91Ó°ÊÓ

Emf$\mathrm{Î\mu }$=12V

Use the formula for work done relating with emf of battery to calculate its value. Then write the formula for the power in terms of current and voltage. Then, putting the current in terms of charge and charge in terms of no. of electrons passing per second, find an expression for power. Inserting the given values in it will give its value.

Formulae are as follow:

$W=qV$

Power$P=iV$

Where,

W isWork done, P is Power,Where, i is current, V is voltage,q is charge.

Work done by an ideal battery:

Work done is related to potential difference as,

W= qV

For an ideal battery $\mathrm{Î\mu }=V$

role="math" localid="1662555209307" $W=q\mathrm{Î\mu }$

For electron,

$$\begin{gathered} W=e\mathrm{Î\mu } \\ Substitutingallvalues, \\ W=e(12V)W=12.0eV \end{gathered}$$

Hence,the work done by an ideal battery is$W=12.0eV$

The power of the battery in watts :

P=iV

But the current is,

$i=\frac{q}{t}$

And,

q=Ne

$i=\frac{Ne}{t}$

Hence,

$P=\frac{NeV}{t}=neV$

Since , $n=3.40×{10}^{18}/s$=12V and$e=1.6×{10}^{-19}C$

$$\begin{gathered} P=(3.40×{10}^{18}{s}^{-1})(1.6×{10}^{-19}C)\left(12V\right) \\ P=6.53W \end{gathered}$$

Hence, the power of the battery in watts P=6.53W

Therefore, by using the formula for work done relating with emf of battery, thw work done and power of battery can be calculated.