Question

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:

• A. 8.00 V, 28.00 V, 26.36 V
• B. 8.00 V, 26.36 V, 28.00 V
• C. 10.00 V, 26.36 V, 28.00 V
• D. 10.00 V, 28.00 V, 26.36 V

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Identify the circuit and label the node.
The \(10\) V source connects through a \(25\) k\(\Omega\) resistor to a node that has the \(10\ \mu\)F capacitor, the \(1\ u(t)\) mA current source (switched in at \(t=0\)), and the \(100\) k\(\Omega\) resistor, all in parallel to ground. Let \(v_C\) be the voltage at this node, across the capacitor.

Step 2: Find \(v_C(0^{-})\).
Before \(t=0\), the current source is not connected (\(u(t)=0\) for \(t<0\)), and the circuit has been in steady state, so the capacitor carries no current and behaves as an open circuit. The only path is \(10\) V through \(25\) k\(\Omega\) in series with \(100\) k\(\Omega\) to ground.
\[ v_C(0^{-}) = 10\times\frac{100\text{k}}{25\text{k}+100\text{k}} = 10\times\frac{100}{125} = 8\text{ V} \]

Step 3: Apply continuity of capacitor voltage.
The voltage across a capacitor cannot jump instantly, so
\[ v_C(0^{+}) = v_C(0^{-}) = 8.00\text{ V} \]
This already rules out options (C) and (D), which start from \(10.00\) V.

Step 4: Find the final value \(v_C(\infty)\).
For \(t>0\), the \(1\) mA source is active and injects current into the node. Writing KCL at the node with all quantities in mA and k\(\Omega\) (so that products of mA and k\(\Omega\) directly give volts), and calling the node voltage \(V\):
\[ \frac{10-V}{25}+1 = \frac{V}{100} \]
Multiply through by \(100\):
\[ 4(10-V)+100 = V \]
\[ 40-4V+100 = V \implies 140 = 5V \implies V = 28\text{ V} \]
So \(v_C(\infty) = 28.00\) V.

Step 5: Find the Thevenin resistance seen by the capacitor.
To get the time constant, turn off the independent sources: the \(10\) V source becomes a short, and the \(1\) mA current source becomes an open circuit (it does not affect the resistance calculation since it is an ideal current source). Looking into the node from the capacitor's terminals, the resistance is \(25\) k\(\Omega\) in parallel with \(100\) k\(\Omega\):
\[ R_{th} = \frac{25\times100}{25+100}\text{k}\Omega = \frac{2500}{125}\text{k}\Omega = 20\text{ k}\Omega \]

Step 6: Find the time constant and write \(v_C(t)\).
\[ \tau = R_{th}C = 20\text{k}\Omega\times10\ \mu\text{F} = 0.2\text{ s} \]
\[ v_C(t) = v_C(\infty)+\big[v_C(0^{+})-v_C(\infty)\big]e^{-t/\tau} = 28-20e^{-t/0.2} \]

Step 7: Evaluate at \(t=0.50\) s.
\[ \frac{t}{\tau} = \frac{0.5}{0.2} = 2.5,\qquad e^{-2.5}\approx0.0821 \]
\[ v_C(0.5) = 28-20(0.0821) = 28-1.64 \approx 26.36\text{ V} \]

Step 8: Collect the three values in the order asked.
\[ v_C(0^{+})=8.00\text{ V},\quad v_C(0.5)=26.36\text{ V},\quad v_C(\infty)=28.00\text{ V} \]
This matches option (B), and rules out option (A), which places the values \(28\) and \(26.36\) in the wrong order.

Final Answer:
\[ \boxed{8.00\text{ V},\ 26.36\text{ V},\ 28.00\text{ V}} \]