1) In the circuit shown here, the point C is kept connected to point A till the current flowing through the circuit becomes constant. Afterwards, suddenly point C is disconnected from point A and connected to point B at time t=0, Ration of the voltage across resistance and the inductor at t=L/R will be equal to A) $\frac{e}{1-e}$ B) 1 C) -1 D) $\frac{1-e}{e}$ Answer: Option CExplanation: After connecting C to B hanging the switch, the circuit will act like an L-R discharging circuit. Applying Kirchhoff's loop equation, $V_{R}+V_{L}=0\Rightarrow V_{R}=-V_{L}$ $\frac{V_{R}}{V_{L}}=-1$
