91影视

According to Ohm鈥檚 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$\left(惟\right)$and$V$is the potential difference volt$\left(V\right)$.

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

So, formula for the terminal potential difference is:

$V_{ab}=蔚-Ir$

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$惟-m,L$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$L_c=4.6脳{10}^7\mathrm{J}/\mathrm{kg}$

Now, the volume of gasoline is:

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

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

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