Two Resistors, One Power Supply: Where Does the Voltage Go?
A student builds a circuit with a 9.0 V battery of internal resistance 0.50 ohm, connected in series with a 6.0 ohm resistor and a parallel combination of a 12 ohm and a 4.0 ohm resistor. A voltmeter across the parallel pair reads 3.1 V. The student expected the parallel pair to read closer to 4.0 V and suspects the battery is 'weak.' The ammeter in the main line reads 0.85 A. All meter leads are confirmed connected correctly.
1. Compute the theoretical equivalent resistance, main current, and voltage across the parallel pair. Compare with the meter readings and decide whether the battery is defective or the readings are consistent with the model.
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Parallel pair: R_p = (12 x 4)/(12 + 4) = 48/16 = 3.0 ohm. Total external R = 6.0 + 3.0 = 9.0 ohm. Total including internal: 9.5 ohm. Main current I = 9.0/9.5 = 0.947 A (vs. 0.85 A measured). Terminal/parallel voltage: V_p = I x R_p = 0.947 x 3.0 = 2.84 V.
Measured 0.85 A and 3.1 V are internally consistent with each other (3.1/3.0 = 1.03 A, not 0.85 - so actually they are not perfectly consistent). Recompute: if V_p = 3.1 V across 3.0 ohm, branch current total = 1.03 A, which exceeds the 0.85 A main-line reading. That inconsistency points to a meter/connection issue (e.g., voltmeter loading is negligible here, but a poor contact adds resistance) rather than a 'weak battery.' A weak battery would show lower voltage under load, not a current mismatch. The model predicts ~2.8-2.9 V; the student's 4.0 V expectation is simply wrong because they forgot the 6.0 ohm series drop and internal resistance.
2. Explain why the student's intuition of 'about 4 V' was wrong, and identify what measurement would most directly test for a bad battery.
Read the worked answer
The student likely computed the parallel voltage as if the 9 V dropped only across the parallel pair. But the 6.0 ohm series resistor and the 0.50 ohm internal resistance both drop voltage proportional to current. Voltage division: V_p = 9.0 x R_p/(R_total) = 9.0 x 3.0/9.5 = 2.84 V. The series resistor takes 6.0/9.5 of the supply, leaving only ~32% for the parallel pair.
Most direct battery test: measure the battery's terminal voltage both open-circuit and under the known load. A good 9 V cell should read near 9 V open-circuit and drop only by I x r_int (about 0.5 V at ~1 A). A large drop (e.g., to 7 V) or a high internal resistance inferred from (V_oc - V_load)/I would confirm degradation. Also verify the ammeter is in series and the voltmeter in parallel; a misranged or misconnected meter explains the 0.85 A vs 1.03 A inconsistency better than battery failure.