<^< Energy Stored in a Capacitor | Course Index | The Unit of Inductance >^>
Example 1:
A voltage is changing at a uniform rate from 10V to 20V in a time of 1s.
If this voltage is applied to a capacitor of 100µF, determine the current that will flow.
Now the current flowing will be given by:
I = C x (rate of change of voltage), thus:
= 
= 100 x 10-6 x 10 = 1 x 10-3 = 1mA
Example 2:
A potential difference of 150V appears across the terminal of a 2µF capacitor. Determine the amount of charge stored.
Q = C V = 2 x 10-6 x 150 = 300µC
Example 3:
Determine the energy stored in a 100µF capacitor when a potential difference of 20V is applied to it.
Using W = ½ C V2 gives:
W = ½ x 100 x 10-6 x (202)
= 20mJ
<^< Energy Stored in a Capacitor | Course index | The Unit of Inductance >^>
