With the first equation, you can find the percentage of charge (Q/Q_max) X (100%), by substituting the time elapsed, resistance of charging circuit and capacitance of capacitor. Assuming that your cap is at zero charge before charging.
The product RC is also known as the time constant.
For example, with a circuit resistance of 10â¦ and a 6F capacitor, the time constant is 60 seconds.
Lets assume our circuit is as follows:
- 10â¦ resistance
- 6F, 5V capacitor
- 5V DC source
After one time constant (60s) has passed, the charge ratio would be 0.632, hence 63.2% of maximum charge has been attained.
The thing about capacitors is that as the charge in the caps increase, the rate of charge decreases, this decrease becomes more significant as the cap approaches full charge.
Technically, the time taken for a full charge would be ∞, but in practice 5 time constants is the time taken to reach full charge.
Next, how to calculate energy stored in a capacitor.
Plug in the values into the equation given above, couldn't be simpler.
Example: 1.5F, 5.5V cap would have energy = 22.6875J.
One last useful equation would be to calculate for instantaneous voltage in capacitor when charging.
Using the first equation, substitute Q for capacitor voltage and Q_max for voltage of DC source.
V_cap = V_source *[1-e^(-t/RC)]
*note that capacitor voltage will never rise above that of the source, and supply voltage used for charging should not be greater than the capacitor's voltage rating. DO NOT attach supercapacitors in reverse polarity to a source, it probably won't explode but it hurts to see an expensive component go up in flames.
Hope this has been of use.
- To learn more, visit hyperphysics.
- To get your hands on some supercaps, order samples from Cooper Bussmann.