In a circuit, a 20 V battery is connected in series with a 2 resistor. The circuit then splits into two parallel branches: the first branch contains a 3 resistor, and the second branch contains a 4 resistor in series with a 5 F capacitor. What is the charge stored on the capacitor at steady state?

Step-by-step solution

At steady state, the capacitor is fully charged and acts as an open circuit, so no current flows through the branch containing the 4 resistor and the 5 F capacitor. The steady-state current flows only through the 2 and 3 resistors, which are effectively in series. The current in the circuit is: I = V R_ eq = 20 2 + 3 = 4 A The potential difference across the parallel section is the voltage drop across the 3 resistor: V_ p = I 3 = 4 3 = 12 V Since there is no current in the 4 resistor, there is no voltage drop acros