Battery Internal Resistance Under Parallel Load
A student is testing a battery pack labeled 9.0 V. A high-impedance voltmeter reads 9.00 V with no load. When a 12.0 Ω resistor is connected, a digital ammeter in series reads 0.600 A. When an unknown resistor is connected in series with the 12.0 Ω resistor, the current falls to 0.380 A. The student assumes the battery’s internal resistance is constant and predicts the current if the unknown resistor is connected in parallel with the 12.0 Ω resistor. The measured current is 0.96 A, lower than predicted. The ammeter has a 0.50 Ω shunt on its 2 A range; the resistors are rated 5% tolerance; the battery becomes noticeably warm during the parallel test. The student wonders whether the ammeter or the battery is responsible and wants to know which result to report.
1. Determine the battery’s internal resistance and the unknown resistance, accounting for the ammeter. State how the internal resistance would change if the ammeter were ignored.
Read the worked answer
With no load, the high-impedance voltmeter gives (\varepsilon=9.00\ \mathrm{V}). In the first trial the total resistance is
(R_{\mathrm{tot}}=\frac{9.00}{0.600}=15.0\ \Omega).
This includes the 12.0 Ω resistor and the 0.50 Ω ammeter, so
(r=15.0-12.0-0.50=2.5\ \Omega).
In the second trial,
(R_{\mathrm{tot}}=\frac{9.00}{0.380}=23.7\ \Omega),
so the unknown resistor is
(R_x=23.7-12.0-0.50-2.5=8.7\ \Omega).
If the ammeter’s 0.50 Ω were ignored, the first trial would give (r=3.0\ \Omega), a 0.5 Ω error. The unknown resistance happens to come out nearly the same because the ammeter affects both trials similarly.
2. Show the constant-internal-resistance prediction for the parallel connection, compare it with 0.96 A, and decide whether the ammeter or battery heating is the more likely explanation. What measurement would confirm it?
Read the worked answer
The parallel equivalent of 12.0 Ω and 8.7 Ω is
(R_{\mathrm{eq}}=\frac{(12.0)(8.7)}{12.0+8.7}\approx5.04\ \Omega).
Including the ammeter and (r=2.5\ \Omega), the predicted current is
(I=\frac{9.00}{5.04+0.50+2.5}\approx1.12\ \mathrm{A}).
Even using the larger (r=3.0\ \Omega) gives about 1.05 A. The measured 0.96 A is lower than either prediction. The ammeter is already included in the prediction and cannot explain the remaining drop. The battery warming can: to obtain 0.96 A, the total resistance must be (9.00/0.96=9.38\ \Omega), requiring (r\approx9.38-5.04-0.50=3.84\ \Omega), higher than the small-current value. A confirming measurement is the terminal voltage during the parallel test: (r=(9.00-V)/I). The student should report that (r) is not constant and give the current range over which it was measured.