Question

An inductor of inductance $L=400mH$ and resistors of resistances $R_1=2\Omega$ and $R_2=2\Omega$ are connected to a battery of emf $12$ $V$ as shown in the figure. The intemal resistance of the battery is negligible. The switch $S$ is closed at $t=0$. The potential drop across $L$ as a function of time is :

• A. $6$ $e^{-5t}V$
• B. $\frac{12}{t}{e}^{-3t}V$
• C. $6(1-e^{-t/0.2})V$
• D. $12$ $e^{-5t}$ $V$

Answer 1

Answer: (D)

$12$ $e^{-5t}$ $V$

$I=\frac{F_1}{R}=\frac{12}{2}=6A$
$E=L\frac{d{I}_2}{dt}+R_2{I}_2$
$I_2=I_0(1-e^{\frac{t}{t_c}})\Rightarrow I_0=\frac{E}{R_2}=\frac{12}{2}=6A$
$t_c=\frac{L}{R}=\frac{400\times {10}^{-3}}{2}=0.2$
$I_2=6(1-e^{\frac{t}{0.2}})$
Potential drop across $L=E-R_2{I}_2=12-2\times 6(1-e^{\frac{t}{0.2}})=12e^{-5t}$