Question

An RC circuit is connected to two dc power supplies, as shown in the figure. With switch \( S \) open, the capacitor is fully charged. \( S \) is then closed at time \( t = 0 \). The voltage across the capacitor at \( t = 2.4 \, \text{ms} \) is ................ V (Round off to one decimal place). 

Hint:

In an RC circuit, the capacitor charges according to the formula \( V(t) = V_0 \left(1 - e^{-t/RC}\right) \), with the time constant \( \tau = RC \).

Answer 1

Correct Answer: 18.5

Step 1: Use the charging formula for the capacitor in an RC circuit.
The voltage across the capacitor as a function of time is given by the formula: \[ V(t) = V_0 \left(1 - e^{-t/RC}\right), \] where \( V_0 \) is the initial voltage across the capacitor, \( R \) is the resistance, \( C \) is the capacitance, and \( t \) is the time. Step 2: Calculate the time constant \( \tau = RC \).
The total resistance in the circuit is \( R = 4 \, \text{k}\Omega + 6 \, \text{k}\Omega = 10 \, \text{k}\Omega \), and the capacitance is \( C = 10 \, \mu\text{F} = 10 \times 10^{-6} \, \text{F} \). Therefore, the time constant is \[ \tau = (10 \times 10^3) \times (10 \times 10^{-6}) = 0.1 \, \text{seconds}. \] Step 3: Calculate the voltage across the capacitor at \( t = 2.4 \, \text{ms} = 2.4 \times 10^{-3} \, \text{s} \).
Using the formula for \( V(t) \) with \( V_0 = 20 \, \text{V} \), \[ V(t) = 20 \left(1 - e^{-(2.4 \times 10^{-3}) / 0.1}\right). \] Calculate the exponent: \[ \frac{2.4 \times 10^{-3}}{0.1} = 0.024 \Rightarrow e^{-0.024} \approx 0.976. \] Substitute into the formula: \[ V(t) = 20 \times (1 - 0.976) = 20 \times 0.024 = 0.48 \, \text{V}. \] Final Answer: The voltage across the capacitor at \( t = 2.4 \, \text{ms} \) is \( \boxed{0.5} \, \text{V}. \)