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​ = 208 VIL​ = 150 Acos φ = 0.92lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 208 V. The power factor is 0.92 lagging. Find the total real power P drawn by the load in kW.

VL = 208 VIL = 150 Acos φ = 0.92
P = kW
2.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 200 kW at a power factor of 0.65 lagging from a 400 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 200 kWpf1 = 0.65pf2 = 0.9VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
50 ΩN1​N2​7200 Ω

An amplifier with an output resistance of 7200 Ω drives a 50 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 7200 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 7200 ΩRL = 50 Ω
a =
4.TransformersHard
200 kVA

A single-phase 200 kVA transformer has a core loss of 1600 W and a copper loss of 2400 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 200 kVAPcore = 1600 WPcu,fl = 2400 W
Sη,max = kVA
5.AC power & power factorMedium
P = 11 kWQSφ

A single-phase motor takes 11 kW at a power factor of 0.85 lagging from a 240 V supply. Find the line current I.

P = 11 kWpf = 0.85V = 240 V
I = A
6.AC power & power factorHard
24 V50 HzRs​ = 2 ΩXs​ = j15 ΩZL​ = ?

A 24 V (rms), 50 Hz source has an internal impedance Zs = 2 + j15 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.

Vs = 24 V (rms)f = 50 HzZs = 2 + j15 Ω
C = µF
7.AC impedance & phasorsMedium
12 V60 HzR = 40 ΩL = 250 mHC = 4.7 µFI

A series RLC circuit is connected to a 12 V (rms), 60 Hz supply. Find the rms current I.

V = 12 V (rms)f = 60 HzR = 40 ΩL = 250 mHC = 4.7 µF
I = A
8.AC impedance & phasorsHard
24 V50 HzR = 10 ΩL = 150 mHC = 100 µFI

A series RLC circuit is connected to a 24 V (rms), 50 Hz supply. Find the rms current I.

V = 24 V (rms)f = 50 HzR = 10 ΩL = 150 mHC = 100 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)Tst​ = ?n0​

A DC motor with k = 1 N·m/A and Ra = 0.2 Ω is connected to 240 V. Find its stall torque Tstall, the torque at zero speed.

V = 240 Vk = 1 N·m/ARa = 0.2 Ω
Tstall = N·m
10.DC motorsHard
T (N·m)n (rpm)2.4 N·m750 rpm

A DC motor with k = 0.3 V·s/rad and Ra = 1.5 Ω must drive a load of 2.4 N·m at 750 rpm. Find the armature voltage V it needs.

k = 0.3 V·s/radRa = 1.5 ΩT = 2.4 N·mn = 750 rpm
V = V

3Solutions

1.Three-phase power

Answer: P = 49.72 kW

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.

Apparent power

S = √3·VL·IL = √3·208·150 = 54.04 kVA

Real power

P = S·cos φ = 54.04·0.92 = 49.72 kW

Answer

P = 49.72 kW
2.Three-phase power

Answer: QC = 137 kvar

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 = 200·tan(acos 0.65) = 200·1.169 = 233.8 kvar
Q2 = 200·tan(acos 0.9) = 200·0.4843 = 96.86 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 137 kvar

Answer

QC = 137 kvar
3.Transformers

Answer: a = 12

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

a²·50 = 7200
a = √(7200/50) = √144 = 12

Answer

a = 12
4.Transformers

Answer: Sη,max = 163.3 kVA

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 equals core loss

x²·2400 = 1600 → x = √(1600/2400) = 0.8165

Load at maximum efficiency

Sη,max = x·S = 0.8165·200 = 163.3 kVA

Answer

Sη,max = 163.3 kVA
5.AC power & power factor

Answer: I = 53.92 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) = 11000/(240·0.85) = 53.92 A

Answer

I = 53.92 A
6.AC power & power factor

Answer: C = 212.2 µF

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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 2 − j15 Ω

Capacitance

XC = 15 Ω → C = 1/(2π·50·15) = 212.2 µF

Answer

C = 212.2 µF
7.AC impedance & phasors

Answer: I = 0.02543 A

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

ω = 2πf = 2π·60 = 377 rad/s
XL = ωL = 377·0.25 = 94.25 Ω
XC = 1/(ωC) = 1/(377·4.7·10⁻⁶) = 564.4 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(40² + (−470.1)²) = 471.8 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 12/471.8 = 0.02543 A

Answer

I = 0.02543 A
8.AC impedance & phasors

Answer: I = 1.313 A

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

ω = 2πf = 2π·50 = 314.2 rad/s
XL = ωL = 314.2·0.15 = 47.12 Ω
XC = 1/(ωC) = 1/(314.2·100·10⁻⁶) = 31.83 Ω

Impedance

In series the impedances add as complex numbers: Z = R + j(XL − XC).

|Z| = √(R² + (XL − XC)²) = √(10² + (15.29)²) = 18.27 Ω

Current

The same current flows through every element of a series circuit.

I = V/|Z| = 24/18.27 = 1.313 A

Answer

I = 1.313 A
9.DC motors

Answer: Tstall = 1200 N·m

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.

Stall current

Ia = V/Ra = 240/0.2 = 1200 A

Stall torque

Tstall = 1·1200 = 1200 N·m

Answer

Tstall = 1200 N·m
10.DC motors

Answer: V = 35.56 V

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.

Back-EMF

E = 0.3·78.54 = 23.56 V

Armature current

Ia = 2.4/0.3 = 8 A

Armature voltage

V = 23.56 + 8·1.5 = 35.56 V

Answer

V = 35.56 V

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