Capacitive reactance is the opposition to alternating current produced by capacitance. The magnitude of reactance depends upon the rate of change of voltage.

Consider a capacitor connected to an alternating voltage source. If i, v, and q are the instantaneous current, voltage and charge respectively, then:

C=Q/V C=q/v q=Cv dq/dt=Cdv/dt i=Cdv/dt

The instantaneous current is at its maximum when the rate of change of voltage is also at its maximum. When the capacitor is fully charged, the voltage across it is at its maximum and the current is zero. The graph below shows the relationship between current and voltage across the capacitor. Here ω is the angular frequency. Note that the current leads the voltage by π/2.

Let V_{0} be the maximum voltage, and I_{0} be the maximum current:

v=V_{0}sin(ωt) i=C(dV_{0}sin(ωt))/dt i=ωCV_{0}cos(ωt) =ωCV_{0}sin(ωt + π/2) I_{0}=ωCV_{0}[i=I_{0}when sin(ωt + π/2)=1]

Capacitive reactance is measured in ohms and written X_{C}.
It is calculated as follows

X_{C}=V_{0}/I_{0}=V_{0}/ωCV_{0}=1/ωC

At high frequencies the current will not have a chance to reach zero before the current alternates which means the reactance will be less than that at lower frequencies where the current will have a chance to reach zero.

Consider the following circuit.
V_{1} is measured by channel 1 on an oscilloscope, and V_{2} is measured by channel 2.
The difference between them (V_{1} - V_{2}) should give us the voltage across the
resistor and
according to Ohm's law,
this is proportional to the current (in our case we just divide by 1000).

The screen shot from the oscilloscope shows the voltage across the resistor in red and the voltage across the capacitor in yellow. We see that the current leads the voltage by π/2 and this is in agreement with the graph above.

Fischer-Cripps. A.C., *The Electronics Companion.* Institute of Physics, 2005.

Copyright © 2014 Barry Watson. All rights reserved.