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:

Transient analysis problems follow a two-step process:
1. Analyze the circuit in DC steady state just before the switching event (at \( t = 0^- \)) to find initial inductor currents \( i_L(0^-) \) and capacitor voltages \( v_C(0^-) \). Inductors are treated as shorts and capacitors as opens.
2. Analyze the new circuit configuration for \( t > 0 \) using the initial conditions, since \( i_L(0^+) = i_L(0^-) \) and \( v_C(0^+) = v_C(0^-) \).

Answer 1

Correct Answer: 4

1. Steady state before opening (\(t=0^-\)).
In steady state the inductor behaves like a short circuit. The total DC resistance seen by the 2 V source (with switch closed) is just the \(1 \,\Omega\) resistor. Thus the inductor current just before opening is \[ i_L(0^-) = \frac{V}{R} = \frac{2 \,\text{V}}{1 \,\Omega} = 2 \,\text{A}. \] Since the inductor current cannot change instantaneously, \[ i_L(0^+) = i_L(0^-) = 2 \,\text{A}. \]
2. Circuit for \(t > 0\) (switch opened).
When the switch opens, the source is disconnected and the inductor current flows through the \(1 \,\Omega\) resistor only. The inductor current decays exponentially with time constant \[ \tau = \frac{L}{R} = \frac{1 \,\text{H}}{1 \,\Omega} = 1 \,\text{s}, \] so \[ i_R(t) = i_L(t) = 2e^{-t/\tau} = 2e^{-t} \quad (t \ge 0). \]
3. Voltage across the resistor.
The resistor voltage (with the polarity shown, current entering the marked terminal) is \[ V_R(t) = -\,i_R(t)\,R = -\,(2e^{-t})(1) = -2e^{-t} \,\text{V}. \] (The negative sign indicates the actual polarity is opposite the marked sign; the magnitude is \(2e^{-t}\).)
4. Maximum magnitude.
For \(t \ge 0\), the magnitude \(|V_R(t)| = 2e^{-t}\) is largest at \(t = 0\), giving \[ |V_R|_{\max} = 2e^{0} = 2 \,\text{V}. \]
Answer: \(\boxed{2 \,\text{V}}\)