4)
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Answer:
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