Electromagnetic Induction

JEE Main level, single correct, medium

144 of 153

Question 144 of 153

An inductor of inductance 1010 mH having resistance of 100 Ω100\,\Omega is connected to battery of E.M.F. 1.01.0 V through a switch as shown in the figure below. After switch is closed, the ratio of instantaneous voltages across the inductor when the current passing through it is 22 mA and 44 mA is _______.

Show answer and solution

Answer: (A) 4/34/3

In an LR circuit, the equation for the voltage is given by Kirchhoff's loop law:

E=Ldidt+iRE = L \dfrac{di}{dt} + iR

The instantaneous voltage across the ideal inductor is the induced EMF, which is given by:

VL=Ldidt=E−iRV_L = L \dfrac{di}{dt} = E - iR

Given values are:
E=1.0E = 1.0 V
R=100 ΩR = 100\,\Omega

For current i1=2i_1 = 2 mA =2×10−3= 2 \times 10^{-3} A, the voltage across the inductor is:

VL1=1.0−(2×10−3×100)=1.0−0.2=0.8V_{L1} = 1.0 - (2 \times 10^{-3} \times 100) = 1.0 - 0.2 = 0.8 V

For current i2=4i_2 = 4 mA =4×10−3= 4 \times 10^{-3} A, the voltage across the inductor is:

VL2=1.0−(4×10−3×100)=1.0−0.4=0.6V_{L2} = 1.0 - (4 \times 10^{-3} \times 100) = 1.0 - 0.4 = 0.6 V

The ratio of the instantaneous voltages is:

VL1VL2=0.80.6=43\dfrac{V_{L1}}{V_{L2}} = \dfrac{0.8}{0.6} = \dfrac{4}{3}

Answer: 4/34/3

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electromagnetic induction simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Electromagnetic Induction (LR Circuit), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.