Question

In the circuit shown below, switch S was closed for a long time. If the switch is opened at $t=0$, the maximum magnitude of the voltage $V_R$, in volts, is _____________ (rounded off to the nearest integer). 

Hint:

At $t=0^+$, inductor current is continuous and equals its $t=0^-$ value. In RL transients, the peak resistor voltage right after switching is simply $R\,i_L(0^-)$.

Answer 1

Step 1: Initial steady state before switch opens (t < 0)
When the switch is closed for a long time, the circuit is in DC steady state. Under DC, an inductor acts as a short circuit. - The branch with the inductor shorts out the 2 Ω resistor. - Therefore, the 2 V source drives current only through the 1 Ω resistor into the inductor short. - The inductor current just before opening the switch is: \[ i_L(0^-) = \frac{2\ \text{V}}{1\ \Omega} = 2\ \text{A}. \] So at t = 0⁻, the inductor carries 2 A steady current.

Step 2: Immediately after switch opens (t = 0⁺)
When the switch opens, the source is disconnected. The inductor current cannot change instantaneously, so it must remain 2 A at t = 0⁺. Now, the current flows through the 2 Ω resistor, since the inductor releases its stored energy.

Step 3: Equivalent decay circuit
Now the inductor current decays in an RL circuit with resistance \(R = 2\ \Omega\) and inductance \(L = 1\ \text{H}\). The time constant is \[ \tau = \frac{L}{R} = \frac{1}{2}\ \text{s}. \] Thus, the current for t ≥ 0 is \[ i_L(t) = i_L(0^+) e^{-t/\tau} = 2 e^{-2t}. \]

Step 4: Voltage across the 2 Ω resistor
The resistor voltage is proportional to the inductor current: \[ V_R(t) = (2\ \Omega)\, i_L(t) = 2 \times 2 e^{-2t} = 4 e^{-2t}. \] At t = 0⁺, this reaches its maximum value: \[ V_{R,\max} = 4\ \text{V}. \]

Final Answer:
\[ \boxed{4} \]