First slide

Capacitance

Question

A 40 μF capacitor in a defibrillator is charged to 3000 V. 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 :

Easy

A

45 kW
B

90 kW
C

180 kW
D

360 kW

Solution

$P = \frac{W}{t} = \frac{\frac{1}{2} {CV}^{2}}{t}$
$= \frac{\frac{1}{2} x 40 x 10^{- 6} x {(300)}^{2}}{2 x 10^{- 3}} = 9 x 10^{4} W = 90 kW$

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