Answer to Question #126014 in Electric Circuits for Meet Darji

Question

A fully charged defibrillator contains 1.20 kJ of energy stored in a 1.10 x10-4 F capacitor. In
a discharge through a patient, 6.00 x102 J of electrical energy is delivered in 3.50 ms. (a) Find the
voltage needed to store 1.20 kJ in the unit. (b) What average power is delivered to the patient?

Answer 1

Since we know the energy and capacitance, we can write he equation that ties these quantities and voltage together. In other words, the energy of a capacitor in terms of voltage and capacitance is

$$E=\frac{1}{2}CV^2,\\\space\\ V=\sqrt{\frac{2E}{C}}=\sqrt{\frac{2\cdot1200}{1.1\cdot10^{-4}}}=4670\text{ V}.$$

The average power is energy divided by time required to transfer of convert this energy. The average power shows how quickly a device can perform work (i.e., deliver or convert energy of one type to another):

$$P=\frac{E}{t}=\frac{600}{3.5\cdot10^{-3}}=171430\text{ W}.$$