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​ = 415 VIL​ = ?M22 kW, η = 95%cos φ = 0.82

A three-phase induction motor delivers 22 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.

Pout = 22 kWη = 95%cos φ = 0.82VL = 415 V
IL = A
2.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 80 kW at a power factor of 0.7 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 80 kWpf1 = 0.7pf2 = 0.95VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
RL​ = 5 Ω2014600 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4600 V applied to its primary winding. A 5 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 20:1V1 = 4600 VRL = 5 Ω
I1 = A
4.TransformersHard
150 kVA

A single-phase 150 kVA transformer has a core loss of 450 W and a copper loss of 1350 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 150 kVAPcore = 450 WPcu,fl = 1350 W
Sη,max = kVA
5.AC power & power factorMedium
230 VR + jXP = 368 W2 A

An inductive load takes 2 A and 368 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 230 VI = 2 AP = 368 W
R = Ω
6.AC power & power factorHard
P1​ = 16 kWQ1​S1​φ

A 50 kVA single-phase transformer supplies a load of 16 kW at a power factor of 0.65 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 = 50 kVAP1 = 16 kWpf1 = 0.65
ΔP = kW
7.AC impedance & phasorsMedium
Vf0​ = 1000 HzRL = ?C = 0.1 µF

A series RLC circuit with C = 0.1 µF must resonate at 1000 Hz. Find the inductance L needed.

f0 = 1000 HzC = 0.1 µF
L = mH
8.AC impedance & phasorsHard
12 V50 HzR = 10 ΩL = 150 mHC = 100 µFI

A series RLC circuit is connected to a 12 V (rms), 50 Hz supply. Find the rms current I.

V = 12 V (rms)f = 50 HzR = 10 ΩL = 150 mHC = 100 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)11.25 N·mn0​nFL​

A DC motor (k = 0.75 V·s/rad, Ra = 0.8 Ω) runs from 220 V. At full load it draws 15 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 220 Vk = 0.75 V·s/radRa = 0.8 ΩIFL = 15 A
SR = %
10.DC motorsHard
+−240 VRst​ = ?Ra​ = 0.25 Ω100 AM+−E = 0n = 0T

A 240 V DC motor has an armature resistance of 0.25 Ω. Find the series starting resistance Rst that limits the starting current to 100 A.

V = 240 VRa = 0.25 ΩImax = 100 A
Rst = Ω

3Solutions

1.Three-phase power

Answer: IL = 39.29 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 = 22/0.95 = 23.16 kW

Line current

IL = Pin/(√3·VL·cos φ) = 23158/(√3·415·0.82) = 39.29 A

Answer

IL = 39.29 A
2.Three-phase power

Answer: QC = 55.32 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 = 80·tan(acos 0.7) = 80·1.02 = 81.62 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 55.32 kvar

Answer

QC = 55.32 kvar
3.Transformers

Answer: I1 = 2.3 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.

Secondary voltage and current

V2 = 4600/20 = 230 V
I2 = 230/5 = 46 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/20 = 2.3 A

Answer

I1 = 2.3 A

Check: the load seen from the primary is a²·RL = 2000 Ω, and 4600/2000 gives the same current.

4.Transformers

Answer: Sη,max = 86.6 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

x²·1350 = 450 → x = √(450/1350) = 0.5774

Load at maximum efficiency

Sη,max = x·S = 0.5774·150 = 86.6 kVA

Answer

Sη,max = 86.6 kVA
5.AC power & power factor

Answer: R = 92 Ω

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.

Resistance from the real power

R = P/I² = 368/2² = 92 Ω

Answer

R = 92 Ω
6.AC power & power factor

Answer: ΔP = 30.37 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 = 16·tan(arccos 0.65) = 18.71 kvar

Largest total real power

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

P = √(Smax² − Q1²) = √(50² − 18.71²) = 46.37 kW

Extra real power

ΔP = 46.37 − 16 = 30.37 kW

Answer

ΔP = 30.37 kW
7.AC impedance & phasors

Answer: L = 253.3 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·1000 = 6283 rad/s
L = 1/(ω₀²·C) = 1/(6283²·0.1·10⁻⁶) = 253.3 mH

Answer

L = 253.3 mH
8.AC impedance & phasors

Answer: I = 0.6567 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| = 12/18.27 = 0.6567 A

Answer

I = 0.6567 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 = 60·220/(2π·0.75) = 2801 rpm
nFL = 60·(220 − 15·0.8)/(2π·0.75) = 2648 rpm

Speed regulation

SR = (2801 − 2648)/2648·100 = 5.769 %

Answer

SR = 5.769 %
10.DC motors

Answer: Rst = 2.15 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 = 240/100 = 2.4 Ω

Starting resistor

Rst = 2.4 − 0.25 = 2.15 Ω

Answer

Rst = 2.15 Ω

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

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