12 min readConcept sheet3-mark numericals and a 5-mark derivation of impedance with a phasor diagram; also a NEET favourite

Series LCR circuit: phasor diagram and impedance

In a series LCR circuit the same current flows through all three. Adding voltage phasors gives Z = √(R² + (X_C − X_L)²) and the phase angle φ.

ImpedanceZ = √(R² + (X_C − X_L)²), in ohm

Peak currentiₘ = vₘ/Z; with rms values, I = V/Z

Phase angletan φ = (X_C − X_L)/R, with i = iₘ sin(ωt + φ)

Nature of the circuitX_L > X_C: inductive, current lags; X_C > X_L: capacitive, current leads

The circuit

A resistor R, an inductor L and a capacitor C are joined in series to a source v = vₘ sin ωt. Being in series, the same current i = iₘ sin(ωt + φ) flows through all three. φ is the phase difference between the source voltage and the current. The voltages across the three elements are out of phase, so their peak values cannot simply be added. We add them as phasors. The total opposition of the circuit to AC is its impedance Z, measured in ohm.

Derivation with the phasor diagram

V_R, V_L and V_C below are the peak voltages across R, L and C.

1
Draw the current phasor I as the reference. The same current flows through R, L and C.
2
V_R = iₘR is in phase with I, so draw it along I.
3
V_L = iₘX_L leads I by π/2, so draw it a quarter turn anticlockwise from I.
4
V_C = iₘX_C lags I by π/2, so draw it a quarter turn clockwise from I.
5
V_L and V_C point in opposite directions. Their resultant has size (V_C − V_L), taking V_C > V_L.
6
The source voltage is the phasor sum: vₘ² = V_R² + (V_C − V_L)² = iₘ²[R² + (X_C − X_L)²].
7★
So, iₘ = vₘ/Z with impedance Z = √(R² + (X_C − X_L)²), and tan φ = (V_C − V_L)/V_R = (X_C − X_L)/R.

★ key step

How to draw the phasor diagram and impedance triangle

Draw I as a horizontal arrow from the origin, and V_R along it. Label both.
Draw V_L vertically up and V_C vertically down from the origin, with V_C longer (for X_C > X_L).
Draw (V_C − V_L) downwards. Complete the rectangle with V_R to get the resultant voltage vₘ.
Mark the angle φ between vₘ and I.
Impedance triangle: a right triangle with base R, height (X_C − X_L) and hypotenuse Z. The angle between R and Z is φ, so cos φ = R/Z.

Worked example: from L, C and ω

Problem

A series LCR circuit has R = 80 Ω, L = 0.1 H and C = 25 μF. It is connected to a source of peak voltage 200 V and angular frequency 1000 rad/s. Find X_L, X_C, Z, the peak current and the phase angle.

Working

X_L = ωL = 1000 × 0.1 = 100 Ω.
X_C = 1/(ωC) = 1/(1000 × 25 × 10⁻⁶) = 1/0.025 = 40 Ω.
Z = √(R² + (X_C − X_L)²) = √(80² + (40 − 100)²) = √(6400 + 3600) = √10000 = 100 Ω.
iₘ = vₘ/Z = 200 V ÷ 100 Ω = 2 A.
tan φ = (X_C − X_L)/R = −60/80 = −0.75, so φ ≈ −37°.
φ is negative: X_L > X_C, so the circuit is inductive and the current lags the voltage by about 37°.

Answer

X_L = 100 Ω, X_C = 40 Ω, Z = 100 Ω, iₘ = 2 A; the current lags by about 37°

Worked example: voltages across each part

Problem

In a series LCR circuit, R = 30 Ω, X_L = 80 Ω and X_C = 40 Ω. It is joined to a 200 V (rms) supply. Find Z, the rms current and the rms voltage across each element. Show that they add as phasors to 200 V.

Working

Z = √(30² + (80 − 40)²) = √(900 + 1600) = √2500 = 50 Ω.
I = V/Z = 200 V ÷ 50 Ω = 4 A.
V_R = IR = 4 × 30 = 120 V; V_L = IX_L = 4 × 80 = 320 V; V_C = IX_C = 4 × 40 = 160 V.
Phasor sum: √(V_R² + (V_L − V_C)²) = √(120² + 160²) = √(14400 + 25600) = √40000 = 200 V.
V_L (320 V) is more than the supply voltage. This is possible because V_L and V_C partly cancel each other.

Answer

Z = 50 Ω, I = 4 A, V_R = 120 V, V_L = 320 V, V_C = 160 V; the phasor sum is 200 V

In real life

A ceiling fan motor winding behaves like R and L in series. The fan's capacitor, in a second winding, shifts the phase of the current there, so the two windings together make the motor start turning by itself.

Exam angle

5 marks: “Using a phasor diagram, derive an expression for the impedance of a series LCR circuit and the phase angle between the current and the voltage.” Phasors: V_R along I, V_L ahead by π/2, V_C behind by π/2; vₘ² = V_R² + (V_C − V_L)²; Z = √(R² + (X_C − X_L)²); tan φ = (X_C − X_L)/R.

3 marks: “A series LCR circuit with R = 40 Ω, X_L = 50 Ω and X_C = 20 Ω is connected to a 100 V (rms) supply. Find the impedance, the current and the phase angle.” Z = √(40² + 30²) = √2500 = 50 Ω; I = 100 V ÷ 50 Ω = 2 A; tan φ = (20 − 50)/40 = −0.75, φ ≈ −37°, so the current lags.

2 marks: “In a series LCR circuit, X_L > X_C. Is the circuit inductive or capacitive? Does the current lead or lag the voltage?” Inductive; the current lags the voltage.

Remember it

R along I, L up, C down: combine V_L and V_C first (they cancel), then join the result to V_R by Pythagoras.