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 = (1)/(2) CV^2`

Here,

C = Capacitance

V = Voltage,

Rearranging to find the voltage,

`V = \sqrt{(2E)/(C)}`

Replacing with our values,

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

`V= 2846.97V`

`V = 2.84kV`

Therefore the voltage is 2.84kV