Question

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?

Short answer

1. The total energy supplied by a$12-V,60-\mathrm{A}$battery is$2.60\mathrm{MJ}$.
2. The volume of gasoline islocalid="1655719323933" $0.063$litres.
3. $1.60\mathrm{h}$required for generator to charge the battery fully.

Step-by-step solution

Define the ohm’s law, resistance(R)and power (P).

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$\left(I\right)$.

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

Determine the energy.

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$0.60\mathrm{MJ}$.

Determine the volume of gasoline.

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

Determine the time.

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.