Two-Load Test to Find a Cell’s EMF and Internal Resistance
A student is testing an old dry cell. A moving-coil voltmeter marked 0–3 V and 1000 Ω/V is used on its 3 V range, so the meter resistance is 3000 Ω. When the cell is connected across a 10.0 Ω resistor, the voltmeter across the resistor reads 1.20 V. When the cell is connected across a 20.0 Ω resistor, the voltmeter reads 1.32 V. The student concludes that the cell’s emf is 1.32 V because that is the largest reading. The teacher says both readings are terminal voltages and that the cell has internal resistance. The 10 Ω and 20 Ω resistors are each rated 0.25 W. The student needs to decide whether this cell can run a torch bulb rated 0.25 A at 1.2 V. Ignore lead resistance. Treat the voltmeter as having finite resistance but much larger than either load resistor.
1. Estimate the cell’s emf E and internal resistance r from the two load readings. State the equations and the assumption about the voltmeter. Is the student’s conclusion E = 1.32 V correct?
Read the worked answer
For a load R, the terminal voltage is V = ER/(R + r), assuming the voltmeter draws negligible current compared with the load. Thus:
1.20 = E(10.0)/(10.0 + r) 1.32 = E(20.0)/(20.0 + r)
Divide the first equation by the second:
1.20/1.32 = [10.0(20.0 + r)]/[20.0(10.0 + r)] 0.9091 = (200 + 10r)/(200 + 20r) 200 + 10r = 181.8 + 18.18r 18.2 = 8.18r r ≈ 2.22 Ω
Then E = 1.20(10.0 + 2.22)/10.0 = 1.47 V. Check with the second load: E = 1.32(20.0 + 2.22)/20.0 = 1.47 V.
The student is wrong: 1.32 V is only the terminal voltage at the larger load. The terminal voltage approaches the emf as the current approaches zero, so the emf must be greater than both readings. The 3000 Ω voltmeter is in parallel with 10 Ω or 20 Ω; that lowers the effective load by less than 0.7%, so ignoring its loading is reasonable here.
2. Check whether either resistor exceeds its 0.25 W rating during these tests.
Read the worked answer
For the 10.0 Ω resistor, P = V²/R = (1.20)²/10.0 = 0.144 W. For the 20.0 Ω resistor, P = (1.32)²/20.0 = 0.0871 W. Both are below 0.25 W, so the resistors are safe during these measurements. The currents are also modest: about 0.120 A through the 10 Ω resistor and 0.066 A through the 20 Ω resistor.
3. Can the cell deliver 0.25 A at 1.2 V to the torch bulb? Support your answer with a calculation, and state what terminal voltage it would actually provide at 0.25 A.
Read the worked answer
Using E = 1.47 V and r = 2.22 Ω, the terminal voltage at I = 0.25 A would be:
V = E − Ir = 1.47 − (0.25)(2.22) = 1.47 − 0.555 = 0.915 V.
That is well below the 1.2 V the bulb needs at 0.25 A, so the cell cannot meet the rating. To supply 0.25 A at 1.2 V, the cell would need E = 1.2 + 0.25r. With r = 2.22 Ω this requires E ≈ 1.76 V, not 1.47 V. Equivalently, a 1.2 V, 0.25 A bulb has resistance 4.8 Ω; with this cell the current would be 1.47/(4.8 + 2.22) ≈ 0.209 A, and the terminal voltage would be about 1.00 V. The bulb would be dim.