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 VNZ = 7 + j24 Ωper phase

A balanced Y-connected load with an impedance of 7 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 480 V. Find the total real power P absorbed by the load in kW.

VL = 480 VZ = 7 + j24 Ω per phase
P = kW
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 75 kW at a power factor of 0.8 lagging from a 690 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 = 75 kWpf1 = 0.8pf2 = 0.9VL = 690 Vf = 50 Hz
QC = kvar
3.TransformersMedium
4800 V240 VIHV​ = ?ILV​250 kVA

A single-phase 250 kVA, 4800/240 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 250 kVAV1 = 4800 VV2 = 240 V
IHV = A
4.TransformersHard
pf 0.9500 kVA, 50% 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 50% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.

S = 500 kVAPcore = 3000 WPcu,fl = 6000 Wx = 0.5pf = 0.9
Ploss = W
5.AC power & power factorMedium
120 VZ16 + j12 ΩI

A load with impedance Z = 16 + j12 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.

V = 120 VZ = 16 + j12 Ω
Q = var
6.AC power & power factorHard
P1​ = 83 kWQ1​S1​φ

A 150 kVA single-phase transformer supplies a load of 83 kW at a power factor of 0.85 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?

Smax = 150 kVAP1 = 83 kWpf1 = 0.85
ΔP = kW
7.AC impedance & phasorsMedium
Vf0​ = 50 HzRL = ?C = 2.2 µF

A series RLC circuit with C = 2.2 µF must resonate at 50 Hz. Find the inductance L needed.

f0 = 50 HzC = 2.2 µF
L = mH
8.AC impedance & phasorsHard
12 V60 HzR = 50 ΩL = 250 mHC = 47 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 = 50 ΩL = 250 mHC = 47 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)3 N·mn0​nFL​

A DC motor (k = 0.6 V·s/rad, Ra = 1.2 Ω) runs from 110 V. At full load it draws 5 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 110 Vk = 0.6 V·s/radRa = 1.2 ΩIFL = 5 A
SR = %
10.DC motorsHard
T (N·m)n (rpm)6 N·m500 rpm

A DC motor with k = 1.5 V·s/rad and Ra = 2 Ω must drive a load of 6 N·m at 500 rpm. Find the armature voltage V it needs.

k = 1.5 V·s/radRa = 2 ΩT = 6 N·mn = 500 rpm
V = V

3Solutions

1.Three-phase power

Answer: P = 2.58 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 = 480/√3 = 277.1 V

Phase current

|Z| = √(7² + 24²) = 25 Ω
Iph = 277.1/25 = 11.09 A

Total real power

Only the resistive part of each impedance absorbs real power.

P = 3·Iph²·R = 3·11.09²·7 = 2.58 kW

Answer

P = 2.58 kW

Check: cos φ = R/|Z| = 0.28, and √3·VL·IL·cos φ gives the same power.

2.Three-phase power

Answer: QC = 19.93 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 = 75·tan(acos 0.8) = 75·0.75 = 56.25 kvar
Q2 = 75·tan(acos 0.9) = 75·0.4843 = 36.32 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 19.93 kvar

Answer

QC = 19.93 kvar
3.Transformers

Answer: IHV = 52.08 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.

IHV = S/V = 250000/4800 = 52.08 A

Answer

IHV = 52.08 A
4.Transformers

Answer: Ploss = 4500 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 = 0.25·6000 = 1500 W

Total losses

The core loss is the same at any load.

Ploss = 3000 + 1500 = 4500 W

Answer

Ploss = 4500 W

The efficiency at this load is η = 98.04%.

5.AC power & power factor

Answer: Q = 432 var

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.

Current

|Z| = √(16² + 12²) = 20 Ω
I = V/|Z| = 120/20 = 6 A

Reactive power

Q = I²·X = 6²·12 = 432 var

Answer

Q = 432 var
6.AC power & power factor

Answer: ΔP = 57.9 kW

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 the existing load

Q1 = 83·tan(arccos 0.85) = 51.44 kvar

Largest total real power

At full rating S = Smax, and Q is unchanged by the resistive load.

P = √(Smax² − Q1²) = √(150² − 51.44²) = 140.9 kW

Extra real power

ΔP = 140.9 − 83 = 57.9 kW

Answer

ΔP = 57.9 kW
7.AC impedance & phasors

Answer: L = 4606 mH

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·50 = 314.2 rad/s
L = 1/(ω₀²·C) = 1/(314.2²·2.2·10⁻⁶) = 4606 mH

Answer

L = 4606 mH
8.AC impedance & phasors

Answer: I = 0.1914 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/(377·47·10⁻⁶) = 56.44 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(50² + (37.81)²) = 62.69 Ω

Current

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

I = V/|Z| = 12/62.69 = 0.1914 A

Answer

I = 0.1914 A
9.DC motors

Answer: SR = 5.769 %

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 = 60·110/(2π·0.6) = 1751 rpm
nFL = 60·(110 − 5·1.2)/(2π·0.6) = 1655 rpm

Speed regulation

SR = (1751 − 1655)/1655·100 = 5.769 %

Answer

SR = 5.769 %
10.DC motors

Answer: V = 86.54 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 = 1.5·52.36 = 78.54 V

Armature current

Ia = 6/1.5 = 4 A

Armature voltage

V = 78.54 + 4·2 = 86.54 V

Answer

V = 86.54 V

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