The electrical network shown has an independent voltage source (10 V) and a current source (1 u(t) mA). The voltage across the capacitor at time instants (in seconds) \(t=0^{+}\), \(t=0.50\), and \(t=\infty\), respectively, is:
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Find the pre-switching capacitor voltage first (it cannot jump), then the new final value with the current source active, and use \(\tau=R_{th}C\) with \(R_{th}\) as the 25k-100k parallel combination.
Step 1: Split the problem into before and after switching.
Before $t=0$, the 1 mA source is absent, so only the 10 V source drives the circuit. With the capacitor fully charged (open circuit), the loop current is $10/(25\text{k}+100\text{k})=0.08$ mA, and the capacitor voltage equals the drop across the $100$ k$\Omega$ resistor:
\[ v_C(0^-)=0.08\text{ mA}\times100\text{k}\Omega = 8\text{ V} \]
Since capacitor voltage is continuous, $v_C(0^+)=8.00$ V.
Step 2: Use superposition for the final value.
As $t\to\infty$, the capacitor is again open. Find $v_C(\infty)$ by superposition of the two sources:
Due to the 10 V source alone (current source open): $v_1=10\times\dfrac{100}{125}=8$ V.
Due to the 1 mA source alone (voltage source shorted, so $25$k and $100$k appear in parallel to the current source): $R_{eq}=\dfrac{25\times100}{125}=20$ k$\Omega$, so $v_2=1\text{ mA}\times20\text{k}\Omega=20$ V.
Adding, $v_C(\infty)=v_1+v_2=8+20=28$ V.
Step 3: Get the time constant from the deactivated network.
Killing both sources, the capacitor sees $25$k$\parallel$100k$=20$k$\Omega$, so $\tau=20\text{k}\Omega\times10\,\mu\text{F}=0.2$ s.
Step 4: Write the exponential response directly from initial and final values.
Every first-order response takes the form $v_C(t)=v_C(\infty)+[v_C(0^+)-v_C(\infty)]e^{-t/\tau}$, so
\[ v_C(t)=28+(8-28)e^{-t/0.2}=28-20e^{-5t} \]
Step 5: Substitute $t=0.5$ s.
\[ v_C(0.5)=28-20e^{-2.5}=28-20(0.0821)=28-1.64=26.36\text{ V} \]
Step 6: State the three values in order.
\[ \boxed{v_C(0^+)=8.00\text{ V},\ v_C(0.5\text{ s})=26.36\text{ V},\ v_C(\infty)=28.00\text{ V}} \]
