Series LCR Circuit & Power Factor

PHYAC Circuits, Power & Resonance

In a series LCR circuit the resistor, inductor and capacitor carry the same current but their voltages differ in phase, so they combine vectorially into an impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ . Students compute each reactance, the impedance, the RMS current, the phase angle, power factor and average power.

The phasor reasoning and $\cos\phi = R/Z$ are central exam skills.

Impedance

Z = \sqrt{R^2 + (X_L - X_C)^2}

R

= resistance,

X_L = \omega L

X_C = 1/\omega C

(all in

\Omega

);

(X_L - X_C)

is the NET reactance.

RMS current

I_{rms} = \dfrac{V_{rms}}{Z}

the same current flows through all three series elements;

V_{rms}

= applied RMS voltage.

Phase angle & power factor

\tan\phi = \dfrac{X_L - X_C}{R}, \qquad \cos\phi = \dfrac{R}{Z}

\phi

= phase of the voltage relative to the current;

\cos\phi

= power factor (between 0 and 1).

Average power

P_{av} = V_{rms} I_{rms} \cos\phi = I_{rms}^2 R

only

R

dissipates power; the simplest form is

I_{rms}^2 R

All three series elements share the SAME current; it is the VOLTAGES that add as phasors, not as a simple algebraic sum.
If $X_L > X_C$ the circuit is net inductive (voltage leads current, $\phi > 0$ ); if $X_C > X_L$ it is net capacitive (current leads).
The voltages across $L$ and $C$ are $180^{\circ}$ out of phase, so they partially cancel — that is why only the DIFFERENCE $(X_L - X_C)$ enters $Z$ .
Power factor $\cos\phi = R/Z$ ranges from 1 (purely resistive / at resonance) down toward 0 (purely reactive).
Compute the reactances with $\omega = 2\pi f$ , and convert $\mu$ F to F and any mH to H first.
Average power equals $I_{rms}^2R$ — the reactive elements contribute zero, so $V_{rms}I_{rms}\cos\phi$ reduces to $I_{rms}^2R$ .
Near resonance the voltages across the individual $L$ and $C$ can each exceed the source voltage, yet their phasor sum equals the source — not a contradiction.

Where the marks go

Adding reactances arithmetically, e.g. $Z = R + X_L + X_C$ , instead of the phasor form $\sqrt{R^2 + (X_L - X_C)^2}$ .
Adding $X_L + X_C$ rather than subtracting; the net reactance is the DIFFERENCE $(X_L - X_C)$ .
Using $\cos\phi = X/Z$ instead of $R/Z$ for the power factor.
Forgetting $2\pi$ in $\omega$ or leaving $C$ in $\mu$ F when finding $X_C$ .

How the board asks it

Numericalimpedance, rms current, power factor and average power
A series LCR circuit with $R = 30\,\Omega$ , $L = 0.2\,\text{H}$ and $C = 50\,\mu\text{F}$ is connected to a $220\,\text{V}$ , $50\,\text{Hz}$ AC supply. Calculate (i) the impedance, (ii) the rms current, (iii) the power factor and (iv) the average power consumed.
Derive / provephasor diagram for impedance and power factor
With the help of a labelled phasor diagram, derive an expression for the impedance $Z = \sqrt{R^2 + (X_L - X_C)^2}$ of a series LCR circuit, and hence write the expression for the power factor $\cos\phi = R/Z$ .
Define / statepower factor and its limiting values
Define the power factor of a series LCR circuit. State its value when the circuit is (i) purely resistive and (ii) purely reactive.
Give reasonsaverage power dissipated only in resistance
Explain why the average power consumed in a series LCR circuit is $V_{rms}\,I_{rms}\cos\phi$ and not $V_{rms}\,I_{rms}$ , even though the same current flows through $L$ , $C$ and $R$ .
Assertion–Reasonvoltage across L or C exceeding the source near resonance
Assertion: In a series LCR circuit near resonance, the voltage across the inductor can exceed the applied source voltage. Reason: The voltages across $L$ and $C$ are $180^{\circ}$ out of phase and partially cancel, so their phasor sum can be small while each is individually large. Select the correct option.

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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.