Question

An inductor of inductance $L=200\text{mH}$ and the resistors $R_1$ and $R_2$, each of which is equal to $4\Omega$, are connected to a battery of emf $12\text{V}$ through a switch $S$ as shown in the Figure. The switch $S$ is closed and after the steady state is reached, the switch $S$ is opened. The current in $R_1$ then is $($in $A)$

• A. $6e^{-20t}\text{from B to A}$
• B. $3e^{-40t}\text{from A to B}$
• C. $3e^{-40t}\text{from B to A}$
• D. $6e^{-20t}\text{from A to B}$

Answer 1

Answer: (C)

$3e^{-40t}\text{from B to A}$

When S is closed at t = 0 s, current in $L{R}_2$ branch

$i_2=3(1-e^{-\frac{R_2}{L}t}),\frac{R_2}{L}=\frac{4}{2\times {10}^{-1}}=20s^{-1}$

$=3(1-e^{-20t})A$

Current i, through $R_1=3A$

When S is opened, current in R${}_1$ reduces to zero just after opening.

But in $L-R_2$, it decreases exponentially

$\therefore$ Current through $R_1=3e^{-40t}$ A, from B to A