Question

The value of parameters of the circuit shown in the figure are $R_{1 = 2\Omega, R_{2} = 2\Omega, R_{3} = 3\Omega, L = 10 \,\text{mH}, C = 100 \,\mu\text{F}$. For time $t < 0$, the circuit is at steady state with the switch $K$ in closed condition. If the switch is opened at $t = 0$, the value of the voltage across the inductor ($V_{L}$) at $t = 0^{+}$ in Volts is \underline{\hspace{3cm}} (Round off to 1 decimal place).} \begin{center} \includegraphics[width=0.5\textwidth]{12.jpeg} \end{center}

Hint:

For $t=0^{+}$ analysis in RLC circuits: - Inductor current is continuous ($i_{L}(0^{+})=i_{L}(0^{-})$). - Capacitor voltage is continuous ($v_{C}(0^{+})=v_{C}(0^{-})$). Use these to quickly compute transient initial voltages and currents.

Answer 1

Step 1: Steady state before $t=0$.
At steady DC, the inductor acts as a short circuit and the capacitor acts as open circuit. - Current source = 10 A. - With switch $K$ closed, $R_{3}$ is directly across the current source, bypassing the main branch. Thus, full $10$ A flows through $R_{3}$.

Step 2: Current through the inductor just before $t=0$.
Since $R_{3}$ is bypassing, no current flows through $R_{1}-L-R_{2}-C$ branch. Hence, inductor current $i_{L}(0^{-}) = 0 \,\text{A}$.

Step 3: Behavior just after $t=0$.
When switch opens at $t=0$, the 10 A current source must split between $R_{1}-L$ and $R_{2}-C$ branch. \[ i_{L}(0^{+}) = i_{L}(0^{-}) = 0 \,\text{A} \] (because inductor current cannot change instantaneously). Therefore, at $t=0^{+}$, the entire 10 A from the source flows into the $R_{2}-C$ branch.

Step 4: Voltage across the capacitor at $t=0^{+$.}
Capacitor voltage cannot change instantaneously. At $t=0^{-}$, capacitor had no current (open circuit) $\Rightarrow v_{C}(0^{-})=0$. So, $v_{C}(0^{+}) = 0$. Thus, the $R_{2}-C$ branch initially looks like just a resistor $R_{2}=2\Omega$ to the current source. So, voltage across $R_{2}$ is: \[ v_{R2} = i \cdot R_{2} = 10 \times 2 = 20 \,\text{V}. \]

Step 5: Apply KCL at node across $R_{1$ and $R_{2}$.}
Let node voltage (top of $R_{1}, R_{2}$) be $V_{x}$. - Through $R_{2}$ branch: $\frac{V_{x} - v_{C}}{R_{2}} = \frac{V_{x}-0}{2}$. This current must equal 10 A (since inductor branch current = 0). \[ \frac{V_{x}}{2} = 10 \Rightarrow V_{x} = 20 \,\text{V}. \]

Step 6: Voltage across inductor.
At $t=0^{+}$, inductor current is zero, but voltage across it is: \[ v_{L}(0^{+}) = V_{x} - (i_{L}(0^{+}) \cdot R_{1}) = 20 - 0 = 20 \,\text{V}. \] However, note $i_{L}(0^{+})=0$ $\Rightarrow$ drop across $R_{1}=0$, so indeed full node voltage appears across inductor. % Final Answer \[ \boxed{V_{L}(0^{+}) = 20.0 \,\text{V}} \]