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

Answer:

The energy E is 1603.008 J.

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

$E=\dfrac{CV^2}{2}$

Put the value into the formula

$E=\dfrac{18.4\times10^{-6}\times(13.2\times10^{3})^2}{2}$

$E=1603.008\ J$

Hence, The energy E is 1603.008 J.