Q68 / 75 · JEE Main 2026 (06 April Shift 1) · Physics
An inductor of inductance $10$ mH having resistance of $100\,\Omega$ is connected to battery of E.M.F. $1.0$ 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 $2$ mA and $4$ mA is _______.
- Correct answerA$4/3$
- B$3/4$
- C$5/3$
- D$3/5$
Solution
In an LR circuit, the equation for the voltage is given by Kirchhoff's loop law:
$E = L \dfrac{di}{dt} + iR$
The instantaneous voltage across the ideal inductor is the induced EMF, which is given by:
$V_L = L \dfrac{di}{dt} = E - iR$
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:
$\dfrac{V_{L1}}{V_{L2}} = \dfrac{0.8}{0.6} = \dfrac{4}{3}$
Answer: $4/3$
$E = L \dfrac{di}{dt} + iR$
The instantaneous voltage across the ideal inductor is the induced EMF, which is given by:
$V_L = L \dfrac{di}{dt} = E - iR$
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:
$\dfrac{V_{L1}}{V_{L2}} = \dfrac{0.8}{0.6} = \dfrac{4}{3}$
Answer: $4/3$
Hence, the correct answer is option: 'A'.