For a certain load, the true power is 100 W and the reactive power is 100 VAR. What is the apparent power?

Question

Prepp Team · Accepted Answer

Understanding Power in AC Circuits

In alternating current (AC) circuits, power is described in three ways:

True Power (P): This is the power actually consumed or utilized by the load to perform useful work. It is measured in Watts (W). It is the power dissipated in the resistive part of the load.
Reactive Power (Q): This is the power that is stored and then returned to the source by reactive components like inductors and capacitors. It does not do any useful work but is necessary for the operation of devices that use magnetic or electric fields. It is measured in Volt-Ampere Reactive (VAR).
Apparent Power (S): This is the total power delivered to the load from the source. It is the vector sum of true power and reactive power. It is measured in Volt-Amperes (VA).

These three types of power are related through the power triangle, where true power and reactive power form the two legs of a right-angled triangle, and apparent power is the hypotenuse. The relationship is given by the Pythagorean theorem.

Relating True Power, Reactive Power, and Apparent Power

The relationship between true power (P), reactive power (Q), and apparent power (S) is fundamental in AC circuit analysis. It is expressed by the formula derived from the power triangle:

\[S^2 = P^2 + Q^2\]

Or, to find the apparent power (S):

\[S = \sqrt{P^2 + Q^2}\]

Here:

P is the true power (in Watts, W)
Q is the reactive power (in Volt-Ampere Reactive, VAR)
S is the apparent power (in Volt-Ampere, VA)

Calculating Apparent Power

The question provides the following information:

True Power, \(P = 100 \text{ W}\)
Reactive Power, \(Q = 100 \text{ VAR}\)

We need to calculate the apparent power, \(S\). Using the formula \(S = \sqrt{P^2 + Q^2}\), we substitute the given values:

\[S = \sqrt{(100 \text{ W})^2 + (100 \text{ VAR})^2}\]

\[S = \sqrt{10000 + 10000}\]

\[S = \sqrt{20000}\]

To simplify \(\sqrt{20000}\), we can write it as \(\sqrt{2 \times 10000}\):

\[S = \sqrt{10000} \times \sqrt{2}\]

\[S = 100 \times \sqrt{2}\]

The value of \(\sqrt{2}\) is approximately 1.414.

\[S \approx 100 \times 1.414\]

\[S \approx 141.4 \text{ VA}\]

Comparing Calculated Apparent Power with Options

The calculated apparent power is approximately 141.4 VA. Let's compare this with the given options:

Option	Apparent Power Value
1	100 VA
2	120 VA
3	141.4 VA
4	200 VA

The calculated value of 141.4 VA matches Option 3.

Revision Table: AC Power Concepts

Power Type	Symbol	Unit	Description	Relationship
True Power	P	Watts (W)	Actual power used by the load for work (dissipated in resistance)	\(P = V_{rms} I_{rms} \cos(\phi)\)
Reactive Power	Q	Volt-Ampere Reactive (VAR)	Power exchanged between source and reactive components (inductors/capacitors)	\(Q = V_{rms} I_{rms} \sin(\phi)\)
Apparent Power	S	Volt-Ampere (VA)	Total power delivered from the source (vector sum of P and Q)	\(S = V_{rms} I_{rms}\), \(S = \sqrt{P^2 + Q^2}\)

Additional Information: Power Factor

The power factor (PF) is another important concept in AC circuits. It is defined as the ratio of true power to apparent power:

\[\text{Power Factor} = \frac{P}{S} = \cos(\phi)\]

where \(\phi\) is the phase angle between the voltage and current. The power factor indicates how effectively the apparent power delivered to the load is being converted into useful work (true power). A power factor close to 1 (or unity) indicates efficient power usage, while a low power factor indicates that a significant portion of the apparent power is reactive power, which increases losses in the system.

In this specific problem, with \(P=100\) W and \(S \approx 141.4\) VA, the power factor would be:

\[\text{PF} = \frac{100}{141.4} \approx 0.707\]

This value corresponds to a phase angle \(\phi = \arccos(0.707)\), which is approximately 45 degrees. Since P and Q are equal, this represents a load with equal resistance and reactance (or a load with a total impedance angle of 45 degrees).