Question

In the given circuit, let $i_{1}$ be the current drawn from battery at time $t=0$ and $i_{2}$ be steady current at $t=\infty$, then the ratio $\frac{i_{1}}{i_{2}}$ is

• A. 1.0
• B. 0.8
• C. 1.2
• D. 1.5

Answer 1

Answer: (B)

At $t=0$, inductor behaves as open circuit, so
$i_{1}=\frac{10}{10}=1 A$
At $t=\infty$, inductor behave as short circuit, so
$i_{2}=\frac{10}{8}=\frac{5}{4} A$
Hence, $\frac{i_{1}}{i_{2}}=\frac{1}{5 / 4}$
$=\frac{4}{5}=0.8$