Question

In the circuit, the switch is closed at $t=0.$ Then-

• A. At $t=0,I_1=I_2=0$
• B. At any time $t,\frac{I_1}{I_2}=\frac{L_2}{L_1}$
• C. At any time $t=\infty ,I_1+I_2=\frac{E}{R}$
• D. At$t=\infty ,I_1$ and $I_2$ are independent of $L_1$ and $L_2$ respectively.

Answer 1

Answer: (A), (B), (C), (D)

At$t=\infty ,I_1$ and $I_2$ are independent of $L_1$ and $L_2$ respectively.

As we know that at $t=0$, just after the connection, inductor acts like an open switch,

Thus, $I_1=I_2=0$

Since coils are in parallel. Hence, potential difference across them will be the same.

At any instant time $t,$

$\left|\begin{array}{c}{E}_1\end{array}\right|=\left|\begin{array}{c}{E}_2\end{array}\right|$

$L_1\frac{d{I}_1}{dt}=L_2\frac{d{I}_2}{dt}$

Integrating both sides with proper limit, we get,

$\Rightarrow {\int }_0^{I_1}{L}_{1}d{I}_1={\int }_0^{I_2}{L}_{2}d{I}_2$

$\Rightarrow L_1{I}_1=L_2{I}_2$

$\frac{L_1}{L_2}=\frac{I_2}{I_1}$

At $t=\infty$, inductor acts like closed switch,

Hence, $I_1$ and $I_2$ will be independent of $L_1$ and $L_2$ respectively.

Also, $I_1+I_2=\frac{E}{R}$ only at $t=\infty$.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, all the options are correct.

Why this question ? Key concept: You can remember that an inductor acts like an open switch $(R=\infty )$ initially at $(t=0)$ and like a closed switch $(R=0)$ finally at $(t=\infty )$.