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Q.

A 40μF 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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a

45 kW

b

180 kW

c

360 k

d

90 kW

answer is B.

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Detailed Solution

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 U=12CV2=12×(40μF)×(3000V)2=180J

This energy is given to the patient in 0.002 second.
∴ power given  =1800.002=90kW 

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