Car headlights dimming: internal resistance and a non-ohmic filament
A student is asked why a car’s headlights dim when the starter motor is switched on. The battery has emf 12.0 V and internal resistance 0.050 Ω. With the engine off, one headlamp (rated 12.0 V, 55 W) is connected and draws its normal current. When the starter is engaged, it draws 150 A from the battery for a short time while the headlamp remains connected. The student models the headlamp as a fixed resistor and ignores the resistance of the connecting wires. The battery’s emf is assumed constant during the short starting pulse. The student measures the terminal voltage during starting and finds it much lower than 12.0 V, then claims the headlamp should go out completely. Use the ratings to estimate the lamp resistance, and treat the lamp as ohmic for the first calculation. Discuss how the real filament changes the estimate.
1. Calculate the terminal voltage during starting and the power dissipated in the headlamp, assuming the lamp resistance is constant.
Read the worked answer
First find the lamp resistance from its rating:
P = V²/R → R = 12.0² / 55 = 2.62 Ω
Take the 150 A as the total current drawn from the battery during the starting pulse. The terminal voltage is
V_terminal = ε − I r = 12.0 − 150 × 0.050 = 4.5 V
The lamp is now connected across 4.5 V, so its current is
I_lamp = 4.5 / 2.62 = 1.72 A
and its power is
P_lamp = V²/R = 4.5² / 2.62 = 7.7 W
This is about 14% of the normal 55 W, so the lamp becomes very dim but does not go out completely. The student’s claim that it should go out is not supported by the model.
2. Explain in circuit terms why the terminal voltage falls when the starter is engaged, and why the headlamp and starter are not independent.
Read the worked answer
The battery has an internal resistance r in series with its emf. Any current I from the battery causes an internal voltage drop Ir, so the terminal voltage is ε − Ir. The starter draws a very large current, about 150 A, which makes Ir = 150 × 0.050 = 7.5 V. The terminal voltage therefore falls from 12.0 V to about 4.5 V.
The headlamp is connected to the same terminals, so it sees this reduced terminal voltage. That is why the two are not independent: the starter current changes the terminal voltage, and the lamp current changes the total battery current. Strictly, if 150 A is only the starter current, the total battery current is slightly larger because the lamp also draws current; this lowers the terminal voltage a little further. The main effect, however, is the large starter current through the internal resistance.
3. Evaluate the constant-resistance assumption. In reality, a cooler filament has lower resistance. Does that make the actual lamp power larger or smaller than 7.7 W? Justify your answer.
Read the worked answer
The constant-resistance value 7.7 W is only an estimate. A tungsten filament is strongly non-ohmic: its resistance falls when it is cooler. During starting, the lamp is far below normal operating temperature, so its actual resistance is lower than 2.62 Ω.
For a given terminal voltage, P = V²/R. If R is lower than 2.62 Ω, the power is larger than 7.7 W, and the lamp current is larger too. The actual lamp may therefore be a little brighter than the simple estimate, but it will still be far below 55 W because the terminal voltage has collapsed to about 4.5 V.
A second-order effect is that the extra lamp current increases the total battery current, lowering the terminal voltage slightly and reducing the lamp power. This does not overturn the main conclusion: the headlamp dims severely but does not extinguish.