Question

In the circuit shown in figure, $E=120V,R_1=3.0\Omega ,R_2=5.0\Omega$ and $L=0.200H$. Switch $S$ is closed at $t=0$. The switch is closed for a long time and then opened. Just after the switch is opened. What is the potential difference $V_{ab}$ across the resistor $R_1$?

Answer 1

When the switch is opened, current through $R_1$ will immediately become zero. While through $R_2$, will decrease to zero from the value $\frac{E}{R_2}=2.4A=i_0$(say), exponenetially. Path of this decay of current will be $cdbac$.
$\therefore$ Just after the switch is opened,
$V_{ab}=-i_0{R}_1=-2.4\times 3.0=-7.2volt$