In one sense, capacitive reactance behaves like a reflection of inductive reactance. But looked at another way, XC is an extension of XL into negative values.
If the frequency of an ac source (in hertz) is given as f, and the capacitance (in farads) is given as C, then the capacitive reactance in ohms, XC, is calculated as follows:
XC = −1/(2πfC )
Again, we meet our friend π! And again, for most practical purposes, we can take 2π to be equal to 6.28. Thus, the preceding formula can be expressed like this:
XC = −1/(6.28fC )
This same formula applies if the frequency, f, is in megahertz and the capacitance, C, is in microfarads. Capacitive reactance varies inversely with the frequency. This means that the function XC versus f appears as a curve when graphed, and this curve “blows up” as the frequency gets close to zero. Capacitive reactance also varies inversely with the actual value of capacitance, given a fixed frequency. Therefore, the function of XC versus C also appears as a curve that blows up as the capacitance approaches zero.
The negative of XC is inversely proportional to frequency, and also to capacitance. Relative graphs of these functions are shown in following figure.
Capacitive reactance is negatively, and inversely, proportional to capacitance. Capacitive reactance is also negatively, and inversely, proportional to frequency.
Suppose a capacitor has a value of 0.00100 μF at a frequency of 1.00 MHz. What is the capacitive reactance?
Use the formula and plug in the numbers. You can do this directly, because the data is specified in microfarads (millionths) and in megahertz (millions):
XC = −1/(6.28 × 1.0 × 0.00100) = −1/(0.00628) = −159 Ω
This is rounded to three significant figures, because all the data is given to that many digits.
What is the capacitive reactance of the preceding capacitor if the frequency decreases to zero (that is, if the voltage source is pure dc)?
In this case, if you plug the numbers into the formula, you get a zero denominator. Mathematicians will tell you that such a quantity is undefined. But we can say that the reactance is negative infinity for all practical purposes.
Suppose a capacitor has a reactance of −100 Ω at a frequency of 10.0 MHz. What is its capacitance?
In this problem, you need to put the numbers in the formula and solve for the unknown C. Begin with this equation:
−100 = −1/(6.28 × 10.0 × C )
Dividing through by −100, you get:
1 = 1/(628 × 10.0 × C )
Multiply each side of this by C, and you obtain C = 1/(628 × 10.0). This can be worked out with a calculator. You should find that C = 0.000159 to three significant figures. Because the frequency is given in megahertz, the capacitance comes out in microfarads. That means C = 0.000159 μF. You can also say it is 159 pF. (Remember that 1 pF = 0.000001 μF.)