How do capacitors effect ohm's law? How do you calculate the value on a capacitor to correspond with your circuit?

Just to expand on the previous answer.

When you first turn on your DC circuit, you are going from zero volts to something volts (lets just say ten) so this is sort of like an AC voltage because it takes time to rise to 10 volts. as such, the capacitor looks sort of like a short circuit the instant we turn on power. Then, the capacitor will charge and after five time constants (One time constant being R x C)

E.G. if you have a 1k resistor and a 10uF capacitor, one time constant would be

1000 x 0.00001 = 10mS

therefore your capacitor will be charged after 50mS

as for using a capacitor in an AC circuit, the capacitive reactance (which is like resistance) will vary with frequency and also the value of capacitor.

It is found by the formula


so if we had an ac frequency of 1KHz and a 10uF capacitor then the 'resistance' (correctly termed reactance) of the capacitor is:

1/( 2 x π x 1000 x 0.00001) = 15.9 ohms

So in this case, it is quite close to being a short circuit.
UziMonkey8 years ago
Capacitors change with time. Connected to a direct current, they will appear to conduct current for a short period of time, slowly tapering off as the capacitor charges. Connected to an alternating current, depending on the value of the capacitor and the frequency of the AC, a capacitor will be nearly invisible.