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​ = ?M110 kW, η = 91%cos φ = 0.86

A three-phase induction motor delivers 110 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.

Pout = 110 kWη = 91%cos φ = 0.86VL = 480 V
IL = A
2.Three-phase powerHard
P = 80 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 80 kW at a power factor of 0.72 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 = 80 kWpf1 = 0.72pf2 = 0.98VL = 480 Vf = 50 Hz
Cph = µF
3.TransformersMedium
2 Ω81

A 2 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 8:1. Find the resistance R1′ seen looking into the primary winding.

N1:N2 = 8:1RL = 2 Ω
R1′ = Ω
4.TransformersHard
pf 0.8250 kVA, 75% load

A single-phase 250 kVA transformer has a core loss of 1000 W and a copper loss of 2500 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.

S = 250 kVAPcore = 1000 WPcu,fl = 2500 Wx = 0.75pf = 0.8
Ploss = W
5.AC power & power factorMedium
P = 4 kWQSφ

A single-phase motor takes 4 kW at a power factor of 0.6 lagging from a 230 V supply. Find the line current I.

P = 4 kWpf = 0.6V = 230 V
I = A
6.AC power & power factorHard
400 V60 Hzload21 kW, 0.65C100 µFI = ?

A single-phase load takes 21 kW at a power factor of 0.65 lagging from a 400 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new line current I.

P = 21 kWpf1 = 0.65V = 400 Vf = 60 HzC = 100 µF
I = A
7.AC impedance & phasorsMedium
Vf0​ = 500 HzRL = 20 mHC = ?

A series RLC circuit with L = 20 mH must resonate at 500 Hz. Find the capacitance C needed.

f0 = 500 HzL = 20 mH
C = µF
8.AC impedance & phasorsHard
1 Vf0​R = 50 ΩL = 10 mHC = 0.1 µF

A series RLC circuit is driven by a 1 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 1 V (rms)R = 50 ΩL = 10 mHC = 0.1 µF
VL = V
9.DC motorsMedium
T (N·m)n (rpm)1.5 N·mn0​n = ?

A DC motor (k = 0.25 N·m/A, Ra = 0.2 Ω) runs from 24 V and drives a load torque of 1.5 N·m. Find its speed n in rpm.

V = 24 Vk = 0.25 N·m/ARa = 0.2 ΩT = 1.5 N·m
n = rpm
10.DC motorsHard
T (N·m)n (rpm)1.44 N·m750 rpm

A DC motor with k = 0.12 V·s/rad and Ra = 0.4 Ω must drive a load of 1.44 N·m at 750 rpm. Find the armature voltage V it needs.

k = 0.12 V·s/radRa = 0.4 ΩT = 1.44 N·mn = 750 rpm
V = V

3Solutions

1.Three-phase power

Answer: IL = 169.1 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 = 110/0.91 = 120.9 kW

Line current

IL = Pin/(√3·VL·cos φ) = 120879/(√3·480·0.86) = 169.1 A

Answer

IL = 169.1 A
2.Three-phase power

Answer: Cph = 280.3 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 = 80·tan(acos 0.72) = 80·0.9639 = 77.11 kvar
Q2 = 80·tan(acos 0.98) = 80·0.2031 = 16.24 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 60.86 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) = 20288/(314.2·480²) = 280.3 µF

Answer

Cph = 280.3 µF
3.Transformers

Answer: R1′ = 128 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 = 8²·2 = 128 Ω

Answer

R1′ = 128 Ω
4.Transformers

Answer: Ploss = 2406 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 = 0.5625·2500 = 1406 W

Total losses

The core loss is the same at any load.

Ploss = 1000 + 1406 = 2406 W

Answer

Ploss = 2406 W

The efficiency at this load is η = 98.42%.

5.AC power & power factor

Answer: I = 28.99 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) = 4000/(230·0.6) = 28.99 A

Answer

I = 28.99 A
6.AC power & power factor

Answer: I = 70 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 before and after

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

Q1 = 21·tan(arccos 0.65) = 24.55 kvar
QC = V²·2πf·C = 400²·2π·60·100·10⁻⁶ = 6032 var = 6.032 kvar
Q2 = 24.55 − 6.032 = 18.52 kvar

New apparent power

S2 = √(21² + 18.52²) = 28 kVA

New line current

I = S2/V = 28000/400 = 70 A

Answer

I = 70 A
7.AC impedance & phasors

Answer: C = 5.066 µF

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 condition

At resonance the inductive and capacitive reactances cancel: ωL = 1/(ωC).

ω₀ = 2π·500 = 3142 rad/s
C = 1/(ω₀²·L) = 1/(3142²·0.02) = 5.066 µF

Answer

C = 5.066 µF
8.AC impedance & phasors

Answer: VL = 6.325 V

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 reactances cancel, so the impedance is just R.

ω₀ = 1/√(LC) = 1/√(0.01·0.1·10⁻⁶) = 31623 rad/s
I = V/R = 1/50 = 0.02 A

Voltage across the inductor

XL = ω₀L = 31623·0.01 = 316.2 Ω
VL = I·XL = 0.02·316.2 = 6.325 V

Answer

VL = 6.325 V

This is Q = 6.325 times the supply voltage.

9.DC motors

Answer: n = 870.9 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 = 1.5/0.25 = 6 A

Back-EMF and speed

E = 24 − 6·0.2 = 22.8 V
ω = E/k = 91.2 rad/s → n = 870.9 rpm

Answer

n = 870.9 rpm
10.DC motors

Answer: V = 14.22 V

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

E = 0.12·78.54 = 9.425 V

Armature current

Ia = 1.44/0.12 = 12 A

Armature voltage

V = 9.425 + 12·0.4 = 14.22 V

Answer

V = 14.22 V

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