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}

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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}

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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: $$ \boxed{1.67 \, ms} $$

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