₂

(B) r₁ + r₂

(C) (r₁/2) + r₂

(D) (r₁/2) – r₂

Answer (D)

Solution

\(\begin{array}{l}i = \frac{3E}{R+r_{1}+r_{2}}\end{array} \)

Current Electricity JEE Advanced Previous Year Solved Questions

If potential difference across terminals of first cell is zero

V_A = V_B

2 E = i r₁

\(\begin{array}{l}2E = \frac{3E}{R+r_{1}+r_{2}}r_{1}\end{array} \)

2R + 2r₁ + 2r₂ = 3r₁

R = (r₁/2)- r₂

Question 2) The four arms of a Wheatstone bridge have resistances as shown in the figure. A galvanometer of 15 resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

(A) 4.87 mA

(B) 4.87 μA

(C) 2.44 μA

(D) 2.44 mA

Answer (A)

Solution:

Applying KCL for point B,

\(\begin{array}{l}\frac{V_{B}-10}{100}+\frac{V_{B}-V_{D}}{15}+\frac{V_{B}-0}{10}=0\end{array} \)

\(\begin{array}{l}\frac{V_{B}-10}{20}+\frac{V_{B}-V_{D}}{3}+\frac{V_{B}-0}{2}=0\end{array} \)

3V_B-30+20V_B-20V_D+30V_B=0

53V_B– 20V_D = 30 ____(1)

Similarly applying KCL for point D,

\(\begin{array}{l}\frac{V_{D}-10}{60}+\frac{V_{D}-V_{B}}{15}+\frac{V_{D}-0}{5}=0\end{array} \)

V_D – 10 + 4V_D – 4V_B+ 12V_D= 0

– 4V_B + 17V_D = 10 _____(2)

after solving equation (1) & (2)

V_D = 0.792 volt

V_B = 0.865 volt

Then the current through the galvanometer

\(\begin{array}{l}=\frac{V_{B}-V_{D}}{R}\end{array} \)

\(\begin{array}{l}=\frac{0.865-0.792}{15}\end{array} \)

= 4.67 mA

Question 3) What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?

(A) Both , including reactance and current will be doubled

(B) Both, inductive reactance and current will be halved

(C) Inductive reactance will be halved and current will be doubled

(D) Inductive reactance will be doubled and current will be halved.

Answer: (C)

Solution

X_L = ωL

If frequency is halved, X’_L=(X_L/2)

[inductive reactance is halved]

Therefore, I = V/X_L

& I’ = 2V/X_L=2I [current will be doubled]

Question 4) The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60 W at the same distance is √(x/5)E. Where the value of x = ________

Answer (3)

Solution:

The intensity of electromagnetic radiation

I = 1/2(C∈₀E²)

where E is electric field intensity at a point

E² ∝ I

I = Power/Area

E² ∝ (P/A)

E ∝ √P

[at the same distance, A will be the same]

\(\begin{array}{l}\frac{E’}{E}=\sqrt{\frac{60}{100}}\end{array} \)

\(\begin{array}{l}E’=\sqrt{\frac{3}{5}}E\end{array} \)

So the value of x = 3

Question 5) A 2μF capacitor C₁ is first charged to a potential difference of 10V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C₂ of 8 μF. The charge in C₂ on equilibrium conditions is ________ μC. (Round off to the Nearest Integer)

JEE Advanced Previous Year Paper with Answers Current Electricity

Answer (16)

Solution

After capacitor C₁ is fully charged,

JEE Advanced Past Year Paper with Answers Current Electricity

When battery is removed & the capacitor is connected

At equilibrium condition, let voltage across each capacitor be V.

Then, using conservation of charge

2V + 8V = 20

10V = 20

V = 2 volt

Q = CV

Q = 8×2 = 16 μc

Question 6) The electric field in a region is given by

\(\begin{array}{l}\vec{E}=\frac{2}{3}E_{0}i+\frac{3}{5}E_{0}j\end{array} \)

with E₀=4.0×10³N/C. The flux of this field through a rectangular surface are 0.4 m² parallel to Y–Z plane is ________Nm²C^-1.

Answer (640)

Solution:

From Gauss’ law

\(\begin{array}{l}\phi =\oint \vec{E}.d\vec{A}\end{array} \)

= (⅔)E₀ x (0.4)

= (⅔) x 4 x 10³ x 0.4

Φ = 640 Nm² c^-1

Question 7) In the given circuit find the current through 6Ω resistance

(A) 10 A

(B) 7 A

(C) 25 A

(D) 30 A

Answer: (A) 10 A

Solution:

Applying Kirchoff’s law,

\(\begin{array}{l}\frac{V-140}{20}+\frac{V-0}{6}+\frac{V-90}{5}=0\end{array} \)

3(V-140)+10V+(V-90) =0

3V-420+10V+12V-1080 = 0

25V= 1500

V = 60

Current through 6Ω resistance, I = V/R = 60/6 = 10 A

Question 8) An AC circuit consists of a series combination of an inductance L 1 mH, a resistance R = 1Ω and a capacitance C. It is observed that the current leads the voltage by 45°. Find the value of capacitance ‘C’ if the angular frequency of applied AC is 200 rad/s.

(A) 5.6 mF

(B) 3.92 mF

(C) 4.16 mF

(D) 5.2 mF

Answer: (C) 4.16 mF

Solution

\(\begin{array}{l}tan \Phi =\frac{X_{c}-X_{L}}{R}=\frac{\frac{1}{\omega C}-\omega L}{R}\end{array} \)

\(\begin{array}{l}1(1)=\frac{1}{200C}-200(10^{-3})\end{array} \)

\(\begin{array}{l}\frac{1}{200C}= 1+0.2 = 1.2\end{array} \)

\(\begin{array}{l}C = \frac{1}{200\times 1.2}=0.00416 F = 4.16 mF\end{array} \)

Question 9) In a magnesium rod of area 3m², current I = 5𝐴 is flowing angle of 60⁰ from the axis of the rod. The resistivity of the material is 44 × 10⁻² ohm x m. Find an electric field inside the rod

(A) 0.567

(B) 0.367

(C) 0.667

(D) 0.767

Answer: (B) 0.367

Solution

\(\begin{array}{l}\frac{I}{A_{effective}}=\frac{E}{\rho }\end{array} \)

\(\begin{array}{l}E = \frac{\rho I}{A}cos 60^{0}=\frac{44\times 10^{-2}\times 5}{3\times 2}\end{array} \)

E = 0.367

Question 10) A RLC circuit is in its resonance condition. Its circuit components have value

R = 5Ω, L = 2H, C = 0.5mF, V= 250V

Then find the power in the circuit.

(A) 6 kW

(B) 10 kW

(C) 12 kW

(D) 12.5 kW

Answer: (D)

Solution

A circuit is in resonance. Thus

Also Read:

Current Electricity JEE Main Previous Year Questions With Solutions

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