Battery Internal Resistance and Load Prediction
A battery of unknown emf and internal resistance is connected across a 10.0 Ω resistor, and an ideal voltmeter reads 8.40 V across the resistor. The resistor is replaced by a 4.70 Ω resistor, and the voltmeter reads 7.90 V. A student uses only the first measurement to conclude that the battery’s emf is 8.40 V and that its internal resistance is negligible. The voltmeter is later found to have a finite resistance of 1.0 kΩ, although the student’s readings were taken with it connected in parallel with each resistor. The resistors have 1% tolerance, and the connecting wires have negligible resistance. The student needs to predict the terminal voltage for a 2.0 Ω load and decide whether the battery can deliver 2.0 A continuously without dropping below 6.0 V.
1. Calculate the battery’s emf and internal resistance using the two measurements, treating the voltmeter as ideal. Is the student’s conclusion from the first measurement alone justified?
Read the worked answer
For the 10.0 Ω resistor, the current is
I₁ = 8.40/10.0 = 0.840 A.
For the 4.70 Ω resistor,
I₂ = 7.90/4.70 = 1.681 A.
Terminal voltage is V = ε − Ir, so the two measurements give
8.40 = ε − 0.840r 7.90 = ε − 1.681r.
Subtracting the second equation from the first:
0.50 = (1.681 − 0.840)r = 0.841r,
so
r = 0.595 Ω.
Then
ε = 8.40 + 0.840(0.595) = 8.90 V.
The first measurement alone gives one equation with two unknowns, ε and r. Setting r = 0 is an extra assumption, not a result. The second load shows that terminal voltage falls when current increases, so the internal resistance is not negligible. The student’s conclusion is not justified.
2. Predict the terminal voltage and current for a 2.0 Ω load. Can the battery deliver 2.0 A while keeping the terminal voltage at least 6.0 V?
Read the worked answer
For R = 2.0 Ω, the total resistance is R + r = 2.0 + 0.595 = 2.595 Ω. The current is
I = ε/(R + r) = 8.90/2.595 = 3.43 A.
The terminal voltage is
V = IR = (3.43)(2.0) = 6.86 V.
So with a 2.0 Ω load, the terminal voltage is above 6.0 V, but the current is 3.43 A, not 2.0 A.
At exactly 2.0 A, the terminal voltage would be
V = ε − Ir = 8.90 − (2.0)(0.595) = 7.71 V,
which is also above 6.0 V. The voltage criterion is met at 2.0 A. However, the internal power dissipation at 2.0 A is
P = I²r = (2.0)²(0.595) = 2.38 W,
so continuous operation may heat the battery. The given data address voltage, not thermal limits.
3. The voltmeter has a finite resistance of 1.0 kΩ. Does including it materially change the emf and internal-resistance result? Estimate the effect.
Read the worked answer
With the voltmeter in parallel, the effective load is slightly smaller than the resistor alone.
For the 10.0 Ω case:
R_eq = (10.0 × 1000)/(10.0 + 1000) = 9.90 Ω.
For the 4.70 Ω case:
R_eq = (4.70 × 1000)/(4.70 + 1000) = 4.68 Ω.
The battery currents are therefore
I₁ = 8.40/9.90 = 0.848 A I₂ = 7.90/4.68 = 1.688 A.
Solving the same two equations gives
r = (8.40 − 7.90)/(1.688 − 0.848) = 0.50/0.840 = 0.595 Ω,
and
ε = 8.40 + 0.848(0.595) = 8.90 V.
The correction is below 1% and does not change the conclusion. The main error in the student’s reasoning was neglecting internal resistance, not voltmeter loading. A 1.0 kΩ voltmeter is large compared with the 10 Ω and 4.7 Ω loads, so treating it as ideal is reasonable here.