The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours .$\left(A‐h\right)\mathbf{‐}\mathit{A}\mathbf{}\mathbf{50}\mathbf{}\mathit{A}\mathbf{‐}\mathit{h}$A battery can supply a current of$\mathbf{50}\mathbf{}\mathbf{A}$for $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{h}\mathbf{,}\mathbf{}\mathbf{or}\mathbf{}\mathbf{25}\mathbf{}\mathbf{A}\mathbf{}\mathbf{for}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{h}$or for and so on. (a) What total energy can be supplied by a $\mathbf{12}\mathbf{-}\mathbf{v}\mathbf{,}\mathbf{}\mathbf{60}\mathbf{-}\mathbf{A}\mathbf{‐}\mathbf{h}$battery if its internal resistance is negligible? (b) What volume (in litres) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is $\mathbf{900}\mathbf{}\mathbf{kg}\mathbf{/}{\mathbf{m}}^{\mathbf{3}}$.) (c) If a generator with an average electrical power output ofrole="math" localid="1655719210000" $\mathbf{0}\mathbf{.}\mathbf{45}\mathbf{}\mathit{k}\mathit{W}$ is connected to the battery, how much time will be required for it to charge the battery fully?

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

$V=IR$

Where, $I$is current in ampere$\left(A\right)$,$R$is resistance in ohms$\left(\Omega \right)$and$V$is the potential difference volt$\left(V\right)$.

If real source of emf $\epsilon$has internal energy $r,$then its terminal potential difference$V_{ab}$depends upon current.

So, formula for the terminal potential difference is:

$V_{ab}=\epsilon -Ir$

The ratio of$\mathit{V}$to $\mathit{I}$for a particular conductor is called its resistance $\mathit{R}$

$R=\frac{V}{I}or\frac{pL}{A}$

Where,$p$is resistivity$\Omega -m,L$is length in $m$and $A$is area in${\mathrm{m}}^2$.

The power$\left(P\right)$is the product of potential difference$\left(V\right)$and the current$\left(I\right)$.

$P=VIor{I}^{2}Ror\frac{V^2}{R}$

Given that,

$$\begin{gathered} I=60A \\ V=12V \\ t=3600s \end{gathered}$$

The energy produced in $1$hour is:

$$\begin{gathered} U=IVt \\ =\left(60\right)\left(12\right)\left(3600\right) \\ =2.60\mathrm{MJ} \end{gathered}$$

Hence, the total energy supplied by a$12-V,60-\mathrm{A}.\mathrm{h}$battery is$0.60\mathrm{MJ}$.

Given that,$p=900\mathrm{kg}/{\mathrm{m}}^3$

And heat of combustion of gasoline$L_c=4.6\times {10}^7\mathrm{J}/\mathrm{kg}$

Now, the volume of gasoline is:

$$\begin{gathered} \mathrm{Volume}=\frac{U}{p{L}_c} \\ =\frac{2.60}{\left(900\right)\left(4.6\times {10}^7\right)}\times \frac{1000}{1} \\ =0.063\mathrm{L} \end{gathered}$$

Hence, the volume of gasoline is$0.063$litres

Given that,$P=0.45\mathrm{kW}$

Now, the time required to charge battery is

role="math" localid="1655718908147" $$\begin{gathered} t=\frac{U}{P} \\ =\frac{2.60\times {10}^4}{0.45\times 1000}\times \frac{1}{3600} \\ =1.60\mathrm{h} \end{gathered}$$

Hence, $1.60\mathrm{h}$required for generator to charge the battery fully.