Junior Lineman / Energy Assistant — ITI Electrical Quick Book · Chapter 3

AC Fundamentals, Single and Three Phase

What to remember

Supply frequency in India is 50 Hz. For a sine wave: RMS = 0.707 × peak; average = 0.637 × peak; form factor = 1.11; peak factor = 1.414.
Power in AC: P = V I cosφ (watt); apparent power S = V I (VA); reactive power Q = V I sinφ (VAr). Power factor = cosφ = P / S.
Star: V_L = √3 V_ph, I_L = I_ph. Delta: V_L = V_ph, I_L = √3 I_ph. Three-phase power = √3 V_L I_L cosφ in both connections.

Alternating current and the sine wave

Direct current (DC) flows in one direction only. Alternating current (AC) changes its magnitude and direction regularly. AC is used for power supply because voltage can be raised or lowered easily with transformers, and it is cheaper to generate and transmit over long distances. An AC generator (alternator) produces a sine wave by rotating a coil in a magnetic field.

Terms for an alternating wave:

Cycle: one complete set of positive and negative values.
Time period (T): time for one cycle, in seconds. Frequency (f): cycles per second, in hertz. f = 1 / T. For 50 Hz, T = 0.02 s (20 ms).
Amplitude (peak value): the maximum value reached in a half cycle.
Angular frequency: ω = 2πf radian per second.
Instantaneous value: e = E_m sin(ωt).
Phase difference: the angle between two waves of the same frequency. If one wave reaches its peak before the other, it is said to lead.

Average value of a full sine cycle is zero, so the half-cycle average is used: I_av = 0.637 I_m (that is, 2/π × I_m).

RMS (effective) value is the DC value that gives the same heating effect: I_rms = I_m / √2 = 0.707 I_m. Ordinary AC meters and the rated 230 V of the household supply are RMS values.

Form factor = RMS / average = 1.11. Peak factor = peak / RMS = 1.414.

Worked example 1: A 230 V RMS supply has a peak value of 230 × 1.414 = 325 V (about).

Worked example 2: A current has a peak of 10 A. Its RMS value is 7.07 A and its half-cycle average value is 6.37 A.

R, L and C in an AC circuit

Element	Opposition	Phase of current relative to voltage
Pure resistance R	R (ohm), independent of frequency	In phase (0°)
Pure inductance L	Inductive reactance X_L = 2πfL	Current lags voltage by 90°
Pure capacitance C	Capacitive reactance X_C = 1 / (2πfC)	Current leads voltage by 90°

X_L increases with frequency; X_C decreases with frequency. A capacitor blocks DC and passes AC. An inductor passes DC easily but opposes AC. A memory aid: ELI the ICE man (voltage E before current I in an inductor L; current I before voltage E in a capacitor C).

Series R-L circuit: Impedance Z = √(R² + X_L²). Current I = V / Z. Current lags the voltage by φ, where tanφ = X_L / R and cosφ = R / Z.

Series R-C circuit: Z = √(R² + X_C²). Current leads the voltage.

Series R-L-C circuit: Z = √(R² + (X_L − X_C)²).

Worked example 3: R = 3 Ω and X_L = 4 Ω in series give Z = 5 Ω. On 100 V, I = 20 A, power factor = 3/5 = 0.6 lagging, power = 100 × 20 × 0.6 = 1200 W (also I²R = 400 × 3 = 1200 W).

Worked example 4: A 0.1 H inductor at 50 Hz has X_L = 2 × 3.14 × 50 × 0.1 = 31.4 Ω.

Resonance

In a series R-L-C circuit, resonance occurs when X_L = X_C. Then the impedance equals R only (minimum) and the current is maximum. Power factor is unity. Resonant frequency f₀ = 1 / (2π√(LC)). The quality factor Q = X_L / R = ω₀L / R measures the sharpness of resonance; voltages across L and C can be Q times the supply voltage (voltage magnification).

In a parallel resonant circuit the line current is minimum and the impedance is maximum at resonance (the branch currents are Q times the line current, so it gives current magnification; it is called a rejector circuit).

Power and power factor

Active (true) power P = V I cosφ, in watt (W). Measured by a wattmeter.
Reactive power Q = V I sinφ, in volt-ampere reactive (VAr).
Apparent power S = V I, in volt-ampere (VA). S² = P² + Q².
Power factor cosφ = P / S = R / Z. It is lagging for inductive loads (motors, transformers, chokes) and leading for capacitive loads. It is unity for a purely resistive load (heater, filament lamp).

Why a low power factor is bad: For the same power and voltage, current is I = P / (V cosφ). Lower cosφ means more current, higher copper loss, larger cable and equipment sizes, poor voltage regulation and lower capacity of generators and transformers. Supply companies often penalise low power factor.

