**Reactance And Frequency**

Capacitive reactance behaves in many ways like a mirror image of inductive reactance. In another sense, however, *X* _{C} is an extension of *X* _{L} into negative values—below zero—with its own peculiar set of characteristics.

If the frequency of an ac source is given in hertz as *f* and the capacitance of a capacitor in farads is given as *C* , then the capacitive reactance is

*X* _{C} = −1/(2π/ *fC* ) = −(2π *fC* ) ^{−1} ≈ −(6.2832 *fC* ) ^{−1}

This same formula applies if the frequency is in megahertz (MHz) and the capacitance is in microfarads (μF). Remember that if the frequency is in millions, the capacitance must be in millionths. This formula also would apply for frequencies in kilohertz (kHz) and millifarads (mF), but for some reason, you’ll almost never see millifarads used in practice. Even millifarads are large units for capacitance; components of more than 1,000 μF (which would be 1 mF) are rarely found in real-world electrical systems.

Capacitive reactance varies inversely with the frequency. This means that the function *X* _{C} versus *f* appears as a curve when graphed, and this curve “blows up negatively” as the frequency nears zero. Capacitive reactance also varies inversely with the actual value of capacitance given a fixed frequency. The function of *X* _{C} versus *C* also appears as a curve that “blows up negatively” as the capacitance approaches zero. The negative of *X* _{C} is inversely proportional to frequency, as well as to capacitance. Relative graphs of these functions are shown in Fig. 15-9.

**Reactance And Frequency Practice Problem **

**Reactance And Frequency Practice Problem**

**Problem 1**

A capacitor has a value of 0.00100 μF at a frequency of 1.00 MHz. What is the capacitive reactance?

**Solution 1**

Use the formula and plug in the numbers. You can do this directly because the data are specified in microfarads (millionths) and in megahertz (millions):

*X* _{C} = −1/(6.2832 × 1.00 × 0.00100) = −1/(0.0062832) = −159 ohms

This is rounded to three significant figures because the data are specified only to this many digits.

**Problem 2**

What will be the capacitive reactance of the preceding capacitor if the frequency decreases to zero—that is, if the power source is dc?

**Solution 2**

In this case, if you plug the numbers into the formula, you’ll get zero in the denominator. Division by zero is not defined. In reality, however, there is nothing to prevent you from connecting a dc battery to a capacitor! You might say, “The reactance is extremely large negatively and, for all practical purposes, is negative infinity.” More appropriately, you should call the capacitor an open circuit for dc.

**Problem 3**

Suppose that a capacitor has a reactance of −100 ohms at a frequency of 10.0 MHz. What is its capacitance?

**Solution 3**

In this problem you need to put the numbers in the formula and solve for the unknown *C* . Begin with this equation:

−100 = −(6.2832 × 10.0 × *C* ) ^{−1}

Dividing through by −100:

1 = (628.32 × 10.0 × *C* ) ^{−1}

Multiply each side of this by *C:*

*C* = (628.32 × 10.0) ^{−1}

= 6283.2 ^{−1}

This can be solved easily enough. Divide out *C* = 1/6283.2 on your calculator, getting *C* = 0.00015915. Because the frequency is given in megahertz, this capacitance comes out in microfarads, so *C* = 0.00015915 μF. This must be rounded to *C* = 0.000159 μF in this scenario. You also can say *C* = 159 pF (remember that 1 pF = 0.000001 μF).

The arithmetic for dealing with capacitive reactance is a little messier than that for inductive reactance for two reasons. First, you have to work with reciprocals, and therefore, the numbers sometimes get awkward. Second, you have to watch those negative signs. It’s easy to leave them out or to put them in when they should not be there. They are important when looking at reactances in the coordinate plane because the minus sign tells you that the reactance is capacitive rather than inductive.

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