Steady-State Circuit with Internal Resistance and a Capacitor
A battery of emf 12.0 V and internal resistance 0.50 Ω is connected in series with a 4.0 Ω resistor R1. The circuit then splits into two parallel branches: R2 = 6.0 Ω and R3 = 3.0 Ω, which rejoin and return to the battery. A 10 µF capacitor is connected across R2. The circuit has been closed for a long time. A student measures the terminal voltage of the battery and the voltage across R2, and finds they are not equal; the student concludes that Kirchhoff's loop rule has failed. You are asked to resolve the measurement and find the steady-state quantities.
1. Find the current through the battery and the terminal potential difference at steady state.
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The three resistors form R1 in series with the parallel pair R2||R3. The parallel equivalent is (6.0×3.0)/(6.0+3.0) = 2.0 Ω. The total resistance in the loop is R_total = 4.0 + 2.0 + 0.50 = 6.5 Ω. The battery current is I = emf/R_total = 12.0/6.5 = 1.85 A. The terminal potential difference is V_terminal = emf − Ir = 12.0 − 1.85×0.50 = 12.0 − 0.92 = 11.1 V. This is also the voltage across the external circuit: I(4.0+2.0) = 1.85×6.0 = 11.1 V. The voltage across the parallel pair alone is I×2.0 = 3.69 V, and the voltage across R1 is I×4.0 = 7.38 V; their sum is 11.1 V.
2. Find the charge stored on the capacitor and the power dissipated in the internal resistance.
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At steady state the capacitor is fully charged and no current flows through its branch, so it does not alter the resistor currents. It is connected across R2, so its voltage is the parallel-pair voltage, 3.69 V. The charge is Q = CV = 10×10^-6 × 3.69 = 3.69×10^-5 C = 36.9 µC. The power dissipated in the internal resistance is P = I²r = (1.85)^2 × 0.50 = 3.42×0.50 = 1.71 W. The power delivered chemically by the battery is emf×I = 12.0×1.85 = 22.2 W; the external resistors dissipate I²R_ext = 3.42×6.0 = 20.5 W, and the remaining 1.7 W appears in the internal resistance, as expected.
3. The student says the capacitor should carry current at steady state because it is part of the circuit. Evaluate this claim and explain why the measured voltages do not violate Kirchhoff's rules.
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Kirchhoff's loop rule has not failed. The terminal voltage is not the same as the voltage across R2; it is the voltage across the whole external circuit, R1 plus the parallel pair. A loop through the battery and the external circuit gives emf − Ir − I(R1 + R_parallel) = 0, or 12.0 − 0.92 − 11.1 = 0. A loop containing the capacitor gives V_C − V_R2 = 0 at steady state, which is why the capacitor voltage equals 3.69 V. The student's error is to compare two different points in the circuit and call the difference a violation. The capacitor branch carries no steady current once charged, but it still has a potential difference; that is consistent with the rules.