In this RC circuit, after the switch $A$ is closed, the capacitor $C$ (100 μF) will start charging through the resistor $R$ (10 kΩ) from the 28V source.

The voltage across a charging capacitor $V_C(t)$ as a function of time $t$ is given by the formula:

$V_C(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right)$

where:

• $V_0$ is the supply voltage (28V),
• $R$ is the resistance (10 kΩ = 10,000 Ω),
• $C$ is the capacitance (100 μF = $100 \times 10^{-6} \, \text{F}$).

The time constant $\tau = RC$ defines how quickly the capacitor charges. For this circuit:

$\tau = R \cdot C = 10,000 \cdot 100 \times 10^{-6} = 1 \, \text{s}$

Since the switch has been closed for 5 minutes (300 seconds), which is much longer than the time constant, the capacitor will be fully charged. When fully charged, the voltage across the capacitor $V_C$ reaches the supply voltage $V_0$.

Therefore, the instantaneous voltage across the capacitor $C$ just before the switch opens is:

$V_C = 28 \, \text{V}$

Answer:

The correct answer is 28V.

Would you like further details or have any additional questions?

Here are some related questions:

1. How does the time constant $RC$ affect the charging speed of the capacitor?
2. What would happen to the voltage across the capacitor if the resistance were increased?
3. How does the voltage across the capacitor behave if the switch remains open?
4. What would be the current through the resistor at the instant the switch is closed?
5. How long would it take for the capacitor to reach 99% of the supply voltage?

Tip: The time constant $\tau = RC$ is a critical factor in determining the charging and discharging rates of capacitors in RC circuits.