CapacitanceMedium
Question
A 40 μ capacitor in a defibrillator is charged to 3000 V. The energy stored in the capacity tor is set through the patient during a pulse of duration 2 ms. The power delivered to the patient is :
Options
A.45 kW
B.90kW
C.180 kW
D.360 kW
Solution
Energy given during his time period
CV2 = × 40 × 10-6 × 3000 × 3000
= 2 × 9 × 10joule
Power =
= 90 × 10-3 W
= 90 kW
CV2 = × 40 × 10-6 × 3000 × 3000 = 2 × 9 × 10joule
Power =
= 90 × 10-3 W= 90 kW
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