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Q.
A $40\, \mu F$ capacitor in a defibrillator is charged to $3000 \,V$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2\, ms$. The power delivered to the patient is
AIIMSAIIMS 2004Electrostatic Potential and Capacitance
Solution:
A capacitor is a device that stores energy in the electric field created between a pair of conductors on which equal but opposite electric charges have been placed.
The energy stored in a capacitor $=\frac{1}{2} C V^{2}$
Given, $C=40\, \mu F =40 \times 10^{-6} F , V=3000\, V$
$\therefore E=\frac{1}{2} \times 40 \times 10^{-6} \times(3000)^{2}$
$=180\, J$
Also $1 \,W =1 \,J / s$
$\therefore 2\, ms =2 \times 10^{-3} s$
Hence, power
$=\frac{180 J }{2 \times 10^{-3} s }$
$=90 \times 10^{3} W =90\, kW$