Why the Headlamps Dim When the Starter Cranks
A motorist complains that the headlamps of an old car become very dim whenever the starter motor is cranked, although they are bright when the engine is off. The battery has an emf of 12 V and an internal resistance of 0.05 ohm. Each headlamp is rated 12 V, 48 W and the two are wired in parallel; treat the lamp filaments as ohmic. The starter motor has a resistance of 0.10 ohm and, when engaged, is in parallel with the lamps across the same battery. The cabling is short and its resistance is negligible. A mechanic tests the battery with a voltmeter with nothing switched on, reads 12 V, and declares the battery 'perfectly good'. He proposes either fitting a second identical 12 V battery in series to boost the supply, or replacing the lamp cable with a thicker one. Decide, from terminal-voltage and power calculations, whether either proposal cures the dimming.
1. Find the resistance of one lamp and calculate the terminal voltage and the power dissipated in each lamp when only the lamps are switched on.
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Rated values give R = V^2/P = (12)^2/48 = 3 ohm per lamp.
Two 3-ohm lamps in parallel: R_p = 1.5 ohm. With the battery's internal resistance in series the total circuit resistance is 1.5 + 0.05 = 1.55 ohm.
Current drawn from the battery: I = 12/1.55 = 7.74 A. Terminal voltage: V_t = E - I r = 12 - 7.74 x 0.05 = 12 - 0.39 = 11.6 V. Power per lamp: P = V_t^2/R = (11.6)^2/3 = 134.6/3 = 44.9 W.
The lamps are close to their rated 48 W, so they look bright. Note how little the voltmeter test proves: with no current the terminal voltage equals the emf whether the battery is healthy or not, because there is no I r drop.
2. Now switch on the starter as well. Calculate the new terminal voltage and the power in each lamp, and explain the mechanism of the dimming.
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The starter (0.10 ohm) is in parallel with the two lamps (1.5 ohm):
R_p = (1.5 x 0.10)/(1.5 + 0.10) = 0.15/1.60 = 0.0938 ohm. Total = 0.0938 + 0.05 = 0.1438 ohm.
Battery current: I = 12/0.1438 = 83.5 A. Terminal voltage: V_t = 12 - 83.5 x 0.05 = 12 - 4.17 = 7.8 V. Current per lamp = 7.8/3 = 2.6 A; power per lamp = (7.8)^2/3 = 20.4 W.
So each lamp drops from about 45 W to about 20 W, less than half its normal output, which is exactly the dimming the motorist reports. The mechanism is the internal resistance, not the lamps: the starter draws a very large current from the same source, and that current must pass through r, producing a large I r drop inside the battery. The terminal voltage, which is what the lamps actually see, falls from 11.6 V to 7.8 V. The lamps are innocent bystanders sharing a supply whose terminal voltage has collapsed.
3. Evaluate both of the mechanic's proposals. Would the series battery or the thicker cable solve the problem? What is the correct diagnosis and remedy?
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Series battery. With two identical batteries the emf is 24 V and the internal resistance is 0.10 ohm. During cranking: total resistance 0.0938 + 0.10 = 0.1938 ohm, I = 24/0.1938 = 124 A, V_t = 24 - 124 x 0.10 = 11.6 V - the lamps would indeed be bright while cranking. But the fix is catastrophic: with the starter off, the lamps see V_t = 24 x 1.5/1.55 = 23.2 V, i.e. roughly four times their rated power (about 180 W each). The filaments would fail immediately. A 12 V lamp system must never be fed from 24 V.
Thicker cable. This attacks a resistance that was explicitly negligible. Even reducing the cable resistance to zero changes nothing, because the voltage lost appears across the battery's internal resistance. Thicker cable is the right answer only when long, thin runs cause the drop; here it is the wrong target.
Correct diagnosis. The battery's internal resistance (0.05 ohm) is too high - it is aged, sulphated or cold. Fitting a healthy battery of the same 12 V rating but r about 0.01 ohm gives, on cranking: total 0.1038 ohm, I = 116 A, V_t = 12 - 1.16 = 10.8 V, and each lamp delivers (10.8)^2/3 = 39 W. Some dip remains - that is unavoidable physics - but the lamps stay usable. The voltmeter reading of 12 V was a meaningless test, since an open-circuit measurement never reveals r.