Physics - EMI/AC Question with Solution | TestHub

For inductors in parallel, the equivalent inductance $L_{eq}$ is given by:

$$\frac{1}{L_{eq}}=\frac{1}{L}+\frac{1}{2L}=\frac{2+1}{2L}=\frac{3}{2L}$$

$$L_{eq}=\frac{2L}{3}$$

The time constant for an RL circuit is given by:

$$\tau =\frac{L_{eq}}{R}$$

Substituting the value of $L_{eq}$:

$$\tau =\frac{2L/3}{R}=\frac{2L}{3R}$$

$$i_{total}(t)=\frac{V}{R}\left(1-e^{-\frac{t}{2L/(3R)}}\right)$$

$$i_{total}(t)=\frac{V}{R}\left(1-e^{-\frac{3Rt}{2L}}\right)$$

This is the total current flowing from the resistor R into the parallel combination of inductors.

The current through inductor L, denoted as $i_L(t)$, and the current through inductor 2L, denoted as $i_{2L}(t)$, will satisfy:

$$i_L(t)+i_{2L}(t)=i_{total}(t)$$

And the voltage across both parallel inductors is the same:

$$L\frac{d{i}_L(t)}{dt}=2L\frac{d{i}_{2L}(t)}{dt}$$

Integrating both sides with respect to time, assuming initial currents are zero:

$$L{i}_L(t)=2L{i}_{2L}(t)$$

$$i_L(t)=2i_{2L}(t)$$

This means the current through inductor L is twice the current through inductor 2L. The ratio of currents is $i_L:i_{2L}=2:1$.

Now, substitute this into the sum equation:

$$2i_{2L}(t)+i_{2L}(t)=i_{total}(t)$$

$$3i_{2L}(t)=i_{total}(t)$$

$$i_{2L}(t)=\frac{1}{3}{i}_{total}(t)$$

And for $i_L(t)$:

$$i_L(t)=2i_{2L}(t)=2\left(\frac{1}{3}{i}_{total}(t)\right)=\frac{2}{3}{i}_{total}(t)$$

Step 6: Substitute the expression for $i_{total}(t)$ into the equation for $i_L(t)$.

$$i_L(t)=\frac{2}{3}\left(\frac{V}{R}\left(1-e^{-\frac{3Rt}{2L}}\right)\right)$$

$$i_L(t)=\left(\frac{2V}{3R}\right)\left[1-e^{-\frac{3Rt}{2L}}\right]$$

This matches option A.

The final answer is $\boxed{A}$