Power Systems and Electrical Machines: cram pack

Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.

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1Formulas and methods

Three-phase power

Y connection
VL = √3·Vph, IL = Iph
Δ connection
VL = Vph, IL = √3·Iph
Total real power
P = √3·VL·IL·cos φ = 3·Iph²·R
Apparent and reactive power
S = √3·VL·IL, Q = P·tan φ
Correction capacitors
QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
Voltage across each capacitor
Δ: VL Y: VL/√3
Motor input
Pin = Pout/η

Transformers

Turns ratio
a = N1/N2 = V1/V2
Current ratio
I1/I2 = N2/N1 = 1/a
Rated current
I = S/V
Referred impedance
Z1′ = a²·ZL
Copper loss at part load
Pcu = x²·Pcu,fl
Efficiency
η = x·S·pf/(x·S·pf + Pcore + x²·Pcu,fl)
Maximum efficiency
x = √(Pcore/Pcu,fl)

AC power & power factor

Real power
P = V·I·cos φ
Reactive power
Q = V·I·sin φ
Apparent power
S = V·I = √(P² + Q²)
Power factor
pf = cos φ = P/S
Correction capacitor
QC = P(tan φ1 − tan φ2), C = QC/(ωV²)

AC impedance & phasors

Impedances
ZR = R, ZL = jωL, ZC = 1/(jωC)
Angular frequency
ω = 2πf
Magnitude and angle
|Z| = √(R² + X²), φ = atan(X/R)
Phasor Ohm’s law
V = I·Z

Method

  1. Convert the frequency: ω = 2πf.
  2. Find each reactance: XL = ωL and XC = 1/(ωC).
  3. Add impedances like resistors, with Z = R + jX.
  4. Find the magnitude |Z| = √(R² + X²) and the angle φ = atan(X/R), then use Ohm’s law with phasors.

DC motors

Armature circuit
V = E + IaRa
Back-EMF and torque
E = k·ω, T = k·Ia
Speed
ω = 2π·n/60, ω = (V − IaRa)/k
No-load speed and stall torque
ω0 = V/k, Tstall = k·V/Ra
Powers
Pin = V·Ia, Pdev = E·Ia = T·ω, Pout = Pdev − Prot
Starting
Istart = V/(Ra + Rst)
Speed regulation
SR = (n0 − nFL)/nFL
Maximum developed power
Pmax = V²/(4Ra)
Chopper drive
Va = D·Vdc

Method

  1. Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
  2. Use V = E + Ia·Ra to link voltage, current and back-EMF.
  3. Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
  4. For power and efficiency, start from V·Ia, subtract the copper loss Ia²·Ra and then the rotational losses.

2Practice problems

1.Three-phase powerMedium
ABCVL​ = 690 VIL​ = ?M7.5 kW, η = 88%cos φ = 0.8

A three-phase induction motor delivers 7.5 kW of mechanical output power at full load. At this load its efficiency is 88% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 7.5 kWη = 88%cos φ = 0.8VL = 690 V
IL = A
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 75 kW at a power factor of 0.7 lagging from a 400 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 75 kWpf1 = 0.7pf2 = 0.98VL = 400 Vf = 60 Hz
Cph = µF
3.TransformersMedium
16 ΩN1​N2​1600 Ω

An amplifier with an output resistance of 1600 Ω drives a 16 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 1600 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 1600 ΩRL = 16 Ω
a =
4.TransformersHard
pf 0.8500 kVA, 75% load

A single-phase 500 kVA transformer has a core loss of 3000 W and a copper loss of 6000 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.

S = 500 kVAPcore = 3000 WPcu,fl = 6000 Wx = 0.75pf = 0.8
Ploss = W
5.AC power & power factorMedium
P = 14 kWQSφ

A single-phase motor takes 14 kW at a power factor of 0.85 lagging from a 230 V supply. Find the line current I.

P = 14 kWpf = 0.85V = 230 V
I = A
6.AC power & power factorHard
P = 40 kWQ1​S1​φ1​QC​

A single-phase load takes 40 kW at a power factor of 0.65 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the new line current Inew.

P = 40 kWpf1 = 0.65pf2 = 0.98V = 230 Vf = 60 Hz
Inew = A
7.AC impedance & phasorsMedium
230 V200 HzR = 60 ΩL = 150 mHC = 100 µFI

A series RLC circuit is connected to a 230 V (rms), 200 Hz supply. Find the real power P taken from the supply.

V = 230 V (rms)f = 200 HzR = 60 ΩL = 150 mHC = 100 µF
P = W
8.AC impedance & phasorsHard
VfR = 50 ΩL = 100 mHC = 0.01 µF

A series RLC circuit has R = 50 Ω, L = 100 mH and C = 0.01 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 50 ΩL = 100 mHC = 0.01 µF
B = Hz
9.DC motorsMedium
T (N·m)n (rpm)80 N·mn0​nFL​

A DC motor (k = 2 V·s/rad, Ra = 0.2 Ω) runs from 220 V. At full load it draws 40 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 220 Vk = 2 V·s/radRa = 0.2 ΩIFL = 40 A
SR = %
10.DC motorsHard
+−240 VRa​ = 0.2 ΩM+−En = ?T = 48 N·m

A separately excited DC motor (k = 1.2 V·s/rad at full field, Ra = 0.2 Ω, V = 240 V) drives a constant load torque of 48 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.

V = 240 Vk = 1.2 V·s/radRa = 0.2 ΩT = 48 N·mΦ = 85 %
n = rpm

3Solutions

1.Three-phase power

Answer: IL = 8.914 A

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ.

