Terminal Voltage and the Parallel-Cell Misconception
A battery of EMF 12.0 V and internal resistance 0.50 Ω feeds a network consisting of a 6.0 Ω resistor in parallel with a 3.0 Ω resistor, and this parallel combination in series with a 2.0 Ω resistor. An ideal voltmeter is connected across the battery terminals. The two parallel resistors combine to 2.0 Ω, so the external resistance is 4.0 Ω. A student calculates the terminal p.d. as 12.0 × 4.0/4.5 = 10.7 V and then claims that connecting a second, identical cell in parallel with the first will double the current through the 2.0 Ω resistor, since 'two cells give twice the current'. The laboratory has a duplicate cell of the same EMF and internal resistance available, and each resistor is rated at 5 W.
1. Calculate the current through each of the three resistors and the reading of the voltmeter.
Read the worked answer
Equivalent resistance. Parallel pair: R_p = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω. External resistance R_ext = 2.0 + 2.0 = 4.0 Ω. Total resistance in circuit = 4.0 + 0.50 = 4.5 Ω.
Total current: I = EMF / (R_ext + r) = 12.0/4.5 = 2.67 A.
Terminal p.d.: V = EMF − I r = 12.0 − (2.67 × 0.50) = 10.7 V (equivalently I R_ext = 2.67 × 4.0 = 10.7 V), so the student's value is correct.
Branch currents. The p.d. across the parallel pair is I R_p = 2.67 × 2.0 = 5.33 V. I₆ = 5.33/6.0 = 0.89 A; I₃ = 5.33/3.0 = 1.78 A. Sum = 2.67 A ✓ (the smaller resistor takes the larger current, as it must). The 2.0 Ω series resistor carries the full 2.67 A.
2. Find the power dissipated inside the battery and the efficiency of the transfer to the external circuit.
Read the worked answer
Power lost in the internal resistance: P_r = I² r = (2.67)² × 0.50 = 7.11 × 0.50 = 3.56 W (about 3.6 W).
Total power generated by the EMF: P_total = EMF × I = 12.0 × 2.67 = 32.0 W. Useful power in the external circuit: P_ext = I² R_ext = 7.11 × 4.0 = 28.5 W, or V I = 10.7 × 2.67 = 28.5 W. Check: 28.5 + 3.56 = 32.0 W ✓.
Efficiency = P_ext/P_total = 28.5/32.0 = 0.89, i.e. about 89% (also V/EMF = 10.7/12.0). The 11% loss is entirely Ohmic heating inside the cell; it grows as I², so drawing more current from the same cell reduces efficiency even though the terminal voltage falls. Each resistor's dissipation (e.g. the 2.0 Ω one takes 7.11 × 2.0 = 14.2 W) shows that the 5 W rating would in fact be exceeded - a practical warning the numbers reveal.
3. Evaluate the claim that adding a second identical cell in parallel doubles the current through the 2.0 Ω resistor.
Read the worked answer
What parallel connection does. With two identical cells in parallel, the EMF stays 12.0 V (parallel sources do not add EMF), but the internal resistance halves: r = 0.50/2 = 0.25 Ω.
New current: I = 12.0/(4.0 + 0.25) = 12.0/4.25 = 2.82 A, compared with 2.67 A before. Ratio = 2.82/2.67 = 1.06, an increase of only about 6% - not a factor of two.
Why the claim fails. The external resistance (4.0 Ω) is much larger than the internal resistance, so halving r removes only half of a small term. The terminal p.d. rises from 10.7 V to 12.0 − (2.82 × 0.25) = 11.3 V, and the wasted power falls from 3.56 W to 2.82² × 0.25 = 1.99 W. Parallel cells are used to share current and extend life, not to raise EMF.
To double the current, the EMF would have to be doubled, for example by connecting the cells in series (24 V with r = 1.0 Ω gives 24/5.0 = 4.8 A), and even that is not exactly double because the internal drop also doubles.