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​ = 480 VIL​ = 150 Acos φ = 0.82lagging

A balanced three-phase load draws a line current of 150 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.

VL = 480 VIL = 150 Acos φ = 0.82
P = kW
2.Three-phase powerHard
P = 150 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 150 kW at a power factor of 0.75 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 0.92 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 150 kWpf1 = 0.75pf2 = 0.92VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
6600 V230 VIHV​ILV​ = ?500 kVA

A single-phase 500 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 500 kVAV1 = 6600 VV2 = 230 V
ILV = A
4.TransformersHard
pf 0.8250 kVA, η = 98.2%

A single-phase 250 kVA transformer has a core loss of 750 W. At rated load and a power factor of 0.8 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 250 kVAPcore = 750 Wpf = 0.8η = 98.2%
Pcu,fl = W
5.AC power & power factorMedium
400 V121 kW, 0.65216 kW, 0.95I

Two loads are connected in parallel to a 400 V single-phase supply: a 21 kW load at a power factor of 0.65 lagging and a 16 kW load at a power factor of 0.95 lagging. Find the overall power factor cos φ seen by the supply.

V = 400 VP1 = 21 kW, pf1 = 0.65 laggingP2 = 16 kW, pf2 = 0.95 lagging
cos φ =
6.AC power & power factorHard
P = 10 kWQ1​S1​φ1​QC​

A single-phase load takes 10 kW at a power factor of 0.8 lagging from a 230 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.

P = 10 kWpf1 = 0.8pf2 = 0.98V = 230 Vf = 50 Hz
C = µF
7.AC impedance & phasorsMedium
24 V60 HzR = 100 ΩL = 20 mHC = 47 µFI

A series RLC circuit is connected to a 24 V (rms), 60 Hz supply. Find the rms current I.

V = 24 V (rms)f = 60 HzR = 100 ΩL = 20 mHC = 47 µF
I = A
8.AC impedance & phasorsHard
12 V500 HzR = 80 ΩL = 50 mHC = 22 µFI

A series RLC circuit is connected to a 12 V (rms), 500 Hz supply. Find the rms current I.

V = 12 V (rms)f = 500 HzR = 80 ΩL = 50 mHC = 22 µF
I = A
9.DC motorsMedium
+−48 VRa​ = 0.8 ΩIa​ = 5 AM+−EωT

A DC motor draws 5 A from a 48 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 13 W. Find its efficiency η.

V = 48 VIa = 5 ARa = 0.8 ΩProt = 13 W
η = %
10.DC motorsHard
+−240 VRst​ = ?Ra​ = 0.5 Ω20 AM+−E = 0n = 0T

A 240 V DC motor has an armature resistance of 0.5 Ω. Find the series starting resistance Rst that limits the starting current to 20 A.

V = 240 VRa = 0.5 ΩImax = 20 A
Rst = Ω

3Solutions

1.Three-phase power

Answer: P = 102.3 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 = √3·480·150 = 124.7 kVA

Real power

P = S·cos φ = 124.7·0.82 = 102.3 kW

Answer

P = 102.3 kW
2.Three-phase power

Answer: QC = 68.39 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 = 150·tan(acos 0.75) = 150·0.8819 = 132.3 kvar
Q2 = 150·tan(acos 0.92) = 150·0.426 = 63.9 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 68.39 kvar

Answer

QC = 68.39 kvar
3.Transformers

Answer: ILV = 2174 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 = 500000/230 = 2174 A

Answer

ILV = 2174 A
4.Transformers

Answer: Pcu,fl = 2916 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 = 250·0.8 = 200 kW
Pin = 200000/0.982 = 203666 W

Total and copper losses

Ploss = 203666 − 200000 = 3666 W
Pcu,fl = 3666 − 750 = 2916 W

Answer

Pcu,fl = 2916 W
5.AC power & power factor

Answer: cos φ = 0.7787

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 = 21·tan(arccos 0.65) = 24.55 kvar
Q2 = 16·tan(arccos 0.95) = 5.259 kvar

Totals

P = 21 + 16 = 37 kW
Q = 24.55 + 5.259 = 29.81 kvar
S = √(37² + 29.81²) = 47.52 kVA

Overall power factor

cos φ = P/S = 37/47.52 = 0.7787 lagging

Answer

cos φ = 0.7787
6.AC power & power factor

Answer: C = 329.1 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.

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 = 10·tan(36.87°) = 7.5 kvar
Q2 = P·tan φ2 = 10·tan(11.48°) = 2.031 kvar
QC = Q1 − Q2 = 5.469 kvar

Capacitance

A capacitor across the supply delivers Q = V²·ωC.

C = QC/(2πf·V²) = 5469/(2π·50·230²) = 329.1 µF

Answer

C = 329.1 µF
7.AC impedance & phasors

Answer: I = 0.2156 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 = 377·0.02 = 7.54 Ω
XC = 1/(ωC) = 1/(377·47·10⁻⁶) = 56.44 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(100² + (−48.9)²) = 111.3 Ω

Current

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

I = V/|Z| = 24/111.3 = 0.2156 A

Answer

I = 0.2156 A
8.AC impedance & phasors

Answer: I = 0.07339 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·500 = 3142 rad/s
XL = ωL = 3142·0.05 = 157.1 Ω
XC = 1/(ωC) = 1/(3142·22·10⁻⁶) = 14.47 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(80² + (142.6)²) = 163.5 Ω

Current

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

I = V/|Z| = 12/163.5 = 0.07339 A

Answer

I = 0.07339 A
9.DC motors

Answer: η = 86.25 %

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.

Powers

Pin = 48·5 = 240 W
Pdev = 44·5 = 220 W
Pout = 220 − 13 = 207 W

Efficiency

η = 207/240·100 = 86.25 %

Answer

η = 86.25 %
10.DC motors

Answer: Rst = 11.5 Ω

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.

Total resistance

Ra + Rst = 240/20 = 12 Ω

Starting resistor

Rst = 12 − 0.5 = 11.5 Ω

Answer

Rst = 11.5 Ω

The resistor is cut out step by step as the motor speeds up and the back-EMF grows.

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