Question

A cardiac defibrillator stores 1275 J of energy when it is charged to 5.6kV What is the capacitance? O 11.3 pF 96.1 pF O 813 F

Answer 1

Answer: $813.13(10)^{-7}F$

Explanation:

The answer is not among the given options. However, this is a good example of the relation between the energy stored in a capacitor $W$and its capacitance $C$, which is given by the following equation:

$W=\frac{1}{2}CV^{2}$   (1)

Where:

$W=1275J$

$V=5.6kV=5.6(10)^{3}V$ is the voltage

$C$ is the capacitance in Farads, the value we want to find

Isolating $C$ from (1):

$C=\frac{2W}{V^{2}}$   (2)

$C=\frac{2(1275J)}{(5.6(10)^{3}V)^{2}}$   (3)

Finally:

$C=0.000081313F=813.13(10)^{-7}F$ This is the capacitance of the cardiac defibrillator