Question

An inductor of inductance L = 400 mH and resistors of resistance $R_1=4\Omega$ and $R_1=4\Omega$ are connected to battery of emf 12V as shown in the figure. The internal resistance of the battery is negligible. The swich 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. $6(1-e^{-t/0.2})V$
• D. $12e^{-5t}V$

Answer 1

The Correct Option is D

$l_1 = \frac{F}{R_1} = \frac{12}{2} = 6A$ $E = L \frac{dl_2}{dt} + R_2 \times l_2$ $I_2 = I_n (1 - e^{-t/t_c} )$ $\Rightarrow$ $I_0 = \frac{E}{R_2} = \frac{12}{2} = 6 A$ $t_c = \frac{L}{R} = \frac{400 \times 10^{-3}}{2} = 0.2$ $I_2 = 6 ( 1 - e^{-t/0.2})$ Potential drop across $L = E - R_2 L_2 = 12 - 2 \times 6 ( 1 - e^{-bt})$ = 12$e^{-5t}$