Question

The flash unit in a camera uses a $3.0\mathrm{V}$battery to charge a capacitor. The capacitor is then discharged through a flashlamp. The discharge takes $10\mu \mathrm{s}$, and the average power dissipated in the flashlamp is $10\mathrm{W}$. What is the capacitance of the capacitor?

Short answer

The capacitance of the capacitor is$22\mu F.$

Step-by-step solution

Given Information 

Camera Voltage$=3.0V$

Time$=10\mu s$

Power$=10W$

Explanation 

The energy $E$provided by a capacitor with capacitance $C$charged to $U$potential difference is given by

$E=\frac{C{U}^2}{2}$

From this we can express the capacitance as

$C=\frac{2E}{U^2}$

We can calculate the capacitance stored in a capacitor by calculating the average power discharged and the time duration is,

$P=\frac{E}{t}\Rightarrow E=Pt$

Substituting this expression at the one for the capacitance we find

$C=\frac{2Pt}{U^2}$

Then the expression will be,

localid="1648632271586" $C=\left(\frac{2\times 10W\times 1·{10}^{-5}\mu s}{3V^2}\right)$

Substitute the values,

$=2.2·{10}^{-5}\mathrm{F}$

Hence, the capacitance of the capacitor is$22\mu F.$