Ratio of Instantaneous Voltages Across Inductor in RL Circuit

To find the ratio of the instantaneous voltages across the inductor, we apply Kirchhoff's Voltage Law (KVL) to the single-loop $RL$ circuit after the switch is closed.

For a battery of EMF $E$, connected to a real inductor having self-inductance $L$ and internal resistance $R$, the circuit equation at any instant $t$ is given by: $E - i(t)R - L\frac{di}{dt} = 0$

The instantaneous voltage across the pure inductive element (induced EMF) is given by: $V_L = L\frac{di}{dt} = E - iR$

Given data:

• $\text{EMF of battery, } E = 1.0\text{ V}$
• $\text{Resistance of inductor, } R = 100\ \Omega$
• $\text{Inductance, } L = 10\text{ mH}$

Case 1: When the current passing through the circuit is $i_1 = 2\text{ mA} = 2 \times 10^{-3}\text{ A}$: $V_{L1} = E - i_1 R = 1.0 - (2 \times 10^{-3}\text{ A} \times 100\ \Omega)$ $V_{L1} = 1.0 - 0.2 = 0.8\text{ V}$

Case 2: When the current passing through the circuit is $i_2 = 4\text{ mA} = 4 \times 10^{-3}\text{ A}$: $V_{L2} = E - i_2 R = 1.0 - (4 \times 10^{-3}\text{ A} \times 100\ \Omega)$ $V_{L2} = 1.0 - 0.4 = 0.6\text{ V}$

Ratio of Instantaneous Voltages: $\text{Ratio} = \frac{V_{L1}}{V_{L2}} = \frac{0.8\text{ V}}{0.6\text{ V}} = \frac{4}{3}$

Thus, the required ratio is $\frac{4}{3}$, which corresponds to Option A.