Power and Power factor

• In a purely resistive AC circuit, the energy delivered by the source will be dissipated as heat by the resistor.
• In a purely capacitive or purely inductive circuit, all of the energy will be stored during a half cycle, and then returned to the source during the other – there will be no net conversion to heat.
• When there is both a resistive component and a reactive component, some energy will be stored, and some will be converted to heat during each cycle.

Power equations

Purely resistive circuit

Suppose a circuit with load $R$ resistance is supplied a voltage of $v(t)=V_m\cos{\omega t}$.

Instantaneous power dissipated by the load is given by:

$$p(t) = \frac{V_m^2}{R}\cos^2{(\omega t)}$$

Always: $p(t)\gt 0$. $$

$$\text{Average power} = \frac{1}{2}\times\text{Peak power}=\cfrac{V_m^2}{2R}$$

Purely inductive circuit

Suppose a circuit with inductor $L$ is supplied a voltage of $v(t)=V_m\cos{\omega t}$.

Instantaneous power dissipated by the load is given by:

$$p(t) = \frac{V_m^2}{2\omega L}\sin{(2\omega t)}$$

The inductive reactive power is given by:

$$Q = \frac{V^2}{\omega L} = I^2 \omega L$$

Purely capacitive circuit

Suppose a circuit with capacitor $C$ is supplied a voltage of $v(t)=V_m\cos{\omega t}$.

Instantaneous power dissipated by the load is given by:

$$p(t) = -\frac{V_m^2 \omega C}{2}\sin{(2\omega t)}$$

The capacitive reactive power is given by:

$$Q = V^2\omega C = \frac{I^2}{\omega L}$$

General load

Consider a general load with both resistive and reactive components. Depending on how inductive or capacitive the reactive component, the phase shift between voltage and current phasor lies between $90°$ and $−90°$.

Suppose the circuit is supplied a voltage of $v(t) = V_m\cos{(\omega t)}$. And the current phasor shifts in $\theta$ phase angle.

$$i(t) = I_m\cos{(\omega t - \theta)}$$

This ends up with:

$$p(t) = \frac{1}{2}V_mI_m\bigg[\cos\theta+\cos (2\omega t - \theta)\bigg]$$

Average over 1 cycle

$$P_\text{avg} =\frac{1}{T}\int_{t_0}^{t_0 + T} p(t)\,\text{d}t = V_{\text{rms}}I_{\text{rms}}\cos{\theta}$$

Types of power

Active power

Aka. true power, resistive power. In all electrical and electronic systems, it is the true power (the resistive power) that does the work.

$$P = V_{\text{rms}}I_{\text{rms}}\cos{\theta}$$

Note

In a question, if “power” is asked to be calculated, that means “active power”.

Reactive Power

Power delivered to/from a pure energy storage element (inductors and capacitors) is known as reactive power.

• Average power consumed by a pure energy storage element is $0$.
• Current associated with it is not $0$. Transmission lines, transformers, fuses, etc. must all be designed to withstand this current.
• Loads with energy storage elements will draw large currents and require heavy duty wiring even though little average power is consumed.
• Shuttles back and forth between the source and the load.

$$Q_\text{reactive} = V_\text{rms}I_\text{rms}\sin\theta$$

Apparent power

Combination of active and reactive power.

$$S = V_\text{rms}I_\text{rms} = \sqrt{P^2 + Q^2}$$

The apparent power is essentially the effective power that the source “sees”.

The Beer Analogy

• Beer - Active power
  Liquid beer is useful power. The power that does the work.
• Foam - Reactive power
  Wasted or lost power.
• Mug - Apparent power
  Demand power, that is being delivered by the utility.

Power factor

If $\theta$ is the phase angle difference between $v$ and $i$, $\cos(\theta)$ is called the power factor. Higher power factor indicates a more efficient use of electrical power.

Power factor appears in the equation of $P_\text{avg}$. $$

$$\cos{\theta} =\frac{\text{Active power}}{\text{Apparent power}} =\frac{\text{Resistance}}{\text{Impedance}}$$

Power factor is:

• leading when $I$ leads $V$
• lagging when $I$ lags $V$

Power triangle

A right triangle that visually represents the relationship between active, reactive and apparent power in an AC circuit.

They are represented as below:

• Active Power (P): On the horizontal axis
• Reactive Power (Q): On the vertical axis
• Apparent Power (S): On the hypotenuse of the triangle

$$S^2 = P^2 + Q^2$$

The angle θ between the active power and the apparent power represents the phase angle, and the cosine of this angle is the power factor:

$$\cos{\theta} = \frac{P}{S}$$