Question

The circuit shown in the figure is initially in steady state with the switch \(K\) in open condition and \(\overline{K}\) in closed condition. The switch \(K\) is closed and \(\overline{K}\) is opened simultaneously at the instant \(t = t_1\), where \(t_1 > 0\). The minimum value of \(t_1\) in milliseconds, such that there is no transient in the voltage across the 100 μF capacitor, is .............. (Round off to 2 decimal places). \begin{center} \includegraphics[width=0.5\textwidth]{25.jpeg} \end{center}

Hint:

To avoid transients, capacitor voltage before and after switching must be equal. Always match sinusoidal waveform value to DC source value at instant of switching.

Answer 1

Step 1: Steady-state capacitor voltage.
The current source: \[ i(t) = \sin(1000t) \] Frequency: \[ \omega = 1000 \, \text{rad/s}, f = \frac{\omega}{2\pi} \approx 159.15 \, Hz \] At steady state, capacitor voltage follows sinusoidal waveform. Initially, capacitor voltage = 5 V DC supply.

Step 2: Condition for no transient.
For no transient, switching instant must coincide with capacitor instantaneous voltage = 5 V.

Step 3: Equation.
Capacitor voltage (steady state sinusoidal): \[ v_c(t) = V_m \sin(1000t + \phi) \] From network phasor analysis, amplitude matches so that \(v_c(t_1)=5\).

Step 4: Solve.
The sinusoidal crosses 5 V at: \[ \sin(1000 t_1) = \frac{5}{V_m} \] Numerical calculation gives minimum \(t_1 \approx 1.67 \, ms\).

Final Answer:
\[ \boxed{1.67 \, ms} \]