Question

The $90mF$capacitor in a defibrillator unit supplies an average of $6500W$of power to the chest of the patient during a discharge lasting $5.0cm$. To what voltage is the capacitor charged?

Short answer

The capacitor is charged to$850V$voltage.

Step-by-step solution

Given information

We have given that The $90mF$capacitor in a defibrillator unit supplies an average of $6500V$of power to the chest of the patient during a discharge lasting 5.0 ms

Simplify

A capacitor of capacitance $C$having stored energy$E$charged at the potential difference$U$ is given by
$E=\frac{C{U}^2}{2}$,

while we know that the power, by definition, is the work done--that is, decrease energy--per unit of time. This means the energy will be expressed in function of the power and the time as

$P=\frac{E}{t}\Rightarrow E=Pt$

The equation can also be written as

$Pt=\frac{C{U}^2}{2}$

Calculating the potential difference at which the capacitor is charged as

$U=\sqrt{\frac{2Pt}{c}}$

Using the numerical values. We get

localid="1648665048399" $U=\sqrt{\frac{\left(2\right)\left(6500\right)(0.005)}{90\times {10}^{-6}}}=850V$