Question

A pure inductor, pure capacitor, and two pure equal resistors are connected as shown in the circuit, such that $L=2R^{2}C$.
At an instant $t=0$, the switch is closed. If $i_1$ and $i_2$ are currents shown in branches at some time $t=t_o$ relation $\frac{d{i}_1}{dt}+\frac{d{i}_2}{dt}=0$ is satisfied, then

• A. $t_o=2RC\ln 2$
• B. $t_o=4RC\ln 2$
• C. $t_o=3RC\ln 2$
• D. $t_o=3RC\ln 4$

Answer 1

The correct option is A $t_o=2RC\ln 2$


$i_1=\frac{\epsilon }{R}[1-e^{-Rt/L}]$
$i_2=\frac{dq}{dt}=\frac{d}{dt}\left[q_o\right(1-e^{-t/RC}\left)\right]i_2=\frac{\epsilon }{R}{e}^{-t/RC}$
So,
$\frac{d{i}_2}{dt}=-\frac{\epsilon }{R^{2}C}{e}^{-t/RC},\frac{d{i}_1}{dt}=\frac{\epsilon }{L}{e}^{-Rt/L}$
$\frac{d{i}_1}{dt}+\frac{d{i}_2}{dt}=\frac{\epsilon }{L}{e}^{-Rt/L}-\frac{2\epsilon }{L}{e}^{-2Rt/L}=0$
$\Rightarrow 2e^{R{t}_o/L}=e^{2R{t}_o/L}\Rightarrow \ln 2+\frac{R{t}_o}{L}=\frac{2R{t}_o}{L}$
$\Rightarrow t_o=2\left(RC\right)\ln 2$