Power in AC Circuits

PHYAC Circuits, Power & Resonance

In AC circuits only the resistive part dissipates energy; the average (real) power is $P_{av} = V_{rms}I_{rms}\cos\phi$ , where $\cos\phi$ is the power factor. The component of current $90^{\circ}$ out of phase with the voltage — the wattless current — carries no net power.

Students compute real power, power factor and the wattless current, which examiners use to test the difference between apparent and real power.

Average (real) power

P_{av} = V_{rms} I_{rms} \cos\phi

V_{rms}, I_{rms}

= RMS voltage and current,

\cos\phi

= power factor; this is the power actually consumed (in watts).

Power factor

\cos\phi = \dfrac{R}{Z}

R

= resistance,

Z

= impedance; ranges from 0 (purely reactive, wattless) to 1 (purely resistive).

Wattless current

I_{wattless} = I_{rms}\sin\phi

the quadrature (

90^{\circ}

out-of-phase) component of current that transfers zero net power; the in-phase part is

I_{rms}\cos\phi

Apparent power $V_{rms}I_{rms}$ (VA) is what the source supplies; REAL power $V_{rms}I_{rms}\cos\phi$ (W) is what is dissipated.
The current splits into an in-phase (power) component $I_{rms}\cos\phi$ and a quadrature (wattless) component $I_{rms}\sin\phi$ .
Wattless current carries zero average power because it is $90^{\circ}$ out of phase with the voltage.
For a pure $L$ or $C$ , $\phi = 90^{\circ}$ , so $\cos\phi = 0$ and ALL the current is wattless — no power is consumed.
Equivalent real-power forms: $P_{av} = I_{rms}^2 R = V_{rms}I_{rms}\cos\phi$ ; use whichever data are given.
From $R$ and $Z$ get the power factor directly ( $\cos\phi = R/Z$ ); then $X = \sqrt{Z^2 - R^2}$ and $\sin\phi = X/Z$ .
A low power factor means a large current for the same real power, causing higher $I^2R$ line losses — hence the practical interest in improving it.

Where the marks go

Computing the apparent power $V_{rms}I_{rms}$ and calling it the real power — you must multiply by $\cos\phi$ .
Using the wattless component $I_{rms}\sin\phi$ in the power formula instead of the in-phase $I_{rms}\cos\phi$ .
Plugging the phase angle in degrees into a calculator set to radians (or vice versa) when finding $\cos\phi$ .
Forgetting that $\cos\phi = R/Z$ (resistance over impedance), not $X/Z$ .

How the board asks it

Numerical $P_{av} = V_{rms}I_{rms}\cos\phi$ and $\cos\phi = R/Z$
A series $LCR$ circuit with $R = 30\ \Omega$ , $X_L = 60\ \Omega$ and $X_C = 20\ \Omega$ is connected to a $220\ V$ , $50\ Hz$ supply. Calculate the impedance, the power factor and the average power dissipated in the circuit.
Define / statepower factor and wattless current
Define the power factor of an AC circuit. What is meant by wattless current, and state the condition under which the current in an AC circuit becomes completely wattless.
Give reasons $\phi = 90^{\circ}$ for a pure $L$ or $C$
Account for the fact that no average power is dissipated in a circuit containing a pure inductor (or a pure capacitor) connected to an AC source, even though a current flows through it.
Derive / proveaverage (real) power expression
An alternating voltage $V = V_0\sin(\omega t)$ is applied to a series $LCR$ circuit in which the current is $I = I_0\sin(\omega t - \phi)$ . Obtain an expression for the average power dissipated in the circuit over one cycle, and hence define the power factor.
Applicationlow power factor causes higher $I^2R$ losses
Give reasons why a low power factor is undesirable in power transmission, and state one method by which the power factor of an inductive load can be improved.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.