Question

An inductor of inductance L = 400 mH and resistors of resistance $R_1=2\Omega$ and $R_2=2\Omega$ areconnected to a battery of emf 12 V as shown in the figure. Theinternal 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.
• B.
• C.
• D.

Answer 1

The correct option is D

$I_2=\frac{E}{R_1}=\frac{12}{2}=6AE=L\frac{d{l}_2}{dt}+R_2\times l_2{l}_2=l_0(1-e^{-t/t_0})\Rightarrow l_0=\frac{E}{R_2}=\frac{12}{2}=6A{t}_0=\frac{L}{R}=\frac{400\times {10}^{-3}}{2}0.2l_2=6(1-e^{-t/0.2})$

potential drop across $L=E-R_2{l}_2=12-2\times 6(1-e^{-bt})=12e^{-5t}$