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 = ∞, respectively, is:

The correct option is 8.00 V, 26.36 V, 28.00 V.

Analysis of the Circuit Diagram:
Based on the provided circuit diagram:

• An independent DC voltage source of $10\text{V}$ is connected in series with a $25\text{k}\mathrm{\Omega }$ resistor.
• A capacitor with capacitance $C=10\mathrm{\mu }\text{F}$ is connected in parallel with a current source and a $100\text{k}\mathrm{\Omega }$ resistor.
• The independent current source is defined as $1u\left(t\right)\text{mA}$, meaning it is $0\text{mA}$ for $t<0$ and steps up to $1\text{mA}$ pointing upwards at $t=0\text{s}$.

Step 1: Finding the initial capacitor voltage at $t=0^{+}$
For $t<0$, the current source is inactive ($0\text{mA}$, acting as an open circuit). Assuming the circuit has reached steady state, the capacitor acts as an open circuit.
The voltage across the capacitor $v_C\left(0^{-}\right)$ is determined using the voltage divider formula between the $10\text{V}$ source, the $25\text{k}\mathrm{\Omega }$ resistor, and the $100\text{k}\mathrm{\Omega }$ resistor:
$$v_C\left(0^{-}\right)=10\times \frac{100\text{k}\mathrm{\Omega }}{25\text{k}\mathrm{\Omega }+100\text{k}\mathrm{\Omega }}=10\times 0.8=8.00\text{V}$$
Since the voltage across a capacitor cannot change instantaneously:
$$v_C\left(0^{+}\right)=v_C\left(0^{-}\right)=8.00\text{V}$$

Step 2: Finding the steady-state voltage at $t=\infty$
As $t\to \infty$, the capacitor once again acts as an open circuit in DC steady state, while the current source supplies a constant $1\text{mA}$.
Applying Kirchhoff's Current Law (KCL) at the node above the capacitor:
$$\frac{v_C(\infty )-10}{25\text{k}\mathrm{\Omega }}+\frac{v_C(\infty )}{100\text{k}\mathrm{\Omega }}=1\text{mA}$$
Multiply the entire equation by $100\text{k}\mathrm{\Omega }$ to eliminate the denominators:
$$4\left(v_C\right(\infty )-10)+v_C(\infty )=100$$
$$5v_C(\infty )-40=100$$
$$5v_C(\infty )=140\Rightarrow v_C(\infty )=28.00\text{V}$$

Step 3: Finding the time constant ($\tau$)
The equivalent resistance $R_{\text{eq}}$ connected across the capacitor is found by deactivating the independent sources (short-circuiting the $10\text{V}$ source and open-circuiting the current source):
$$R_{\text{eq}}=25\text{k}\mathrm{\Omega }\parallel 100\text{k}\mathrm{\Omega }=\frac{25\times 100}{25+100}=20\text{k}\mathrm{\Omega }$$
The time constant $\tau$ of the RC network is:
$$\tau =R_{\text{eq}}C=(20\times {10}^3\mathrm{\Omega })\times (10\times {10}^{-6}\text{F})=0.2\text{s}$$

Step 4: Finding the voltage at $t=0.50\text{s}$
The transient response of the capacitor voltage for $t\ge 0$ is given by:
$$v_C\left(t\right)=v_C(\infty )+\left[v_C\right(0^{+})-v_C(\infty \left)\right]e^{-t/\tau }$$
Substituting the derived values:
$$v_C\left(t\right)=28+(8-28)e^{-t/0.2}=28-20e^{-5t}\text{V}$$
Substitute $t=0.50\text{s}$:
$$v_C\left(0.50\right)=28-20e^{-5\times 0.50}=28-20e^{-2.5}$$
Using the approximation $e^{-2.5}\approx 0.082085$:
$$v_C\left(0.50\right)\approx 28-20\times 0.082085=28-1.6417=26.36\text{V}$$

Thus, the values of the capacitor voltage at $t=0^{+}$, $t=0.50$, and $t=\infty$ are $8.00\text{V}$, $26.36\text{V}$, and $28.00\text{V}$, respectively.