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`

Step-by-step 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=(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=(2W)/(V^(2))` (2)

`C=(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