Improvement: connect a capacitor bank in parallel with the inductive load (or a synchronous condenser in large plants). The capacitor supplies the reactive current, so the line current falls.

Worked example 5: A single-phase motor takes 4 A at 230 V with p.f. 0.8. Input power = 230 × 4 × 0.8 = 736 W. Apparent power = 920 VA.

Single-phase supply

The household supply in India is single phase, 230 V (nominal), 50 Hz, two wires: phase (live) and neutral, with a separate earth conductor. The live wire is at a voltage above earth; the neutral is at about earth potential at the source. Switches and fuses must go in the live wire. A single-phase supply is taken from one phase of a three-phase system and the neutral.

Three-phase supply

A three-phase alternator has three identical windings placed 120° apart in space. They generate three EMFs of equal magnitude and frequency, displaced by 120° in time. Phase sequence is R-Y-B (red, yellow, blue). Three-phase supply is preferred because it gives smoother power, uses less conductor material for the same power, and three-phase motors are self-starting and simple. The sum of the three instantaneous voltages is zero in a balanced system.

Quantity	Star (Y)	Delta (Δ)
Line voltage	√3 × phase voltage	= phase voltage
Line current	= phase current	√3 × phase current
Neutral available	Yes (4-wire)	No
Typical use	Distribution (415 V line, 240 V phase), motor starting	Transformer secondary/primary options, running motors

Three-phase power: P = √3 V_L I_L cosφ = 3 V_ph I_ph cosφ (watt). Apparent power S = √3 V_L I_L (VA).

Worked example 6: Line voltage is 415 V in a star system. Phase voltage = 415 / 1.732 = about 240 V.

Worked example 7: Balanced load with V_ph = 230 V, I_ph = 10 A and p.f. 0.8 takes P = 3 × 230 × 10 × 0.8 = 5520 W.

Worked example 8: Delta load with I_ph = 10 A has line current 10 × 1.732 = 17.32 A.

In a balanced three-phase four-wire star system the neutral carries no current. In an unbalanced load the neutral carries the difference current. A broken neutral in an unbalanced system can raise voltage on some phases and damage appliances.

Power measurement: The two-wattmeter method measures the total power of a three-phase three-wire load. The sum of the two readings gives the power. At unity power factor, both readings are equal; at 0.5 power factor, one reading is zero.

The colours of the three live phase conductors are red, yellow and blue, and the neutral is black. A newer colour scheme exists, so check the latest rules in force before wiring.

Exam traps

RMS is 0.707 of peak; average is 0.637 of peak. Do not swap the two values.
Inductor: current lags. Capacitor: current leads. Resistor: in phase.
X_L rises with frequency; X_C falls with frequency.
Star: line voltage is √3 times phase voltage; delta: line current is √3 times phase current.
Power factor of a heater is unity; of an induction motor it is lagging and less than one.
The power formula √3 V_L I_L cosφ uses line values; the cosφ is the angle of the phase load.
Capacitors are connected in parallel (not series) to improve power factor.
At series resonance the current is maximum; at parallel resonance the line current is minimum.

One-liners

1. India's supply frequency is 50 Hz and time period is 20 ms.
2. Form factor of a sine wave is 1.11.
3. Peak factor of a sine wave is 1.414.
4. Unit of reactive power is VAr; of apparent power, VA.
5. Impedance of a series R-L circuit is √(R² + X_L²).
6. Resonance in a series RLC circuit occurs at X_L = X_C.
7. Resonant frequency is 1 / (2π√LC).
8. Three-phase EMFs are displaced by 120°.
9. Phase sequence is R-Y-B.
10. Neutral current is zero in a balanced star load.
11. A capacitor bank improves the power factor of an inductive load.
12. Two wattmeters can measure three-phase power in a three-wire system.

