JEE Main 2026 — Electromagnetic Induction Question with Solution

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

$E=L\frac{di}{dt}+iR$

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

$V_L=L\frac{di}{dt}=E-iR$

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

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

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

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

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

The ratio of the instantaneous voltages is:

$\frac{V_{L1}}{V_{L2}}=\frac{0.8}{0.6}=\frac{4}{3}$

Answer: $4/3$