Nonideal Battery With Parallel Resistors
A student connects a battery of emf 12.0 V and internal resistance 0.50 Ω to two resistors, 6.0 Ω and 3.0 Ω, wired in parallel. An ideal ammeter in series with the battery reads 4.80 A, and an ideal voltmeter across the battery terminals reads 9.60 V. The student then removes the 3.0 Ω resistor, leaving only the 6.0 Ω resistor in the circuit, and repeats both measurements. A lab partner argues that because the battery is labeled 12.0 V, each parallel resistor must have 12.0 V across it, and that the ammeter reading should equal 12.0 V divided by the equivalent resistance. The student must reconcile the labels with the measurements, calculate the new readings after the 3.0 Ω resistor is removed, and determine how much power is dissipated inside the battery before and after the change.
1. Use the measurements to find the equivalent external resistance and the internal resistance, and explain why the terminal voltage is 9.60 V rather than 12.0 V.
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The parallel external resistance is (6.0 × 3.0)/(6.0 + 3.0) = 18/9 = 2.0 Ω. The ammeter reads 4.80 A, so the total circuit resistance is R_total = emf/I = 12.0/4.80 = 2.5 Ω. Therefore internal resistance r = 2.5 − 2.0 = 0.50 Ω, matching the label. Terminal voltage V_T = emf − Ir = 12.0 − (4.80)(0.50) = 12.0 − 2.40 = 9.60 V. The 12.0 V is the emf, not the terminal voltage; the internal resistance consumes 2.40 V when current flows. The parallel resistors each see 9.60 V, so branch currents are 9.60/6.0 = 1.60 A and 9.60/3.0 = 3.20 A, summing to 4.80 A.
2. After the 3.0 Ω resistor is removed, calculate the new ammeter and voltmeter readings and the power dissipated inside the battery. Compare with the original internal power.
Read the worked answer
With only 6.0 Ω external, total resistance = 6.0 + 0.50 = 6.5 Ω. Current = 12.0/6.5 = 1.846 A, so the ammeter reads about 1.85 A. Terminal voltage = 12.0 − (1.846)(0.50) = 12.0 − 0.923 = 11.08 V, so the voltmeter reads about 11.1 V. Internal power = I^2 r = (1.846)^2 × 0.50 = 1.70 W (or 0.923 × 1.846 = 1.70 W). Originally internal power = (4.80)^2 × 0.50 = 11.52 W. Removing the parallel branch lowers current and cuts internal dissipation sharply, so terminal voltage rises.
3. Evaluate the lab partner's claim that each parallel resistor must have 12.0 V across it and that the ammeter current should be 12.0 V divided by the equivalent resistance.
Read the worked answer
The claim treats the battery as ideal. A real battery has internal resistance in series with the emf, so the terminal voltage is emf − Ir and is always less than 12.0 V when current flows. The parallel resistors are connected across the terminals, so they share the terminal voltage, 9.60 V, not the emf. Using 12.0 V / 2.0 Ω = 6.0 A would be correct only if r were zero. The measured 4.80 A is consistent with the total resistance 2.5 Ω: 12.0/2.5 = 4.80 A. The partner's error is omitting the internal resistance; the battery label gives emf, not terminal voltage under load.