Question

(a) In electron-volts, how much work does an ideal battery with a 12.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.40$\mathbf{\times }{\mathbf{10}}^{\mathbf{18}}$electrons pass through each second, what is the power of the battery in watts?

Short answer

The work done by an ideal battery is $W=12.0eV$

The power of the battery in watts is$P=6.53W$

Step-by-step solution

 Step 1: Given

Emf$\epsilon$=12V

$n=\frac{N}{t}=3.40\times {10}^{18}{s}^{-1}$

Determining the concept

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.

(a) Determining the work done by an ideal battery

Work done by an ideal battery:

Work done is related to potential difference as,

W= qV

For an ideal battery $\epsilon =V$

role="math" localid="1662555209307" $W=q\epsilon$

For electron,

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

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

(b) determining the power of the battery in watts

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

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

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.