The engine starter and a headlight of a car are connected in parallel to the 12.0-V car battery. In this situation, the headlight operates at 38 W and the engine starter operates at 2.40 kW. If the headlight and starter were then rewired to be in series with each other, what total power would they consume when connected to the 12.0-V battery

The total power they will consume in series is approximately 2.257 W

Explanation:

The connection arrangement of the headlight and the engine starter = Parallel to the battery

The voltage of the battery, V = 12.0 V

The power at which the headlight operates in parallel, = 38 W

The power at which the kick starter operates in parallel, = 2.40 kW

We have;

P = V²/R

Where;

R = The resistance

V = The voltage = 12 V (The voltage is the same in parallel circuit)

For the headlight, we have;

R₁ = V²/  = 12²/38 = 72/19

R₁ = 72/19 Ω

For the kick starter, we have;

R₂ = V²/ = 12²/2.4 = 60

R₂ = 60 Ω

When the headlight and kick starter are rewired to be in series, we have;

Total resistance, R = R₁ + R₂

Therefore;

R = ((72/19) + 60) Ω = (1212/19) Ω

The current flowing, I = V/R

∴ I = 12 V/(1212/19) Ω = (19/101) A

We note that power, P = I²R

In the series connection, we have;

= I² × R₁

∴ = ((19/101) A)² × 72/19 Ω = 1368/10201 W ≈ 0.134 W

The power at which the headlight operates in series, ≈ 0.134 W

= ((19/101) A)² × 60 Ω = 21660/10201 W ≈ 2.123 W

The power at which the kick starter operates in series, ≈ 2.123 W

The total power they will consume, = +

Therefore;

≈ 0.134 W + 2.123 W = 2.257 W

this could be newton#'s law of cooling. the cooling curve is exponential

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