Question

In the L-R circuit shown in the figure, the key $K$ is closed at time $t=0$, then

• A. the current through $R_1$ decreases with time $t$
• B. the potential drop across $L$ increases with $t$
• C. the magnetic energy stored in the steady state equals $\frac{1}{2}L\frac{E^2}{R_2^2}$
• D. the total current through the battery is, in the steady state, $\frac{E}{R_1}+\frac{E}{R_2}$

Answer 1

Answer: (C), (D)

The correct options are
C the magnetic energy stored in the steady state equals $\frac{1}{2}L\frac{E^2}{R_2^2}$
D the total current through the battery is, in the steady state, $\frac{E}{R_1}+\frac{E}{R_2}$

A) The current through $R_1$ is constant.
B) At $t=0$, inductor behaves like an infinite resistance and hence potential difference across $L$ is $E$. At $t=\infty$ ( steady state), $L$ behaves like a wire. Hence potential difference across it is zero.
After the steady state is achieved, the circuit will behave as shown


(C) Energy stored in the inductor in steady state is given by
$U=\frac{1}{2}L{i}_2^2=\frac{1}{2}L{\left(\frac{E}{R_2}\right)}^2$

(D) At steady state,
$i=i_1+i_2=\frac{E}{R_1}+\frac{E}{R_2}$