Electrical input power

Pin = Pout/η = 7.5/0.88 = 8.523 kW

Line current

IL = Pin/(√3·VL·cos φ) = 8523/(√3·690·0.8) = 8.914 A

Answer

IL = 8.914 A
2.Three-phase power

Answer: Cph = 338.7 µF

Balanced three-phase systems: in a Y (star) connection the line-to-line voltage is √3 times the phase voltage, VL = √3·Vph, and the line current equals the phase current. In a Δ (delta) connection the line-to-line voltage equals the phase voltage and the line current is √3 times the phase current, IL = √3·Iph. The total power is three times the power of one phase: P = 3·Vph·Iph·cos φ = √3·VL·IL·cos φ. Power factor correction: capacitors supply reactive power locally, QC = P·(tan φ1 − tan φ2). Each of the three capacitors supplies QC/3 = V²·ω·C, where V is the voltage across that capacitor: VL in a Δ bank, VL/√3 in a Y bank.

Reactive power before and after

The real power is unchanged, so each power factor fixes the reactive power: Q = P·tan φ.

Q1 = 75·tan(acos 0.7) = 75·1.02 = 76.52 kvar
Q2 = 75·tan(acos 0.98) = 75·0.2031 = 15.23 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 61.29 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 400 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 20429/(377·400²) = 338.7 µF

Answer

Cph = 338.7 µF
3.Transformers

Answer: a = 10

Ideal transformer: the voltages are in the ratio of the turns and the currents in the inverse ratio: V₁/V₂ = N₁/N₂ = a and I₁/I₂ = N₂/N₁ = 1/a, so the power in equals the power out, V₁·I₁ = V₂·I₂. A load impedance Z on the secondary looks like a²·Z from the primary.

Match the referred load

a²·16 = 1600
a = √(1600/16) = √100 = 10

Answer

a = 10
4.Transformers

Answer: Ploss = 6375 W

Transformer losses: the core loss (hysteresis and eddy currents) depends on the voltage and is the same at any load. The copper loss in the windings goes with the current squared, so at a fraction x of rated load it is x²·Pcu,fl. Efficiency η = Pout/(Pout + Pcore + Pcu), with Pout = x·S·pf. The efficiency is highest at the load where the copper loss equals the core loss.

Copper loss at this load

Pcu = x²·Pcu,fl = 0.5625·6000 = 3375 W

Total losses

The core loss is the same at any load.

Ploss = 3000 + 3375 = 6375 W

Answer

Ploss = 6375 W

The efficiency at this load is η = 97.92%.

5.AC power & power factor

Answer: I = 71.61 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Line current

Only the real part of the power does work, so divide by the power factor.

I = P/(V·cos φ) = 14000/(230·0.85) = 71.61 A

Answer

I = 71.61 A
6.AC power & power factor

Answer: Inew = 177.5 A

AC power: apparent power S = V·I (VA) splits into real power P = S·cos φ (W), which does useful work, and reactive power Q = S·sin φ (var), which moves back and forth between source and load. The power factor is cos φ. A capacitor in parallel supplies reactive power locally, so the line current drops for the same real power.

Reactive power before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = P·tan φ1 = 40·tan(49.46°) = 46.77 kvar
Q2 = P·tan φ2 = 40·tan(11.48°) = 8.122 kvar
QC = Q1 − Q2 = 38.64 kvar

New line current

Inew = P/(V·cos φ2) = 40000/(230·0.98) = 177.5 A

Answer

Inew = 177.5 A
7.AC impedance & phasors

Answer: P = 87.69 W

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Reactances

XL = 188.5 Ω, XC = 7.958 Ω

Impedance

|Z| = √(60² + (180.5)²) = 190.2 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 230/190.2 = 1.209 A
P = I²·R = 1.209²·60 = 87.69 W

Answer

P = 87.69 W
8.AC impedance & phasors

Answer: B = 79.58 Hz

AC circuits with phasors: resistors, inductors and capacitors have impedances R, jωL and 1/(jωC), with ω = 2πf. In series, impedances add like resistances, but as complex numbers. Rms current and voltage follow Ohm’s law with |Z|.

Resonant frequency

f0 = 1/(2π·√(0.1·0.01·10⁻⁶)) = 5033 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/50)·√(0.1/(0.01·10⁻⁶)) = 63.25

Bandwidth

B = f0/Q = 5033/63.25 = 79.58 Hz

Answer

B = 79.58 Hz
9.DC motors

Answer: SR = 3.774 %

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

No-load and full-load speed

n₀ = 60·220/(2π·2) = 1050 rpm
nFL = 60·(220 − 40·0.2)/(2π·2) = 1012 rpm

Speed regulation

SR = (1050 − 1012)/1012·100 = 3.774 %

Answer

SR = 3.774 %
10.DC motors

Answer: n = 2159 rpm

DC motor: the armature is a resistance Ra in series with the back-EMF E, so V = E + Ia·Ra. The back-EMF grows with speed, E = k·ω, and the torque with current, T = k·Ia, with the same motor constant k (V·s/rad = N·m/A) in SI units. Convert speeds with ω = 2π·n/60. The developed power is E·Ia = T·ω; the input power is V·Ia, and the copper loss Ia²Ra and the rotational loss make the difference. The speed falls in a straight line with torque, ω = V/k − Ra·T/k², from the no-load speed V/k to zero at the stall torque k·V/Ra. At standstill E = 0, so the starting current V/Ra is large unless a resistor limits it.

Weakened motor constant

k₂ = 0.85·1.2 = 1.02 V·s/rad

New armature current

Ia = 48/1.02 = 47.06 A

New speed

ω = (240 − 47.06·0.2)/1.02 = 226.1 rad/s → n = 2159 rpm

Answer

n = 2159 rpm

Field weakening is how a DC drive runs above its base speed.

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