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Unit X · Electromagnetic induction and alternating current (lesson 19)
12 min readConcept sheetSection B long answer (5 marks), often with a choice: phasor diagram, impedance and resonance with a graph; short answer (3 marks) on power factor; Section A objective (2 marks, two parts); also a NEET favourite

Series LCR circuit: impedance, resonance, power and power factor

In a series LCR circuit Z = √(R² + (X_L − X_C)²). At resonance X_L = X_C, so Z = R and ν_r = 1/2π√(LC), with the largest current.

ImpedanceZ = √(R² + (X_L − X_C)²), in ohm; I_rms = V_rms/Z
Phase angletan φ = (X_L − X_C)/R; the current lags if X_L > X_C and leads if X_C > X_L
Resonant frequencyν_r = 1/(2π√(LC)); at resonance Z = R, φ = 0 and the current is largest
Average powerP = V_rms I_rms cos φ, with power factor cos φ = R/Z

Derivation with the phasor diagram

R, L and C are joined in series to an AC source. The **same current** flows through all three, so we draw the current phasor first.

  1. 1
    Draw the peak current I₀ as a horizontal phasor.
  2. 2
    V_R = I₀R is in phase with the current: draw it along I₀.
  3. 3
    V_L = I₀X_L leads the current by 90°: draw it straight up.
  4. 4
    V_C = I₀X_C lags the current by 90°: draw it straight down. V_L and V_C are opposite, so their resultant is (V_L − V_C).
  5. 5
    The source voltage is the phasor sum: E₀² = V_R² + (V_L − V_C)² = I₀²[R² + (X_L − X_C)²].
  6. 6★
    So, I₀ = E₀/Z, with impedance Z = √(R² + (X_L − X_C)²), and tan φ = (X_L − X_C)/R.
  7. 7
    Impedance is the total opposition of the circuit to AC. It is measured in ohm.

★ key step

Resonance and tuning

X_L = 2πνL rises with frequency, while X_C = 1/2πνC falls. At one frequency they become equal. This is resonance. Then V_L and V_C cancel, the impedance is least, Z = R, and the current is largest, I₀ = E₀/R. The current is in phase with the voltage (φ = 0). From 2πν_rL = 1/2πν_rC we get ν_r = 1/(2π√(LC)). The resonant frequency does not depend on R. But a smaller R gives a taller, sharper peak of current. Radio and TV sets use this to tune. Turning the knob changes a capacitor, and so changes ν_r. When ν_r matches a station's frequency, that station's signal produces the largest current and is received.

Power in an AC circuit, and the power factor

  1. 1
    The instantaneous power is P = VI. When V and I differ in phase by φ, P changes all through the cycle.
  2. 2★
    Averaged over a cycle, P = V_rms I_rms cos φ. The factor cos φ is the power factor.
  3. 3
    For a series LCR circuit, cos φ = R/Z. So P = I_rms²R: only the resistor uses energy.
  4. 4
    Pure resistor, or LCR at resonance: φ = 0, cos φ = 1, and the power is largest, V_rms I_rms.
  5. 5
    Pure inductor or pure capacitor: φ = 90°, cos φ = 0, and the average power is zero. Such a current is called wattless current.
  6. 6
    So, a low power factor needs more current for the same power, and that wastes energy in the supply lines.

★ key step

Activity: inductance and resistance of a coil, using a series resistance and an AC voltmeter

You need a coil (for example the secondary of a step-down transformer), a resistance box, a 12 V, 50 Hz AC supply from a step-down transformer, an AC voltmeter (0 to 15 V), a key and wires.

  1. 1
    Join the coil (inductance L, resistance r) in series with the resistance box R, the key and the 12 V AC supply.
  2. 2
    Take out about 200 Ω from the box and close the key. Measure V_R across the box, V_L across the coil and V across both.
  3. 3
    Notice that V is less than V_R + V_L. The two voltages are out of phase, so they add as phasors.
  4. 4
    Draw a line AB whose length stands for V_R (say 1 cm per volt). From B, draw an arc of radius V_L; from A, an arc of radius V. They meet at C.
  5. 5
    Extend AB and drop a perpendicular CD onto it. Then BD stands for Ir and CD stands for IX_L.
  6. 6★
    Since AB stands for IR: r = R × BD/AB and X_L = R × CD/AB. Then L = X_L/2πν, with ν = 50 Hz.
  7. 7
    Repeat for 5 values of R, and take the mean of r and of L.
  8. 8
    Result: write the mean inductance L (in henry) and the mean resistance r (in ohm) of the coil.

