Question

Your car's 30.0-W headlight and $\mathbf{2}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{kW}$starter are ordinarily connected in parallel in a$\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{v}$system. What power would one headlight and the starter consume if connected in series to a 12.0-V battery? (Neglect any other resistance in the circuit and any change in resistance in the two devices.)

Short answer

The power for the series combination of headlight and starter will be$29.6W$.

Step-by-step solution

Definition of resistance 

An unwillingness to accept anything, such as a change or a new thought, is known as resistance.

Finding resistance

To begin, we must determine the resistances of both components, therefore,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{Headlight}}=\frac{{\mathrm{V}}^2}{{\mathrm{P}}_{\mathrm{Headligh}}} \\ =\frac{{\left(12.0\mathrm{V}\right)}^2}{30.0\mathrm{W}} \\ =4.8\mathrm{\Omega } \\ {\mathrm{R}}_{\mathrm{Starter}}=\frac{{\mathrm{V}}^2}{{\mathrm{P}}_{\mathrm{starter}}} \\ =\frac{{\left(12.0\mathrm{V}\right)}^2}{2400\mathrm{W}} \\ =0.06\mathrm{\Omega } \end{gathered}$$

The following formula may be used to compute new power, such that,

$$\begin{gathered} {\mathrm{R}}_{\mathrm{Total}}={\mathrm{R}}_{\mathrm{Headlight}}+{\mathrm{R}}_{\mathrm{Starter}} \\ =4.8\mathrm{\Omega }+0.06\mathrm{\Omega } \\ =4.86\mathrm{\Omega } \end{gathered}$$

Finding power

Therefore, we can calculate the power through this resistance,

$$\begin{gathered} {\mathrm{P}}_{\mathrm{Total}}=\frac{{\mathrm{V}}^2}{{\mathrm{R}}_{\mathrm{Total}}} \\ =\frac{{\left(120\mathrm{V}\right)}^2}{4.86\mathrm{\Omega }} \\ =29.6\mathrm{W} \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{power}\mathrm{is}29.6\mathrm{W} \end{gathered}$$