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 $12V$ as shown in the figure. The internal 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. $6e^{-5t}V$
• B. $\frac{12}{t}{e}^{-3t}V$
• C. $6\left(1-e^{-t/0.2}\right)V$
• D. $12e^{-5t}V$

Answer 1

Answer: (D)

$I_1=\frac{F}{R_1}=\frac{12}{2}=6A$
$E=L\frac{d{I}_2}{dt}+R_2\times I_2$
$I_2=I_o\left(1-e^{-t/t_c}\right)\Rightarrow I_o=\frac{E}{R_2}=\frac{12}{2}=6A$
$t_c=\frac{L}{R}=\frac{400\times 10^{-3}}{2}=0.2$
$I_2=6\left(1-e^{t/0.2}\right)$
Potential drop across $L=E-R_2{I}_2=12-2\times 6\left(1-e^{-bt}\right)=12e^{-5t}$
Directions: Question numbers 28, 29 and 30 are based on the following paragraph. Two moles of helium gas are taken over the cycle ABCDA, as shown in the P - T diagram.