★ key step

Worked example: an LCR circuit, and the coil activity

Problem

(a) A series circuit on the 230 V, 50 Hz mains has R = 15 Ω, X_L = 40 Ω and X_C = 20 Ω. Find Z, the rms current, the power factor and the power, and say whether the current leads or lags. (b) In the coil activity, R = 100 Ω, V_R = 6.0 V, V_L = 8.0 V and V = 12.0 V. Find r and L.

Working
  1. (a) Z = √(15² + (40 − 20)²) Ω = √(225 + 400) Ω = √625 Ω = 25 Ω.
  2. I_rms = 230 V ÷ 25 Ω = 9.2 A.
  3. Power factor cos φ = R/Z = 15 ÷ 25 = 0.60.
  4. P = V_rms I_rms cos φ = 230 × 9.2 × 0.60 W ≈ 1270 W. Check: I²R = 9.2² × 15 W ≈ 1270 W.
  5. X_L > X_C, so the current lags the voltage; tan φ = 20/15, so φ ≈ 53°.
  6. (b) Current I = V_R/R = 6.0 V ÷ 100 Ω = 0.060 A.
  7. The coil's voltage has a part Ir in phase with I and a part IX_L at 90°. So (V_R + Ir)² + (IX_L)² = V² and (Ir)² + (IX_L)² = V_L².
  8. Subtracting: V_R² + 2V_R(Ir) = V² − V_L², so 36 + 12(Ir) = 144 − 64 = 80. This gives Ir ≈ 3.67 V.
  9. IX_L = √(64 − 3.67²) V ≈ √50.6 V ≈ 7.11 V.
  10. r = 3.67 ÷ 0.060 ≈ 61 Ω; X_L = 7.11 ÷ 0.060 ≈ 119 Ω; L = X_L/2πν = 119 ÷ 314 H ≈ 0.38 H.
Answer

(a) Z = 25 Ω, I = 9.2 A, cos φ = 0.60, P ≈ 1270 W, and the current lags by about 53°; (b) r ≈ 61 Ω and L ≈ 0.38 H

How to draw the phasor diagram and the resonance curve

  • Draw the current phasor I₀ as a horizontal arrow from a point O, and draw V_R along it.
  • Draw V_L straight up and V_C straight down from O, with V_L longer when X_L > X_C.
  • Draw (V_L − V_C) upwards. Complete the rectangle with V_R to get the source voltage E₀, and mark the angle φ between E₀ and I₀.
  • Resonance curve: draw the horizontal axis for frequency ν and the vertical axis for peak current I₀.
  • Mark ν_r on the horizontal axis. Draw a curve that rises from near zero, reaches a peak at ν_r, and falls again on the other side.
  • Draw a second curve, for a larger resistance, with its peak at the same ν_r but lower. Label the taller curve 'small R' and the lower curve 'large R'.
In real life

Turning the tuning knob of an FM radio changes a capacitor until the circuit resonates with one station. Factories with many motors add capacitors to raise their power factor and cut wasted current.

Exam angle

5 marks: “Draw a phasor diagram for a series LCR circuit and obtain its impedance and phase angle. Under what condition does resonance occur? Find the resonant frequency and state the power factor at resonance.” Phasors: V_R along I, V_L up, V_C down; E₀² = V_R² + (V_L − V_C)²; Z = √(R² + (X_L − X_C)²); tan φ = (X_L − X_C)/R; resonance when X_L = X_C, ν_r = 1/2π√(LC), Z = R, the current is largest and the power factor is 1.

3 marks: “Define power factor. Show that the average power in a pure inductor is zero. What is wattless current?” cos φ in P = V_rms I_rms cos φ (= R/Z); in a pure L, φ = 90°, cos φ = 0, so P = 0; a current that uses no average power is a wattless current.

2 marks: “Find the resonant frequency of a series circuit with L = 0.10 H and C = 10 μF.” ν_r = 1/(2π√(LC)) = 1/(2π√(10⁻⁶)) = 1/(2π × 10⁻³) ≈ 159 Hz.

Remember it

At resonance X_L = X_C: Z falls to R, the current peaks, and the power factor is 1.