MC - Physics

2009-05-04 2:41 pm
A power source for a portable electrical defibrillator contains a capacitor of capacitance 100uF. The potential difference across the plates of the capactor is raised to 1000V and 5 % of its stored energy is released in a 5.0 ms pulse. Estimate the average power of the pulse. Select the option from the key that is closest to your answer,

A) 2.5 uW
B) 150 uW
C) 0.5 uW
D) 150 mW
E) 5 W
F) 500 W
G) 1 kW
H) 150 kW

回答 (1)

2009-05-04 3:44 pm
✔ 最佳答案
Initial energy stored in the capacitor

= 1/2 CV2

= 1/2 (100 X 10-6)(1000)2

= 50 J

Energy released in the pulse = 50 X 5% = 2.5 J

Average power of the pulse = Energy / time

= 2.5 / (5.0 X 10-3)

= 500 W

The answer is F.

參考: Physics king


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