A 40μF capacitor in a defibrillator is charged to 3000V. The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ms. The power delivered to the patient is :
A
30W
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B
90kW
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C
180kW
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D
360kW
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Solution
The correct option is B30W Given, c=40μF=40×10−6FV=3000V
Energy sent through the patient during a pulse of duration =2ms
The power delivered P=VI=Et=12cv2t=12×40×10−6×30002×10−3=12×12×40×10−3×3000=14×40×10−3×3×103=30W