Practice questions

Which of the statements is/are correct? 1. The RMS value of a sine wave is 0.707 times its peak value. 2. The average value of a half cycle of a sine wave is 0.707 times its peak value.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Half-cycle average is 0.637 of peak; 0.707 is the RMS factor.
Which of the statements is/are correct? 1. In a pure capacitive circuit the current lags the voltage by 90°. 2. In a pure inductive circuit the current lags the voltage by 90°.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
In a capacitor the current leads the voltage by 90°.
Which of the statements is/are correct? 1. Inductive reactance increases with frequency. 2. Capacitive reactance increases with frequency.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
X_C = 1/(2πfC) falls as frequency rises.
Which of the statements is/are correct? 1. At series resonance the current is minimum. 2. At series resonance the impedance of an RLC circuit is minimum.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Impedance equals R at resonance so current is maximum.
Which of the statements is/are correct? 1. In a star connection the line current equals the phase current. 2. In a delta connection the line voltage equals √3 times the phase voltage.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
In delta, line voltage equals phase voltage; line current is √3 times phase current.
Which of the statements is/are correct? 1. A balanced three-phase load always makes the neutral carry heavy current. 2. A neutral wire is available in a three-phase four-wire star system.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
In a balanced load the neutral current is zero.
Which of the statements is/are correct? 1. The power factor of a purely resistive load is unity. 2. The power factor of an induction motor is leading.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Inductive loads such as motors have a lagging power factor.
Which of the statements is/are correct? 1. A low power factor reduces copper loss in the cables. 2. A low power factor increases the line current for the same power.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
More current means more I²R loss in cables.
Which of the statements is/are correct? 1. Power factor can be improved by connecting a capacitor in parallel with an inductive load. 2. Power factor improvement requires a resistor in series with the load.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
Capacitors supply the reactive current; series resistors waste power.
Which of the statements is/are correct? 1. A three-phase alternator produces EMFs 90° apart. 2. Two phase-sequence wires are interchanged and a three-phase induction motor reverses its direction.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Three-phase EMFs are 120° apart; swapping any two lines reverses a motor.
Which of the statements is/are correct? 1. The frequency of the household supply in India is 50 Hz. 2. The time period of the household supply is 50 ms.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
A. 1 only
T = 1/50 = 0.02 s = 20 ms.
Which of the statements is/are correct? 1. Active power is measured in VAr. 2. Apparent power is measured in VA.
1. 1 only
2. 2 only
3. Both 1 and 2
4. Neither 1 nor 2
Answer
B. 2 only
Active power is in W; VAr is the unit of reactive power.
The frequency of the AC supply in India is
1. 60 Hz
2. 25 Hz
3. 100 Hz
4. 50 Hz
Answer
D. 50 Hz
Indian grid frequency is 50 Hz.
The time period of a 50 Hz supply is
1. 0.05 s
2. 0.002 s
3. 0.02 s
4. 0.2 s
Answer
C. 0.02 s
T = 1/f = 1/50 = 0.02 s.
The form factor of a sine wave is
1. 0.637
2. 1.414
3. 0.707
4. 1.11
Answer
D. 1.11
Form factor = RMS/average = 0.707/0.637 = 1.11.
The peak factor of a sine wave is
1. 0.707
2. 1.11
3. 2
4. 1.414
Answer
D. 1.414
Peak factor = peak/RMS = 1.414.
A 230 V (RMS) supply has a peak value of nearly
1. 460 V
2. 162 V
3. 325 V
4. 230 V
Answer
C. 325 V
Peak = 230 × 1.414 = 325 V.
A sinusoidal current has a peak value of 10 A. Its RMS value is
1. 6.37 A
2. 14.14 A
3. 5 A
4. 7.07 A
Answer
D. 7.07 A
I_rms = 10 × 0.707 = 7.07 A.
The inductive reactance of a 0.1 H coil at 50 Hz is nearly (take π = 3.14)
1. 314 Ω
2. 15.7 Ω
3. 31.4 Ω
4. 3.14 Ω
Answer
C. 31.4 Ω
X_L = 2πfL = 2 × 3.14 × 50 × 0.1 = 31.4 Ω.
A resistance of 3 Ω is in series with an inductive reactance of 4 Ω. The impedance is
1. 12 Ω
2. 1 Ω
3. 7 Ω
4. 5 Ω
Answer
D. 5 Ω
Z = √(9 + 16) = 5 Ω.
A series circuit of R = 3 Ω and X_L = 4 Ω works at a power factor of
1. 0.6 leading
2. 0.8 lagging
3. 0.6 lagging
4. 1.0
Answer
C. 0.6 lagging
cosφ = R/Z = 3/5 = 0.6; inductive so lagging.
A motor takes 4 A from a 230 V single-phase supply at 0.8 power factor. The active power is
