Question

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

• A. $45 \, kW$
• B. $90 \, kW$
• C. $180 \, kW$
• D. $360 \, kW$

Answer 1

Answer: (B)

Power, $P=\frac{E n e r g y}{t i m e}=\frac{\frac{1}{2} C V^{2}}{t}$
$P=\frac{C V^{2}}{2 t}$
$P=\frac{40 \times 1 6^{- 6} \times 9 \times 1 0^{6}}{2 \times 2 \times 1 0^{- 3}}$
$P=90\times 1000$
$P=90 \, kW$