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​ = ?M15 kW, η = 88%cos φ = 0.84

A three-phase induction motor delivers 15 kW of mechanical output power at full load. At this load its efficiency is 88% and its power factor is 0.84 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 15 kWη = 88%cos φ = 0.84VL = 480 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 0.7 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 = 200 kWpf1 = 0.7pf2 = 0.95VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
2400 V240 VIHV​ILV​ = ?15 kVA

A single-phase 15 kVA, 2400/240 V transformer operates at its rated load. Treat it as ideal. Find the current ILV in the low-voltage winding.

S = 15 kVAV1 = 2400 VV2 = 240 V
ILV = A
4.TransformersHard
pf 0.950 kVA, η = 98.2%

A single-phase 50 kVA transformer has a core loss of 150 W. At rated load and a power factor of 0.9 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 50 kVAPcore = 150 Wpf = 0.9η = 98.2%
Pcu,fl = W
5.AC power & power factorMedium
P = 21 kWQ = ?Sφ

A single-phase motor takes 21 kW at a power factor of 0.65 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 21 kWpf = 0.65V = 230 V
Q = kvar
6.AC power & power factorHard
50 V60 HzRs​ = 5 ΩXs​ = j3 ΩZL​ = ?

A 50 V (rms), 60 Hz source has an internal impedance Zs = 5 + j3 Ω. The load impedance ZL can be chosen freely. The load for maximum power is built from a resistor in series with a capacitor. Find the capacitance C.

Vs = 50 V (rms)f = 60 HzZs = 5 + j3 Ω
C = µF
7.AC impedance & phasorsMedium
120 V50 HzR = 40 ΩL = 100 mHC = 22 µF

A series RLC circuit is connected to a 120 V (rms), 50 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 120 V (rms)f = 50 HzR = 40 ΩL = 100 mHC = 22 µF
φ = °
8.AC impedance & phasorsHard
12 V500 HzR1​ = 10 ΩR2​ = 60 ΩC = 4.7 µFI

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 12 V (rms), 500 Hz supply. Find the rms current I.

V = 12 V (rms)f = 500 HzR1 = 10 ΩR2 = 60 ΩC = 4.7 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)0.5 N·mn0​n = ?

A DC motor (k = 0.1 N·m/A, Ra = 0.4 Ω) runs from 24 V and drives a load torque of 0.5 N·m. Find its speed n in rpm.

V = 24 Vk = 0.1 N·m/ARa = 0.4 ΩT = 0.5 N·m
n = rpm
10.DC motorsHard
+−240 VRst​ = ?Ra​ = 0.4 Ω30 AM+−E = 0n = 0T

A 240 V DC motor has an armature resistance of 0.4 Ω. Find the series starting resistance Rst that limits the starting current to 30 A.

V = 240 VRa = 0.4 ΩImax = 30 A
Rst = Ω

3Solutions

1.Three-phase power

Answer: IL = 24.41 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 = 15/0.88 = 17.05 kW

Line current

IL = Pin/(√3·VL·cos φ) = 17045/(√3·480·0.84) = 24.41 A

Answer

IL = 24.41 A
2.Three-phase power

Answer: QC = 138.3 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 = 200·tan(acos 0.7) = 200·1.02 = 204 kvar
Q2 = 200·tan(acos 0.95) = 200·0.3287 = 65.74 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 138.3 kvar

Answer

QC = 138.3 kvar
3.Transformers

Answer: ILV = 62.5 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 = 15000/240 = 62.5 A

Answer

ILV = 62.5 A
4.Transformers

Answer: Pcu,fl = 674.8 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 = 50·0.9 = 45 kW
Pin = 45000/0.982 = 45825 W

Total and copper losses

Ploss = 45825 − 45000 = 824.8 W
Pcu,fl = 824.8 − 150 = 674.8 W

Answer

Pcu,fl = 674.8 W
5.AC power & power factor

Answer: Q = 24.55 kvar

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.

Angle

φ = arccos(0.65) = 49.46°, tan φ = 1.169

Reactive power

Q = P·tan φ = 21·1.169 = 24.55 kvar

Answer

Q = 24.55 kvar
6.AC power & power factor

Answer: C = 884.2 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.

Conjugate match

The load reactance cancels the source reactance, and the load resistance equals the source resistance.

ZL = Zs* = 5 − j3 Ω

Capacitance

XC = 3 Ω → C = 1/(2π·60·3) = 884.2 µF

Answer

C = 884.2 µF
7.AC impedance & phasors

Answer: φ = −70.55 °

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 = 314.2·0.1 = 31.42 Ω
XC = 1/(ωC) = 1/(314.2·22·10⁻⁶) = 144.7 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(40² + (−113.3)²) = 120.1 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−113.3/40) = −70.55°

Answer

φ = −70.55°
8.AC impedance & phasors

Answer: I = 0.2272 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|.

Reactance

XC = 1/(2π·500·4.7·10⁻⁶) = 67.73 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 = 33.62 − j29.78 Ω

Total impedance

Z = 10 + 33.62 − j29.78 = 43.62 − j29.78 Ω
|Z| = √(43.62² + 29.78²) = 52.81 Ω

Current

I = V/|Z| = 12/52.81 = 0.2272 A

Answer

I = 0.2272 A
9.DC motors

Answer: n = 2101 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.

Armature current

Ia = T/k = 0.5/0.1 = 5 A

Back-EMF and speed

E = 24 − 5·0.4 = 22 V
ω = E/k = 220 rad/s → n = 2101 rpm

Answer

n = 2101 rpm
10.DC motors

Answer: Rst = 7.6 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 = 240/30 = 8 Ω

Starting resistor

Rst = 8 − 0.4 = 7.6 Ω

Answer

Rst = 7.6 Ω

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

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