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​ = 690 VIL​ = ?M55 kW, η = 93%cos φ = 0.85

A three-phase induction motor delivers 55 kW of mechanical output power at full load. At this load its efficiency is 93% and its power factor is 0.85 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 55 kWη = 93%cos φ = 0.85VL = 690 V
IL = A
2.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 200 kW at a power factor of 0.8 lagging from a 480 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.98 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 200 kWpf1 = 0.8pf2 = 0.98VL = 480 Vf = 50 Hz
Cph = µF
3.TransformersMedium
6600 V230 VIHV​ = ?ILV​25 kVA

A single-phase 25 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 25 kVAV1 = 6600 VV2 = 230 V
IHV = A
4.TransformersHard
pf 0.9575 kVA, 75% load

A single-phase 75 kVA transformer has a core loss of 450 W and a copper loss of 1800 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.95 lagging. Find the total losses Ploss in watts at this load.

S = 75 kVAPcore = 450 WPcu,fl = 1800 Wx = 0.75pf = 0.95
Ploss = W
5.AC power & power factorMedium
P = 13 kWQSφ

A single-phase motor takes 13 kW at a power factor of 0.9 lagging from a 240 V supply. Find the line current I.

P = 13 kWpf = 0.9V = 240 V
I = A
6.AC power & power factorHard
120 V50 HzRs​ = 25 ΩXs​ = j6 ΩZL​ = ?

A 120 V (rms), 50 Hz source has an internal impedance Zs = 25 + j6 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.

Vs = 120 V (rms)f = 50 HzZs = 25 + j6 Ω
Pmax = W
7.AC impedance & phasorsMedium
12 V500 HzR = 60 ΩL = 200 mHC = 1 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 = 60 ΩL = 200 mHC = 1 µF
I = A
8.AC impedance & phasorsHard
24 V50 HzR1​ = 20 ΩR2​ = 60 ΩC = 100 µF

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 24 V (rms), 50 Hz supply. Find the magnitude of the impedance |Z|.

V = 24 V (rms)f = 50 HzR1 = 20 ΩR2 = 60 ΩC = 100 µF
|Z| = Ω
9.DC motorsMedium
+−240 VRa​ = 1.5 ΩIa​ = 5 AM+−EωT

A DC motor draws 5 A from a 240 V supply. Its armature resistance is 1.5 Ω and its rotational (friction, windage and iron) losses are 58 W. Find its efficiency η.

V = 240 VIa = 5 ARa = 1.5 ΩProt = 58 W
η = %
10.DC motorsHard
+−24 VRa​ = 0.5 ΩM+−EωT

A small DC motor has an armature resistance of 0.5 Ω and runs from 24 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 24 VRa = 0.5 Ω
Pmax = W

3Solutions

1.Three-phase power

Answer: IL = 58.22 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 = 55/0.93 = 59.14 kW

Line current

IL = Pin/(√3·VL·cos φ) = 59140/(√3·690·0.85) = 58.22 A

Answer

IL = 58.22 A
2.Three-phase power

Answer: Cph = 503.8 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 = 200·tan(acos 0.8) = 200·0.75 = 150 kvar
Q2 = 200·tan(acos 0.98) = 200·0.2031 = 40.61 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 109.4 kvar

Capacitance per phase

In a Δ bank each capacitor is connected between two lines.

V = 480 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 36463/(314.2·480²) = 503.8 µF

Answer

Cph = 503.8 µF
3.Transformers

Answer: IHV = 3.788 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 = 25000/6600 = 3.788 A

Answer

IHV = 3.788 A
4.Transformers

Answer: Ploss = 1463 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 = 0.5625·1800 = 1013 W

Total losses

The core loss is the same at any load.

Ploss = 450 + 1013 = 1463 W

Answer

Ploss = 1463 W

The efficiency at this load is η = 97.34%.

5.AC power & power factor

Answer: I = 60.19 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) = 13000/(240·0.9) = 60.19 A

Answer

I = 60.19 A
6.AC power & power factor

Answer: Pmax = 144 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* = 25 − j6 Ω

Maximum power

I = Vs/(2Rs) = 120/50 = 2.4 A
Pmax = I²·RL = Vs²/(4Rs) = 120²/(4·25) = 144 W

Answer

Pmax = 144 W
7.AC impedance & phasors

Answer: I = 0.038 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 = 3142·0.2 = 628.3 Ω
XC = 1/(ωC) = 1/(3142·1·10⁻⁶) = 318.3 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(60² + (310)²) = 315.8 Ω

Current

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

I = V/|Z| = 12/315.8 = 0.038 A

Answer

I = 0.038 A
8.AC impedance & phasors

Answer: |Z| = 41.45 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

XC = 1/(2π·50·100·10⁻⁶) = 31.83 Ω

Parallel combination

Multiply top and bottom by the complex conjugate to split Zp into real and imaginary parts.

Zp = R₂·(−jXC)/(R₂ − jXC) = (R₂·XC² − jR₂²·XC)/(R₂² + XC²)
Zp = 13.18 − j24.84 Ω

Total impedance

Z = 20 + 13.18 − j24.84 = 33.18 − j24.84 Ω
|Z| = √(33.18² + 24.84²) = 41.45 Ω

Answer

|Z| = 41.45 Ω
9.DC motors

Answer: η = 92.04 %

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 = 240·5 = 1200 W
Pdev = 232.5·5 = 1163 W
Pout = 1163 − 58 = 1105 W

Efficiency

η = 1105/1200·100 = 92.04 %

Answer

η = 92.04 %
10.DC motors

Answer: Pmax = 288 W

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.

Power as a function of the back-EMF

P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 12 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 24²/(4·0.5) = 288 W

Answer

Pmax = 288 W

At this point the efficiency is only 50 %, so real motors run well above half speed.

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