Question

What voltage, in kilovolts, is applied to the 9.5 μF capacitor of a heart defibrillator that stores 38.5 J of energy?

Answer 1

The relation of energy in a circuit with a capacitance and determined potential is given under the relation,

$E = \frac{1}{2} CV^2$

Here,

C = Capacitance

V = Voltage,

Rearranging to find the voltage,

$V = \sqrt{\frac{2E}{C}}$

Replacing with our values,

$V = \sqrt{\frac{2(38.5)}{9.5*10^{-6}}}$

$V= 2846.97V$

$V = 2.84kV$

Therefore the voltage is 2.84kV