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​ = 80 Acos φ = 0.82lagging

A balanced three-phase load draws a line current of 80 A from a supply with a line-to-line voltage of 480 V. The power factor is 0.82 lagging. Find the total real power P drawn by the load in kW.

VL = 480 VIL = 80 Acos φ = 0.82
P = kW
2.Three-phase powerHard
P = 120 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 120 kW at a power factor of 0.65 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 = 120 kWpf1 = 0.65pf2 = 0.95VL = 400 Vf = 50 Hz
QC = kvar
3.TransformersMedium
50 ΩN1​N2​7200 Ω

An amplifier with an output resistance of 7200 Ω drives a 50 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 7200 Ω from the amplifier. Find the turns ratio a = N₁/N₂.

Rs = 7200 ΩRL = 50 Ω
a =
4.TransformersHard
pf 0.95150 kVA, 75% load

A single-phase 150 kVA transformer has a core loss of 450 W and a copper loss of 900 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 = 150 kVAPcore = 450 WPcu,fl = 900 Wx = 0.75pf = 0.95
Ploss = W
5.AC power & power factorMedium
400 V122 kW, 0.8211 kW, 1I = ?

Two loads are connected in parallel to a 400 V single-phase supply: a 22 kW load at a power factor of 0.8 lagging and an 11 kW heater (unity power factor). Find the total current I drawn from the supply.

V = 400 VP1 = 22 kW, pf1 = 0.8 laggingP2 = 11 kW, pf2 = 1
I = A
6.AC power & power factorHard
230 V60 Hzload12 kW, 0.8C150 µFI

A single-phase load takes 12 kW at a power factor of 0.8 lagging from a 230 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 = 12 kWpf1 = 0.8V = 230 Vf = 60 HzC = 150 µF
cos φ =
7.AC impedance & phasorsMedium
230 V1000 HzR = 50 ΩL = 50 mHC = 22 µFI

A series RLC circuit is connected to a 230 V (rms), 1000 Hz supply. Find the rms current I.

V = 230 V (rms)f = 1000 HzR = 50 ΩL = 50 mHC = 22 µF
I = A
8.AC impedance & phasorsHard
50 V200 HzR1​ = 50 ΩR2​ = 40 ΩC = 22 µF

A resistor R1 is connected in series with a parallel combination of R2 and a capacitor C, across a 50 V (rms), 200 Hz supply. Find the magnitude of the impedance |Z|.

V = 50 V (rms)f = 200 HzR1 = 50 ΩR2 = 40 ΩC = 22 µF
|Z| = Ω
9.DC motorsMedium
+−120 VRa​ = 0.5 ΩIa​ = 10 AM+−EωT

A DC motor draws 10 A from a 120 V supply. Its armature resistance is 0.5 Ω and its rotational (friction, windage and iron) losses are 58 W. Find its efficiency η.

V = 120 VIa = 10 ARa = 0.5 ΩProt = 58 W
η = %
10.DC motorsHard
+−120 VRa​ = 1.5 ΩIa​: 8 → 4 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 1.5 Ω) runs at 1500 rpm with an armature current of 8 A. The load changes and the current becomes 4 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 1.5 Ωn1 = 1500 rpmIa1 = 8 AIa2 = 4 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: P = 54.54 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 = √3·480·80 = 66.51 kVA

Real power

P = S·cos φ = 66.51·0.82 = 54.54 kW

Answer

P = 54.54 kW
2.Three-phase power

Answer: QC = 100.9 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 = 120·tan(acos 0.65) = 120·1.169 = 140.3 kvar
Q2 = 120·tan(acos 0.95) = 120·0.3287 = 39.44 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 100.9 kvar

Answer

QC = 100.9 kvar
3.Transformers

Answer: a = 12

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

a²·50 = 7200
a = √(7200/50) = √144 = 12

Answer

a = 12
4.Transformers

Answer: Ploss = 956.3 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 = 0.5625·900 = 506.3 W

Total losses

The core loss is the same at any load.

Ploss = 450 + 506.3 = 956.3 W

Answer

Ploss = 956.3 W

The efficiency at this load is η = 99.11%.

5.AC power & power factor

Answer: I = 92.24 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.

Reactive power of each load

A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).

Q1 = 22·tan(arccos 0.8) = 16.5 kvar
Q2 = 0 = 0 kvar

Totals

P = 22 + 11 = 33 kW
Q = 16.5 + 0 = 16.5 kvar
S = √(33² + 16.5²) = 36.9 kVA

Total current

I = S/V = 36895/400 = 92.24 A

Answer

I = 92.24 A
6.AC power & power factor

Answer: cos φ = 0.8942

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 = 12·tan(arccos 0.8) = 9 kvar
QC = V²·2πf·C = 230²·2π·60·150·10⁻⁶ = 2991 var = 2.991 kvar
Q2 = 9 − 2.991 = 6.009 kvar

New apparent power

S2 = √(12² + 6.009²) = 13.42 kVA

New power factor

cos φ = P/S2 = 12/13.42 = 0.8942

Answer

cos φ = 0.8942
7.AC impedance & phasors

Answer: I = 0.7396 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·1000 = 6283 rad/s
XL = ωL = 6283·0.05 = 314.2 Ω
XC = 1/(ωC) = 1/(6283·22·10⁻⁶) = 7.234 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(50² + (306.9)²) = 311 Ω

Current

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

I = V/|Z| = 230/311 = 0.7396 A

Answer

I = 0.7396 A
8.AC impedance & phasors

Answer: |Z| = 70.85 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·200·22·10⁻⁶) = 36.17 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 = 17.99 − j19.9 Ω

Total impedance

Z = 50 + 17.99 − j19.9 = 67.99 − j19.9 Ω
|Z| = √(67.99² + 19.9²) = 70.85 Ω

Answer

|Z| = 70.85 Ω
9.DC motors

Answer: η = 91 %

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 = 120·10 = 1200 W
Pdev = 115·10 = 1150 W
Pout = 1150 − 58 = 1092 W

Efficiency

η = 1092/1200·100 = 91 %

Answer

η = 91 %
10.DC motors

Answer: n2 = 1583 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.

Back-EMF before and after

E1 = 120 − 8·1.5 = 108 V
E2 = 120 − 4·1.5 = 114 V

New speed

n2 = 1500·114/108 = 1583 rpm

Answer

n2 = 1583 rpm

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