Question

If we instead connect this headlight and starter in series with the 12.0 V battery, how much total power in W would the headlight and starter consume? (You may neglect any other resistance in the circuit and any change in resistance in the two devices.)

Answer 1

Answer:

If $\large{P_{H}}$ and $\large{P_{S}}$ be the power consumed by the headlight and starter respectively, then the total power consumed will be $\large{P = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}$

Explanation:

We know that electric power ($P$) is given by

$\large{P = \dfrac{V^{2}}{R}}$

where '$V$' is the applied voltage and '$R$' is the resistance.

If we consider that '$\large{P_{H}}$' and '$\large{P_{S}}$' be the power consumed by the headlight and starter respectively, then the resistance ($\large{R_{H}}$) of the headlight and the resistance ($\large{R_{S}}$) of the starter can be written as

$&&\large{R_{H} = \dfrac{V^{2}}{P_{H}} = \dfrac{12^{2}}{P_{H}}}\\&and,& \large{R_{S} = \dfrac{V^{2}}{P_{S}} = \dfrac{12^{2}}{P_{S}}}$

If the headlight and the starter in connected in series, then equivalent resistance ($\large{R_{eq}}$) will be

$\large{R_{eq} = R_{H} + R_{S} = 12^{2}(\dfrac{1}{P_{H} + P_{S}})}$

So the total power ($P$) consumed will be given by

$\large{P = \dfrac{12^{2}}{R_{eq}} = \dfrac{1}{(\dfrac{1}{P_{H}} + \dfrac{1}{P_{S}})} = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}$