Question

In the circuit shown below, the switch S is closed at \( t = 0 \). The magnitude of the steady state voltage, in volts, across the 6 Ω resistor is ________. (round off to two decimal places)

Hint:

In steady state, a fully charged capacitor behaves like an open circuit. Use Ohm's law to calculate the voltage across resistors in a simple series circuit.

Answer 1

Correct Answer: 4.95

To find the steady state voltage across the 6 Ω resistor, we need to analyze the circuit at \( t = \infty \), which corresponds to the steady-state condition in an R-C circuit. At steady state, the capacitor behaves like an open circuit because it is fully charged. Therefore, the current through the capacitor is zero, and the circuit reduces to a simple series circuit with the resistors.
The total resistance in the circuit is the sum of the resistances: \[ R_{\text{total}} = 6 \, \Omega + 3 \, \Omega + 10 \, \Omega + 2 \, \Omega = 21 \, \Omega. \] Now, apply Ohm's law to find the total current in the circuit, using the total applied voltage of 10 V: \[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{10}{21} \approx 0.476 \, \text{A}. \] Next, calculate the voltage drop across the 6 Ω resistor: \[ V_{6 \, \Omega} = I \times 6 = 0.476 \times 6 \approx 2.86 \, \text{V}. \] Thus, the magnitude of the steady-state voltage across the 6 Ω resistor is \( \boxed{2.86} \) V.