Question

In the circuit shown in fig, the switch is closed at $t=0$

• 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, $I_1+I_2=\frac{\epsilon }{R}$
• D. at $t=\infty ,I_1$ and $I_2$ are independent of $L_1$ and $L_2$

Answer 1

Answer: (A), (B)

At $t=0,I_1=I_2=0$
at any time t, $\frac{I_1}{I_2}=\frac{L_2}{L_1}$

The current through the inductor is given by

$I_{ind}=I_o\left({1-e}^{\frac{-tL}{R}}\right)$

thus at $t=0$ : the inductor will have zero current

at $t=\infty$ : the inductor will have maximum current $I_0.$

Therefore at $t=0$ , current $I_1$ and $I_2$ are zero

also, since the potential across both inductors will be same for any time $t$.

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

$\frac{L_1}{L_1}=\frac{d{I}_2}{d{I}_1}=\frac{I_{2t}-I_{20}}{I_{1t}-I_{10}}=\frac{I_2}{I_1}$ since $I_{10}=I_{20}=0.$