Question

The flash unit in a camera uses a 3.0 V battery to charge acapacitor. The capacitor is then discharged through a flashlamp.The discharge takes 7 µs, andthe average power dissipated in the flashlamp is 8 W. What is the capacitance of the capacitor?

Answer 1

Answer:

$C=2.54\cdot 10^5\ F$

Explanation:

Energy Stored on a Capacitor

This energy is stored in the electric field present when a voltage V is applied to a capacitor C. It can be computed by the formula

$\displaystyle E=\frac{CV^2}{2}$

We can compute the energy by knowing the power dissipated in the flashlamp is P=8 W in a time of $7 \mu s=7\cdot 10^{-6}\ sec$

$\displaystyle E=\frac{P}{t}=\frac{8\ W}{7\cdot 10^{-6}\ sec}=1.14\cdot 10^6\ J$

Solving the first equation for C

$\displaystyle C=\frac{2E}{V^2}$

$\displaystyle C=\frac{2\times 1.14\cdot 10^6}{3^2}$

$\boxed{C=2.54\cdot 10^5\ F}$