Question

A photographic flash unit consists of a xenon-filled tube. It gives a flash of average power $2000W$ for $0.04s$. The flash due to discharge of a fully charged capacitor of $40\mu F$. The voltage to which it is charged before a flash is given by the unit is :

• A. $1500V$
• B. $2000V$
• C. $2500V$
• D. $3000V$

Answer 1

Answer: (B)

$2000V$

Total energy stored in capacitor,

$power\times time=2000\times 0.04J$ (i)

Energy stored in capacitor can also be written as,

$E=\frac{C{V}^2}{2}=\frac{1}{2}\times 40\times {10}^{-6}{V}^2$ (ii)

using equation (i) and (ii)
$2000\times 0.04=\frac{1}{2}\times 40\times {10}^{-6}{V}^2$
or $V^2=4\times {10}^{-6}orV=2000V$