Question

A defibrillator containing a 18.4 μF capacitor is used to shock the heart of a patient by holding it to the patient's chest. Just prior to discharging, the capacitor has a voltage of 13.2 kV across its plates. How much energy E is released into the patient, assuming no energy losses? E = _____________ J

Answer 1

The energy E is 1603.008 J.

Step-by-step explanation:

Given that,

Capacitor = 18.4 μF

Voltage = 13.2 kV

We need to calculate the energy

Using formula of energy

`E=W=\int{Q}\ dV`.....(I)

We know that,

`Q=CV`

Put the value of Q in equation (I)

`E=\int{CV}dV`

On integration

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

Put the value into the formula

`E=(18.4*10^(-6)*(13.2*10^(3))^2)/(2)`

`E=1603.008\ J`

Hence, The energy E is 1603.008 J.