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Q.

A 40 $μ$ F capacitor is a defibrillator is charged to 3000 V The energy stored in the capacitance is sent through the patient during a pulse of duration 3 ms. The power delivered to the patient is

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a

45 kW

b

360 kW

c

180 kW

d

90 kW

answer is B.

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Detailed Solution

$\begin{array}{l} P = \frac{W}{t} = \frac{electrostatic potential energy}{time} \\ ∴ P = \frac{\frac{1}{2} {CV}^{2}}{t} = \frac{1}{2} \times \frac{(40 \times 10^{- 6}) \times (3000)^{2}}{(2 \times 10^{- 3})} \\ = 90 \times 10^{3} W = 90 kW . \end{array}$

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