Calculating Apparent Power in AC Circuits.

Calculating Apparent Power in AC Circuits.

Apparent power is the power that appears to the source because of the circuit impedance. Since the impedance is the total opposition to ac, the apparent power is that power the voltage source "sees." Apparent power is the combination of true power and reactive power. Apparent power is not found by simply adding true power and reactive power just as impedance is not found by adding resistance and reactance.

To calculate apparent power, you may use either of the following formulas:

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For example, find the apparent power for the circuit shown in figure (22)

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Recall that current in a series circuit is the same in all parts of the circuit.


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Power Factor

The POWER FACTOR is a number (represented as a decimal or a percentage) that represents the portion of the apparent power dissipated in a circuit.

If you are familiar with trigonometry, the easiest way to find the power factor is to find the cosine of the phase angle Ө. The cosine of the phase angle is equal to the power factor.

You do not need to use trigonometry to find the power factor. Since the power dissipated in a circuit is true power, then:

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If true power and apparent power are known you can use the formula shown above.

Going one step further, another formula for power factor can be developed. By substituting the equations for true power and apparent power in the formula for power factor, you get:

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Since current in a series circuit is the same in all parts of the circuit, IR equals IZ. Therefore, in a series circuit,

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For example, to compute the power factor for the series circuit shown in figure (22), any of the above methods may be used.

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Another method:

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If you are familiar with trigonometry you can use it to solve for angleӨ and the power factor by referring to the tables in appendices V and VI.

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NOTE: As stated earlier the power factor can be expressed as a decimal or percentage. In this example the decimal number .6 could also be expressed as 60%.


Power Factor Correction

The apparent power in an ac circuit has been described as the power the source "sees". As far as the source is concerned the apparent power is the power that must be provided to the circuit. You also know that the true power is the power actually used in the circuit. The difference between apparent power and true power is wasted because, in reality, only true power is consumed. The ideal situation would be for apparent power and true power to be equal. If this were the case the power factor would be 1 (unity) or 100 percent. There are two ways in which this condition can exist. (1) If the circuit is purely resistive or (2) if the circuit "appears" purely resistive to the source. To make the circuit appear purely resistive there must be no reactance. To have no reactance in the circuit, the inductive reactance (XL) and capacitive reactance (XC) must be equal.


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The expression "correcting the power factor" refers to reducing the reactance in a circuit.

The ideal situation is to have no reactance in the circuit. This is accomplished by adding capacitive reactance to a circuit which is inductive and inductive reactance to a circuit which is capacitive. For example, the circuit shown in figure 4-10 has a total reactance of 80 ohms capacitive and the power factor was .6 or 60 percent. If 80 ohms of inductive reactance were added to this circuit (by adding another inductor) the circuit would have a total reactance of zero ohms and a power factor of 1 or 100 percent. The apparent and true power of this circuit would then be equal.



The principles and formulas that have been presented in this chapter are used in all ac circuits. The examples given have been series circuits.

This section of the chapter will not present any new material, but will be an example of using all the principles presented so far. You should follow each example problem step by step to see how each formula used depends upon the information determined in earlier steps. When an example calls for solving for square root, you can practice using the square-root table by looking up the values given.

The example series RLC circuit shown in figure (23) will be used to solve for XL, XC, X, Z, IT, true power, reactive power, apparent power, and power factor.

The values solved for will be rounded off to the nearest whole number.

First solve for XL and XC.

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Figure (23). - Example series RLC circuit

Now solve for X

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Use the value of X to solve for Z.

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This value of Z can be used to solve for total current (IT ).

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Since current is equal in all parts of a series circuit, the value of IT can be used to solve for the various values of power.

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The power factor can now be found using either apparent power and true power or resistance and impedance. The mathematics in this example is easier if you use impedance and resistance.

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When dealing with a parallel ac circuit, you will find that the concepts presented in this chapter for series ac circuits still apply. There is one major difference between a series circuit and a parallel circuit that must be considered. The difference is that current is the same in all parts of a series circuit, whereas voltage is the same across all branches of a parallel circuit. Because of this difference, the total impedance of a parallel circuit must be computed on the basis of the current in the circuit.

You should remember that in the series RLC circuit the following three formulas were used to find reactance, impedance, and power factor:


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When working with a parallel circuit you must use the following formulas instead:


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NOTE: If no value for E is given in a circuit, any value of E can be assumed to find the values of IL, IC, IX, IR, and IZ. The same value of voltage is then used to find impedance.

For example, find the value of Z in the circuit shown in figure (24).

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The first step in solving for Z is to calculate the individual branch currents.

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Figure (24). - Parallel RLC circuit.


Using the values for IR, IL, and IC, solve for IX and IZ.

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Using this value of IZ, solve for Z.

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If the value for E were not given and you were asked to solve for Z, any value of E could be assumed. If, in the example problem above, you assume a value of 50 volts for E, the solution would be:

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First solve for the values of current in the same manner as before.

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Solve for IX and IZ.

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Solve for Z.

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When the voltage is given, you can use the values of currents, I R, IX, and IZ, to calculate for the true power, reactive power, apparent power, and power factor. For the circuit shown in figure 4-12, the calculations would be as follows.

To find true power,

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To find reactive power, first find the value of reactance (X).

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To find apparent power,

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The power factor in a parallel circuit is found by either of the following methods.

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