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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