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

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45 kW

180 kW

360 k

90 kW

answer is B.

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Detailed Solution

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times (40 μ F) \times {(3000 V)}^{2} = 180 J$

This energy is given to the patient in 0.002 second.
∴ power given $= \frac{180}{0.002} = 90 k W$

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