Parallel Lamps and the Internal Resistance of a Battery
A student is testing a 12 V battery whose internal resistance is 0.80 Ω. First, the student connects a lamp of resistance 6.0 Ω across the battery and observes a moderately bright glow. Then, without changing the battery, the student connects a second lamp of resistance 3.0 Ω in parallel with the first. Both lamps now look dimmer than the 6.0 Ω lamp did alone, and the student measures a lower terminal voltage across the parallel combination. The student had expected that adding a parallel branch would not affect the first lamp because the battery voltage is 12 V anyway. The teacher asks the student to model the battery as an emf in series with its internal resistance, to calculate the terminal voltage and the current in each lamp for both connections, and then to predict what would happen if the 3.0 Ω lamp were connected in series with the 6.0 Ω lamp instead. The connecting wires have negligible resistance.
1. For the single 6.0 Ω lamp, calculate the total current, terminal voltage, and power in the lamp. Then repeat for the two lamps in parallel.
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Single lamp: R_total = 6.0 + 0.80 = 6.8 Ω; I = 12/6.8 = 1.76 A. Terminal voltage V = 12 − Ir = 12 − (1.76)(0.80) = 10.6 V. Lamp power = V²/R = (10.6)²/6.0 = 18.7 W.
Parallel: R_p = (6.0×3.0)/(6.0+3.0) = 2.0 Ω; R_total = 2.8 Ω; I_total = 12/2.8 = 4.29 A. Terminal voltage V = 12 − (4.29)(0.80) = 8.57 V. Current in 6 Ω lamp = 8.57/6.0 = 1.43 A; current in 3 Ω lamp = 8.57/3.0 = 2.86 A. Power in 6 Ω lamp = (8.57)²/6.0 = 12.2 W, lower than 18.7 W.
2. Explain, using the internal resistance, why adding the parallel lamp dims the first lamp even though the battery emf is unchanged.
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Adding the parallel branch lowers the external resistance from 6.0 Ω to 2.0 Ω. The total current therefore rises from 1.76 A to 4.29 A. The internal drop Ir rises from 1.41 V to 3.43 V, so the terminal voltage falls from 10.6 V to 8.57 V. The first lamp still has 6.0 Ω, but it now receives a smaller voltage; since P = V²/R, its power and brightness decrease.
3. If the 3.0 Ω lamp is connected in series with the 6.0 Ω lamp, calculate the current and terminal voltage, and compare the brightness of the 6.0 Ω lamp with its brightness when alone. Which connection makes the 6.0 Ω lamp dimmer?
Read the worked answer
Series: R_total = 6.0 + 3.0 + 0.80 = 9.8 Ω; I = 12/9.8 = 1.22 A. Terminal voltage = 12 − (1.22)(0.80) = 11.0 V. Power in the 6 Ω lamp = I²R = (1.22)²(6.0) = 8.9 W. This is less than 12.2 W in parallel and much less than 18.7 W alone. The series connection makes the 6 Ω lamp dimmer than the parallel connection; in series both lamps carry the same current, and the larger 6 Ω resistance drops more voltage and dissipates more power than the 3 Ω lamp.