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​ = 40 Acos φ = 0.8lagging

A balanced three-phase load draws a line current of 40 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.8 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 40 Acos φ = 0.8
P = kW
2.Three-phase powerHard
P = 200 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 200 kW at a power factor of 0.65 lagging from a 480 V (line-to-line), 60 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 = 200 kWpf1 = 0.65pf2 = 0.9VL = 480 Vf = 60 Hz
QC = kvar
3.TransformersMedium
2 Ω31

A 2 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 3:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 3:1RL = 2 Ω
R1′ = Ω
4.TransformersHard
pf 0.8150 kVA, 25% load

A single-phase 150 kVA transformer has a core loss of 750 W and a copper loss of 3000 W at rated load. It supplies 25% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.

S = 150 kVAPcore = 750 WPcu,fl = 3000 Wx = 0.25pf = 0.8
Ploss = W
5.AC power & power factorMedium
P = 8 kWQSφ

A single-phase motor takes 8 kW at a power factor of 0.75 lagging from a 400 V supply. Find the line current I.

P = 8 kWpf = 0.75V = 400 V
I = A
6.AC power & power factorHard
P1​ = 14 kWQ1​S1​φ

A 25 kVA single-phase transformer supplies a load of 14 kW at a power factor of 0.75 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 = 25 kVAP1 = 14 kWpf1 = 0.75
ΔP = kW
7.AC impedance & phasorsMedium
120 V200 HzR = 60 ΩL = 200 mHC = 22 µFI

A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the real power P taken from the supply.

V = 120 V (rms)f = 200 HzR = 60 ΩL = 200 mHC = 22 µF
P = W
8.AC impedance & phasorsHard
VfR = 1 ΩL = 100 mHC = 10 µF

A series RLC circuit has R = 1 Ω, L = 100 mH and C = 10 µF. Find its bandwidth B (the width of the band between the half-power frequencies), in Hz.

R = 1 ΩL = 100 mHC = 10 µF
B = Hz
9.DC motorsMedium
+−VRa​M+−En = 750 rpmT = ?

A DC motor delivers 7.5 kW of mechanical power at 750 rpm. Find the shaft torque T.

P = 7.5 kWn = 750 rpm
T = N·m
10.DC motorsHard
+−240 VD = ?Ra​ = 0.25 ΩM+−En = 400 rpmT = 80 N·m

A DC chopper feeds a motor (k = 2 V·s/rad, Ra = 0.25 Ω) from a 240 V DC bus. The average armature voltage is D·240 V. Find the duty cycle D that runs the motor at 400 rpm with a load of 80 N·m.

Vdc = 240 Vk = 2 V·s/radRa = 0.25 ΩT = 80 N·mn = 400 rpm
D = %

3Solutions

1.Three-phase power

Answer: P = 23 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 = √3·415·40 = 28.75 kVA

Real power

P = S·cos φ = 28.75·0.8 = 23 kW

Answer

P = 23 kW
2.Three-phase power

Answer: QC = 137 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 0.65) = 200·1.169 = 233.8 kvar
Q2 = 200·tan(acos 0.9) = 200·0.4843 = 96.86 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 137 kvar

Answer

QC = 137 kvar
3.Transformers

Answer: R1′ = 18 Ω

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.

Referred resistance

The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².

R1′ = a²·RL = 3²·2 = 18 Ω

Answer

R1′ = 18 Ω
4.Transformers

Answer: Ploss = 937.5 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 = 0.0625·3000 = 187.5 W

Total losses

The core loss is the same at any load.

Ploss = 750 + 187.5 = 937.5 W

Answer

Ploss = 937.5 W

The efficiency at this load is η = 96.97%.

5.AC power & power factor

Answer: I = 26.67 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) = 8000/(400·0.75) = 26.67 A

Answer

I = 26.67 A
6.AC power & power factor

Answer: ΔP = 7.738 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 = 14·tan(arccos 0.75) = 12.35 kvar

Largest total real power

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

P = √(Smax² − Q1²) = √(25² − 12.35²) = 21.74 kW

Extra real power

ΔP = 21.74 − 14 = 7.738 kW

Answer

ΔP = 7.738 kW
7.AC impedance & phasors

Answer: P = 17.32 W

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

XL = 251.3 Ω, XC = 36.17 Ω

Impedance

|Z| = √(60² + (215.2)²) = 223.4 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 120/223.4 = 0.5372 A
P = I²·R = 0.5372²·60 = 17.32 W

Answer

P = 17.32 W
8.AC impedance & phasors

Answer: B = 1.592 Hz

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|.

Resonant frequency

f0 = 1/(2π·√(0.1·10·10⁻⁶)) = 159.2 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/1)·√(0.1/(10·10⁻⁶)) = 100

Bandwidth

B = f0/Q = 159.2/100 = 1.592 Hz

Answer

B = 1.592 Hz
9.DC motors

Answer: T = 95.49 N·m

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.

Speed in rad/s

ω = 2π·750/60 = 78.54 rad/s

Torque

T = P/ω = 7500/78.54 = 95.49 N·m

Answer

T = 95.49 N·m
10.DC motors

Answer: D = 39.07 %

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.

Required armature voltage

Va = 2·41.89 + 40·0.25 = 93.78 V

Duty cycle

D = 93.78/240 = 39.07 %

Answer

D = 39.07 %

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