Power and Power Factor | Sahithyan's Notes

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

Also known as 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}

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 $\theta$ 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}