A certain car battery with a 12.0 V emf has an initial charge of 120 A h. Assuming that the potential across the terminals stays constant until the battery is completely discharged, for how many hours can it deliver energy at the rate of 100 W?

$$\begin{gathered} \mathrm{Charge}\mathrm{q}=120\mathrm{A}.\mathrm{h} \\ \mathrm{Rate}\mathrm{P}=100\mathrm{W} \\ \mathrm{Emf}\mathrm{\epsilon }=12\mathrm{V} \end{gathered}$$

Write two different relations of the rate of the energy transfer from the formula for emf and power and equating those two relations, calculate the required time.

Formulae are as follow:

$$\begin{gathered} \mathrm{\epsilon }=\frac{∆\mathrm{E}}{∆\mathrm{q}} \\ ∆\mathrm{E}=\mathrm{P}∆\mathrm{t} \end{gathered}$$

Where,𝜀 is emf, E is energy, t is time, q is charge, P is power.

The emf of the battery is defined as the work done per unit charge and if the q is the charge that passes through the battery in time$∆\mathrm{t}$, then,

$$\begin{gathered} \mathrm{\epsilon }=\frac{∆\mathrm{E}}{∆\mathrm{q}} \\ ∆\mathrm{E}=\mathrm{\epsilon }∆\mathrm{q} \end{gathered}$$...........................................(1)

And if P is the rate at which battery delivers energy in time$∆\mathrm{t}$is,

$∆\mathrm{E}=\mathrm{P}∆\mathrm{t}$............................................(2)

Equating the relations 1) and 2),

$$\begin{gathered} ∆\mathrm{t}=\frac{∆\mathrm{q\epsilon }}{\mathrm{P}} \\ ∆\mathrm{t}=\frac{\left(120\mathrm{A}.\mathrm{h}\right)\left(12.0\mathrm{V}\right)}{100\mathrm{W}} \\ ∆\mathrm{t}=14.4\mathrm{h} \end{gathered}$$

Hence, the no. Of hours it can deliver energy at the rate of 100 W is$∆\mathrm{t}=14.4\mathrm{h}$

Therefore, by using the formula for emf and power and equating those two relations number of hours can be determined.