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 VNZ = 12 + j16 Ωper phase

A balanced Y-connected load with an impedance of 12 + j16 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.

VL = 208 VZ = 12 + j16 Ω per phase
P = kW
2.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 80 kW at a power factor of 0.72 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 80 kWpf1 = 0.72pf2 = 0.95VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
11000 V400 VIHV​ILV​ = ?150 kVA

A single-phase 150 kVA, 11000/400 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 150 kVAV1 = 11000 VV2 = 400 V
ILV = A
4.TransformersHard
pf 0.95250 kVA, η = 99%

A single-phase 250 kVA transformer has a core loss of 1000 W. At rated load and a power factor of 0.95 lagging its efficiency is 99%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 1000 Wpf = 0.95η = 99%
Pcu,fl = W
5.AC power & power factorMedium
240 V127 kW, 0.8210 kW, 1I = ?

Two loads are connected in parallel to a 240 V single-phase supply: a 27 kW load at a power factor of 0.8 lagging and a 10 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 240 VP1 = 27 kW, pf1 = 0.8 laggingP2 = 10 kW, pf2 = 1
I = A
6.AC power & power factorHard
12 V1000 HzRs​ = 10 ΩXs​ = j10 ΩZL​ = ?

A 12 V (rms), 1000 Hz source has an internal impedance Zs = 10 + j10 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 12 V (rms)f = 1000 HzZs = 10 + j10 Ω
Pmax = W
7.AC impedance & phasorsMedium
120 V200 HzR = 60 ΩL = 150 mHI

A resistor and an inductor are connected in parallel across a 120 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.

V = 120 V (rms)f = 200 HzR = 60 ΩL = 150 mH
|Z| = Ω
8.AC impedance & phasorsHard
10 Vf0​R = 5 ΩL = 10 mHC = 0.47 µF

A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 10 V (rms)R = 5 ΩL = 10 mHC = 0.47 µF
VL = V
9.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 1.2 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 120 Vk = 1.2 V·s/rad
n0 = rpm
10.DC motorsHard
+−24 VD = ?Ra​ = 1.2 ΩM+−En = 300 rpmT = 0.24 N·m

A DC chopper feeds a motor (k = 0.12 V·s/rad, Ra = 1.2 Ω) from a 24 V DC bus. The average armature voltage is D·24 V. Find the duty cycle D that runs the motor at 300 rpm with a load of 0.24 N·m.

Vdc = 24 Vk = 0.12 V·s/radRa = 1.2 ΩT = 0.24 N·mn = 300 rpm
D = %

3Solutions

1.Three-phase power

Answer: P = 1.298 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.

Phase voltage

Vph = 208/√3 = 120.1 V

Phase current

|Z| = √(12² + 16²) = 20 Ω
Iph = 120.1/20 = 6.004 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·6.004²·12 = 1.298 kW

Answer

P = 1.298 kW

Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: Cph = 585 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 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 50.81 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 480/√3 = 277.1 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 16938/(377·277.1²) = 585 µF

Answer

Cph = 585 µF
3.Transformers

Answer: ILV = 375 A

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.

Rated current

Both windings carry the same apparent power.

ILV = S/V = 150000/400 = 375 A

Answer

ILV = 375 A
4.Transformers

Answer: Pcu,fl = 1399 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.

Output and input power

Pout = 250·0.95 = 237.5 kW
Pin = 237500/0.99 = 239899 W

Total and copper losses

Ploss = 239899 − 237500 = 2399 W
Pcu,fl = 2399 − 1000 = 1399 W

Answer

Pcu,fl = 1399 W
5.AC power & power factor

Answer: I = 175.7 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 of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 27·tan(arccos 0.8) = 20.25 kvar
Q2 = 0 = 0 kvar

Totals

P = 27 + 10 = 37 kW
Q = 20.25 + 0 = 20.25 kvar
S = √(37² + 20.25²) = 42.18 kVA

Total current

I = S/V = 42179/240 = 175.7 A

Answer

I = 175.7 A
6.AC power & power factor

Answer: Pmax = 3.6 W

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* = 10 − j10 Ω

Maximum power

I = Vs/(2Rs) = 12/20 = 0.6 A
Pmax = I²·RL = Vs²/(4Rs) = 12²/(4·10) = 3.6 W

Answer

Pmax = 3.6 W
7.AC impedance & phasors

Answer: |Z| = 57.17 Ω

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|.

Reactance

XL = ωL = 2π·200·0.15 = 188.5 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 120/60 = 2 A
IX = V/XL = 120/188.5 = 0.6366 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(2² + 0.6366²) = 2.099 A

Impedance

|Z| = V/I = 120/2.099 = 57.17 Ω

Answer

|Z| = 57.17 Ω
8.AC impedance & phasors

Answer: VL = 291.7 V

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|.

Resonance

At resonance the reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.01·0.47·10⁻⁶) = 14586 rad/s
I = V/R = 10/5 = 2 A

Voltage across the inductor

XL = ω₀L = 14586·0.01 = 145.9 Ω
VL = I·XL = 2·145.9 = 291.7 V

Answer

VL = 291.7 V

This is Q = 29.17 times the supply voltage.

9.DC motors

Answer: n0 = 954.9 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.

No-load speed

ω₀ = V/k = 120/1.2 = 100 rad/s
n0 = 60·100/(2π) = 954.9 rpm

Answer

n0 = 954.9 rpm
10.DC motors

Answer: D = 25.71 %

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.

Required armature voltage

Va = 0.12·31.42 + 2·1.2 = 6.17 V

Duty cycle

D = 6.17/24 = 25.71 %

Answer

D = 25.71 %

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