Q.

A 40 μF 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

90 kW

c

180 kW

d

360 kW

answer is B.

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Detailed Solution

P=Wt= electrostatic potential energy  time ∴P=12CV2t=12×40×10−6×(3000)22×10−3=90×103W=90kW.
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