In a circuit, a 20 V ideal battery is connected in series with a 3 , resistor. The circuit then splits into two parallel branches that reconnect at the negative terminal of the battery. The first branch contains an unknown resistor R . The second branch contains a 4 , F capacitor in series with a 6 , resistor. If the energy stored in the capacitor in the steady state is 128 , J , what is the value of the resistance R

Step-by-step solution

In the steady state, the capacitor acts as an open circuit, so no current flows through the second branch. The energy stored in the capacitor is given by U = 1 2 CV_c^2 . Substituting the given values: 128 10⁻⁶ = 1 2 4 10⁻⁶ V_c^2 V_c^2 = 64 V_c = 8 V Since no current flows through the 6 , resistor, there is no voltage drop across it. Thus, the potential difference across the parallel branches is equal to the voltage across the capacitor, which is 8 V . The voltage drop across the 3 , series resistor is: V_ series =