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A $40 \mu \mathrm{F}$ capacitor in a defibrillator is charged to $3000 \mathrm{~V}$. The energy stored in the capacitor is sent through the patient during a pulse of duration $2 \mathrm{~ms}$. The power delivered to the patient is
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The correct answer is:
$90 \mathrm{~kW}$
The energy stored in capacitor $=\frac{1}{2} C V^2$
$$
\begin{aligned}
& =\frac{1}{2} \times 40 \times 10^{-6} \times(3000)^2 \\
& =20 \times 9 \times 10^6 \times 10^{-6}=180 \mathrm{~J} .
\end{aligned}
$$
The power delivered in $2 \mathrm{~ms}$ is
$$
=\frac{180}{2 \times 10^{-3}}=90 \times 10^3 \mathrm{~W}=90 \mathrm{~kW} .
$$
$$
\begin{aligned}
& =\frac{1}{2} \times 40 \times 10^{-6} \times(3000)^2 \\
& =20 \times 9 \times 10^6 \times 10^{-6}=180 \mathrm{~J} .
\end{aligned}
$$
The power delivered in $2 \mathrm{~ms}$ is
$$
=\frac{180}{2 \times 10^{-3}}=90 \times 10^3 \mathrm{~W}=90 \mathrm{~kW} .
$$
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