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A $40\,\mu\text{F}$ capacitor in a defibrillator is charged to $3000\,\text{V}$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2\,\text{ms}$. The power delivered to the patient is

1. 45 kW
2. 90 kW
3. 180 kW
4. 360 kW

Correct answer: 90 kW

Solution

The energy stored is $U=\frac12 CV^2$. With $C=40\times10^{-6}\,\text{F}$ and $V=3000\,\text{V}$, $U=180\,\text{J}$. Over $2\,\text{ms}$, the power is $P=U/t=180/0.002=9.0\times10^4\,\text{W}=90\,\text{kW}$.

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