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 = 0.82lagging

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 0.82 lagging. Find the total real power P drawn by the load in kW.

VL = 415 VIL = 40 Acos φ = 0.82
P = kW
2.Three-phase powerHard
P = 50 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 50 kW at a power factor of 0.72 lagging from a 415 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 50 kWpf1 = 0.72pf2 = 0.92VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
4 ΩN1​N2​100 Ω

An amplifier with an output resistance of 100 Ω drives a 4 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 100 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 100 ΩRL = 4 Ω
a =
4.TransformersHard
pf 0.85250 kVA, 50% load

A single-phase 250 kVA transformer has a core loss of 2000 W and a copper loss of 4000 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.85 lagging. Find the total losses Ploss in watts at this load.

S = 250 kVAPcore = 2000 WPcu,fl = 4000 Wx = 0.5pf = 0.85
Ploss = W
5.AC power & power factorMedium
P = 27 kWQ = ?Sφ

A single-phase motor takes 27 kW at a power factor of 0.75 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 27 kWpf = 0.75V = 230 V
Q = kvar
6.AC power & power factorHard
240 V60 Hzload5 kW, 0.65C150 µFI

A single-phase load takes 5 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A 150 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 5 kWpf1 = 0.65V = 240 Vf = 60 HzC = 150 µF
cos φ =
7.AC impedance & phasorsMedium
50 V100 HzR = 15 ΩL = 150 mHC = 100 µF

A series RLC circuit is connected to a 50 V (rms), 100 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 50 V (rms)f = 100 HzR = 15 ΩL = 150 mHC = 100 µF
φ = °
8.AC impedance & phasorsHard
24 V200 HzR = 60 ΩL = 200 mHC = 47 µF

A series RLC circuit is connected to a 24 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.

V = 24 V (rms)f = 200 HzR = 60 ΩL = 200 mHC = 47 µF
f0 = Hz
9.DC motorsMedium
T (N·m)n (rpm)24 N·mn0​nFL​

A DC motor (k = 2 V·s/rad, Ra = 2 Ω) runs from 240 V. At full load it draws 12 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 240 Vk = 2 V·s/radRa = 2 ΩIFL = 12 A
SR = %
10.DC motorsHard
+−220 VRs​ = ?Ra​ = 0.8 ΩM+−En = 1000 rpmT = 9 N·m

A DC motor (k = 0.75 V·s/rad, Ra = 0.8 Ω) on a fixed 220 V supply drives a load of 9 N·m. Find the series resistance Rs that slows it down to 1000 rpm.

V = 220 Vk = 0.75 V·s/radRa = 0.8 ΩT = 9 N·mn = 1000 rpm
Rs = Ω

3Solutions

1.Three-phase power

Answer: P = 23.58 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.

Apparent power

S = √3·VL·IL = √3·415·40 = 28.75 kVA

Real power

P = S·cos φ = 28.75·0.82 = 23.58 kW

Answer

P = 23.58 kW
2.Three-phase power

Answer: Cph = 138.1 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 = 50·tan(acos 0.72) = 50·0.9639 = 48.19 kvar
Q2 = 50·tan(acos 0.92) = 50·0.426 = 21.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 26.89 kvar

Capacitance per phase

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

V = 415 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 8964/(377·415²) = 138.1 µF

Answer

Cph = 138.1 µF
3.Transformers

Answer: a = 5

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.

Match the referred load

a²·4 = 100
a = √(100/4) = √25 = 5

Answer

a = 5
4.Transformers

Answer: Ploss = 3000 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 = 0.25·4000 = 1000 W

Total losses

The core loss is the same at any load.

Ploss = 2000 + 1000 = 3000 W

Answer

Ploss = 3000 W

The efficiency at this load is η = 97.25%.

5.AC power & power factor

Answer: Q = 23.81 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(0.75) = 41.41°, tan φ = 0.8819

Reactive power

Q = P·tan φ = 27·0.8819 = 23.81 kvar

Answer

Q = 23.81 kvar
6.AC power & power factor

Answer: cos φ = 0.8881

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 before and after

The real power P does not change; only the reactive power drawn from the supply does.

Q1 = 5·tan(arccos 0.65) = 5.846 kvar
QC = V²·2πf·C = 240²·2π·60·150·10⁻⁶ = 3257 var = 3.257 kvar
Q2 = 5.846 − 3.257 = 2.588 kvar

New apparent power

S2 = √(5² + 2.588²) = 5.63 kVA

New power factor

cos φ = P/S2 = 5/5.63 = 0.8881

Answer

cos φ = 0.8881
7.AC impedance & phasors

Answer: φ = 79.16 °

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·100 = 628.3 rad/s
XL = ωL = 628.3·0.15 = 94.25 Ω
XC = 1/(ωC) = 1/(628.3·100·10⁻⁶) = 15.92 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (78.33)²) = 79.76 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(78.33/15) = 79.16°

Answer

φ = 79.16°
8.AC impedance & phasors

Answer: f0 = 51.91 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|.

Resonance

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.2·47·10⁻⁶)) = 51.91 Hz

Answer

f0 = 51.91 Hz
9.DC motors

Answer: SR = 11.11 %

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 = 60·240/(2π·2) = 1146 rpm
nFL = 60·(240 − 12·2)/(2π·2) = 1031 rpm

Speed regulation

SR = (1146 − 1031)/1031·100 = 11.11 %

Answer

SR = 11.11 %
10.DC motors

Answer: Rs = 10.99 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.

Current and back-EMF

Ia = 9/0.75 = 12 A
E = 0.75·104.7 = 78.54 V

Series resistance

Ra + Rs = (220 − 78.54)/12 = 11.79 Ω
Rs = 10.99 Ω

Answer

Rs = 10.99 Ω

The resistor wastes 1582 W; a chopper controls the speed with far less loss.

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