Average Power to Pure Inductive Load

This question asks about the average power delivered by a transformer to a specific type of load: a pure inductive load. To solve this, we need to understand the concepts of different types of power in AC circuits and the power factor.

Understanding Electrical Power in AC Circuits

In alternating current (AC) circuits, there are three main types of power:

• Apparent Power (S): This is the total power that appears to be supplied by the source. It is the product of the RMS voltage and RMS current, measured in Volt-Amperes (VA). For the given transformer, the apparent power rating is 1000 VA. The load is specified as 500 VA, which is its apparent power requirement.
• Real Power (P) or Average Power: This is the actual power consumed by the load and converted into useful work (e.g., heat, light, mechanical energy). It is measured in Watts (W). This is the power we are asked to find.
• Reactive Power (Q): This is the power that flows back and forth between the source and the reactive components (inductors and capacitors) in the circuit. It is not consumed but is necessary for the operation of these components. It is measured in Volt-Amperes Reactive (VAR).

The Concept of Power Factor

The relationship between these three types of power is governed by the power factor. The power factor ($\cos\phi$) is the cosine of the phase angle ($\phi$) between the voltage and current in an AC circuit. It indicates how much of the apparent power is actually real power. The formula for real power is:

$$P = S \times \cos\phi$$

Where:

• $P$ is the real power (average power) in Watts (W).
• $S$ is the apparent power in Volt-Amperes (VA).
• $\cos\phi$ is the power factor.

Average Power for a Pure Inductive Load

A crucial part of the question is "pure inductive load." Let's analyze what this means for the power factor:

• In a purely resistive load, the voltage and current are in phase, so the phase angle $\phi = 0^\circ$. The power factor $\cos(0^\circ) = 1$. In this case, all apparent power is real power.
• In a purely capacitive load, the current leads the voltage by $90^\circ$, so $\phi = -90^\circ$. The power factor $\cos(-90^\circ) = 0$.
• In a pure inductive load, the current lags the voltage by $90^\circ$. Therefore, the phase angle $\phi = 90^\circ$.

Now, let's calculate the power factor for a pure inductive load:

$$\cos\phi = \cos(90^\circ)$$

We know that $\cos(90^\circ) = 0$.

So, for a pure inductive load, the power factor is 0.

Calculating the Average Power Delivered

Now we can use the formula for real power (average power) with the power factor for a pure inductive load:

$$P = S \times \cos\phi$$

Given the load is purely inductive, its power factor is 0. The apparent power drawn by the load is 500 VA (even though the transformer capacity is 1000 VA, the load itself defines the power it draws).

$$P = 500 \text{ VA} \times 0$$

$$P = 0 \text{ W}$$

This means that a pure inductive load consumes no real power. It only draws reactive power, which is stored in its magnetic field during one part of the cycle and returned to the source during another part. Therefore, the average power dissipated or delivered as useful work to a pure inductive load is 0 Watts.

The final answer is $\text{0 W}$.