Question

Consider the circuit shown in figure. The current through the battery a long time after the switch $S$ is closed is

• A. $\frac{E}{R_{1}}$
• B. $\frac{E}{R_{2}}$
• C. $\frac{E}{R_{1}+R_{2}}$
• D. $\frac{E\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$

Answer 1

Answer: (D)

After a long time, steady current flows, so no emf is induced across inductor.
Hence, circuit reduces like shown. So, current flowing,
$i=\frac{E\left(R_{1}+R_{2}\right)}{R_{1} R_{2}}$