A 40µF capacitor in a defibrillator is charged to 3000V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2ms. The power delivered to the patient is

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180kW

360kW

90kW

45kW

answer is B.

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Detailed Solution

$\begin{array}{l} Power = \frac{Energy}{Time} = \frac{\frac{1}{2} C V^{2}}{t} \\ = \frac{1}{2} \frac{40 \times 10^{- 6} \times 3000 \times 3000}{2 \times 10^{- 3}} = 90 k W \end{array}$

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