Answer:
Explanation:
Given :
Power of headlight [tex]P_{1} = 30[/tex] W
Power of starter [tex]P_{2} = 2500[/tex] W
Voltage of headlight and starter [tex]V = 12[/tex] V
From equation of power,
[tex]P = \frac{V^{2} }{R}[/tex]
[tex]R = \frac{V^{2} }{P}[/tex]
For finding the resistance of headlight and starter,
⇒ For headlight,
[tex]R_{1} = \frac{144}{30} = 4.8[/tex] Ω
⇒ For starter,
[tex]R_{2} = \frac{144}{2500} = 0.057[/tex] Ω
Since equivalent resistance,
[tex]R_{eq} = R_{1} + R_{2} + ........[/tex]
[tex]R_{eq} = 4.8 +0.057 = 4.857[/tex] Ω
So power in series is given by,
[tex]P_{s } = \frac{V^{2} }{R_{eq} } = \frac{144}{4.857}[/tex]
[tex]P_{s } = 29.64[/tex] W