1. 575 W
2. 920 W
3. 736 W
4. 1150 W
Answer
C. 736 W
P = 230 × 4 × 0.8 = 736 W.
The apparent power of a load drawing 5 A at 200 V is
1. 5000 VA
2. 200 VA
3. 40 VA
4. 1000 VA
Answer
D. 1000 VA
S = V × I = 200 × 5 = 1000 VA.
A load has 800 W active power and 600 VAr reactive power. The apparent power is
1. 200 VA
2. 700 VA
3. 1400 VA
4. 1000 VA
Answer
D. 1000 VA
S = √(800² + 600²) = 1000 VA.
The line voltage of a star-connected system is 415 V. The phase voltage is nearly
1. 415 V
2. 138 V
3. 240 V
4. 720 V
Answer
C. 240 V
V_ph = 415/1.732 = 239.6 V.
In a delta-connected load the phase current is 10 A. The line current is nearly
1. 10 A
2. 5.8 A
3. 17.3 A
4. 30 A
Answer
C. 17.3 A
I_L = √3 × I_ph = 1.732 × 10 = 17.3 A.
A balanced three-phase load has V_ph = 230 V, I_ph = 10 A and power factor 0.8. The total power is
1. 3187 W
2. 1840 W
3. 2300 W
4. 5520 W
Answer
D. 5520 W
P = 3 × 230 × 10 × 0.8 = 5520 W.
The resonant frequency of a series RLC circuit is given by
1. √(L/C)
2. 2π√(LC)
3. 1 / (2π√(LC))
4. 1 / (2π LC)
Answer
C. 1 / (2π√(LC))
At resonance X_L = X_C, giving f₀ = 1/(2π√LC).
In a series RLC circuit at resonance, the power factor is
1. zero leading
2. 0.5 leading
3. zero lagging
4. unity
Answer
D. unity
Reactances cancel, so the circuit is purely resistive.
The effective (RMS) value of AC is the DC value that produces the same
1. magnetic effect only
2. heating effect
3. chemical effect only
4. voltage drop in a capacitor
Answer
B. heating effect
RMS is defined by equal heat in the same resistor.
In a three-phase supply the three EMFs are displaced from one another by
1. 90°
2. 180°
3. 120°
4. 60°
Answer
C. 120°
Three windings are 120° apart.
The standard phase sequence of the three-phase supply is
1. Y-R-B
2. R-B-Y
3. B-R-Y
4. R-Y-B
Answer
D. R-Y-B
Red-Yellow-Blue is the standard positive sequence.
The unit of reactive power is
1. W
2. kWh
3. VAr
4. VA
Answer
C. VAr
Reactive power is measured in volt-ampere reactive.
The two-wattmeter method is used to measure
1. frequency of the supply
2. resistance of a coil
3. power factor of a DC load
4. power in a three-phase three-wire load
Answer
D. power in a three-phase three-wire load
The sum of the two readings gives total three-phase power.
At 0.5 power factor in the two-wattmeter method, one wattmeter reads
1. the same as the other
2. negative
3. twice the other
4. zero
Answer
D. zero
At p.f. 0.5 (φ = 60°) one wattmeter reading becomes zero.
The main reason for improving the power factor of an industrial load is to
1. reduce the number of phases
2. increase supply frequency
3. reduce line current for the same power
4. raise the load resistance
Answer
C. reduce line current for the same power
Higher cosφ means lower current and lower losses.
At the resonance of a parallel circuit the line current is
1. maximum
2. infinite
3. minimum
4. equal to the supply voltage
Answer
C. minimum
Parallel resonance gives maximum impedance and so minimum current.
A capacitor opposes the flow of AC with a reactance that
1. falls as frequency rises
2. is independent of frequency
3. rises with frequency
4. is zero at 50 Hz
Answer
A. falls as frequency rises
X_C = 1/(2πfC).
Which pair is correct for a three-phase delta connection?
1. Neutral point is always available
2. Line current = phase current
3. Line voltage = √3 × phase voltage
4. Line voltage = phase voltage
Answer
D. Line voltage = phase voltage
In delta the line voltage equals the phase voltage; line current is √3 times phase current.
A 10 Ω resistor and a 10 Ω inductive reactance are in series on 100 V AC. The current is nearly
1. 5 A
2. 7.07 A
3. 14.1 A
4. 10 A
Answer
B. 7.07 A
Z = √(100 + 100) = 14.14 Ω; I = 100/14.14 = 7.07 A.
An inductive coil at 50 Hz has X_L = 20 Ω. At 100 Hz its reactance becomes
1. 40 Ω
2. 10 Ω
3. 80 Ω
4. 20 Ω
Answer
A. 40 Ω
X_L is proportional to frequency, so it doubles.
A capacitor at 50 Hz has X_C = 40 Ω. At 100 Hz its reactance becomes
1. 80 Ω
2. 40 Ω
3. 20 Ω
4. 10 Ω
Answer
C. 20 Ω
X_C is inversely proportional to frequency, so it halves.
A three-phase load takes 10 A line current at 400 V line voltage with unity power factor (take √3 = 1.732). The power is nearly
1. 4 kW
2. 2.31 kW
3. 12 kW
4. 6.93 kW
Answer
D. 6.93 kW
P = 1.732 × 400 × 10 × 1 = 6928 W.
A series RLC circuit has X_L = 25 Ω, X_C = 25 Ω and R = 10 Ω. The impedance is
1. 60 Ω
2. 10 Ω
3. 25 Ω
4. 50 Ω
Answer
B. 10 Ω
At resonance reactances cancel, so Z = R = 10 Ω.
The wire (conductor) in which a single-phase switch or fuse must be connected is the
1. neutral wire
2. any of the three
3. live (phase) wire
4. earth wire
Answer
C. live (phase) wire
Switching the live wire isolates the load from the supply voltage.

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