Why the 9 V Battery Cannot Run Two Loads at Once
A student is investigating why a 9 V battery cannot seem to run two devices at once. With a digital multimeter the battery's open-circuit terminal voltage is 8.95 V. Connected to a single 10.0 Ω resistor (1% tolerance), an ammeter in series reads 0.720 A and a voltmeter across the resistor reads 7.20 V. The student then connects a 15.0 Ω resistor in parallel with the 10.0 Ω one and predicts a total current of about 1.49 A, reasoning that 8.95 V divided by the 6.0 Ω equivalent resistance is 1.49 A. The ammeter instead reads about 1.05 A, the voltmeter across the parallel pair reads 6.3 V, and the battery becomes noticeably warm. The ammeter has a resistance of 0.10 Ω; the voltmeter is 10 MΩ. The student concludes that the battery must be failing and asks whether a fresh cell would fix the problem.
1. Use the single-resistor data to determine the battery's emf and internal resistance, and explain why the open-circuit voltage alone was misleading.
Read the worked answer
The 10 Ω measurement is self-consistent: 0.720 A × 10.0 Ω = 7.20 V across the resistor. The missing 8.95 − 7.20 = 1.75 V is dropped across the battery's internal resistance, so r = 1.75/0.720 = 2.43 Ω. The model ε = I(R + r) reproduces both readings exactly.
The open-circuit voltage reveals only ε, because with no load there is no current and therefore no internal drop; even that requires the voltmeter to draw negligible current, which it does here (10 MΩ against 2.43 Ω). Internal resistance is invisible until current flows, which is why dividing 8.95 V by 6.0 Ω is wrong: it silently assumes r = 0. The better procedure is to take two or more (I, V) pairs and fit V = ε − Ir. The intercept is the emf and the slope magnitude is the internal resistance, a method that also exposes poor contacts if the points fail to lie on a line.
2. Predict the total current with the parallel pair in place, then rank the contributions of internal resistance, ammeter resistance, resistor tolerance, and voltmeter loading to the discrepancy with 1.49 A.
Read the worked answer
Parallel equivalent: R = (10.0 × 15.0)/25.0 = 6.00 Ω. Adding the ammeter gives 6.10 Ω, and total circuit resistance is 6.10 + 2.43 = 8.53 Ω. Therefore I = 8.95/8.53 = 1.05 A, matching the meter, and the terminal voltage is 1.05 × 6.00 = 6.30 V.
Ranking: internal resistance dominates. With r = 0 the current would be 8.95/6.10 = 1.47 A, and the entire fall from 1.49 A to 1.05 A is essentially the 2.43 Ω. The ammeter's 0.10 Ω costs about 1%, resistor tolerance about 1%, and voltmeter loading about 0.001% (10 MΩ across 6 Ω). The physics is a load line: terminal voltage falls linearly with current, so each added parallel branch receives less voltage and less current than an ideal-battery calculation predicts. The student's arithmetic was right; the model behind it was not.
3. Assess the 'failing battery' hypothesis and explain the warming, including what limits the short-circuit current.
Read the worked answer
The hypothesis is plausible but not proven. High internal resistance is the signature of a depleted or cold cell (fresh alkaline 9 V cells are typically 1–2 Ω), so 2.43 Ω is consistent with a worn battery. But corroded terminals, dirty clips or long thin probe leads add resistance in the same place, so the test is to repeat the two-point fit with a second cell and clean contacts and see whether r changes.
The warming is expected, not a fault. At 1.05 A the internal dissipation is I²r = 1.05² × 2.43 = 2.7 W out of a total chemical power of 8.95 × 1.05 = 9.4 W, nearly 30%, which is why the cell feels hot. Short-circuit current would be ε/r = 8.95/2.43 = 3.7 A, at which the internal dissipation reaches about 33 W, the reason 9 V batteries are vented and often fused.