A \(40 \mu \mathrm{F}\) capacitor in a defibrillator is charged to \(3000 \mathrm{~V}\). The energy stored…
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 ____ (in kW)
Solution
As we know that work \(\mathrm{W}=\) Energy stored in it energy as u=
\(\frac{1}{2} \mathrm{CV}^{2}=180 \mathrm{~J}\)
Power as \(=\mathrm{W} / \mathrm{t}\)
$\begin{aligned}
\text{Power} &= \frac{\frac{1}{2} C V^{2}}{t} = \frac{40 \times 10^{-6} \times(3000)^{2}}{2 \times 2 \times 10^{-3}} \\
&=90 \text{~kW}
\end{aligned}$