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​ = 400 VIL​ = ?M45 kW, η = 92%cos φ = 0.86

A three-phase induction motor delivers 45 kW of mechanical output power at full load. At this load its efficiency is 92% and its power factor is 0.86 lagging. The supply has a line-to-line voltage of 400 V. Find the line current IL.

Pout = 45 kWη = 92%cos φ = 0.86VL = 400 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 0.65 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 0.95 lagging. The real power stays the same. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 80 kWpf1 = 0.65pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar
3.TransformersMedium
RL​ = 6 Ω21480 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 2:1 has 480 V applied to its primary winding. A 6 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 2:1V1 = 480 VRL = 6 Ω
I1 = A
4.TransformersHard
200 kVA

A single-phase 200 kVA transformer has a core loss of 1000 W and a copper loss of 1500 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?

S = 200 kVAPcore = 1000 WPcu,fl = 1500 W
Sη,max = kVA
5.AC power & power factorMedium
P = 6 kWQ = ?Sφ

A single-phase motor takes 6 kW at a power factor of 0.9 lagging from a 240 V supply. Find the reactive power Q it draws.

P = 6 kWpf = 0.9V = 240 V
Q = kvar
6.AC power & power factorHard
120 V60 HzRs​ = 8 ΩXs​ = j40 ΩZL​ = ?

A 120 V (rms), 60 Hz source has an internal impedance Zs = 8 + j40 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 = 120 V (rms)f = 60 HzZs = 8 + j40 Ω
C = µF
7.AC impedance & phasorsMedium
Vf0​ = 500 HzRL = ?C = 4.7 µF

A series RLC circuit with C = 4.7 µF must resonate at 500 Hz. Find the inductance L needed.

f0 = 500 HzC = 4.7 µF
L = mH
8.AC impedance & phasorsHard
10 Vf0​R = 10 ΩL = 10 mHC = 0.47 µF

A series RLC circuit is driven by a 10 V (rms) source whose frequency is set to the resonant frequency of the circuit. Find the rms voltage VL across the inductor.

V = 10 V (rms)R = 10 ΩL = 10 mHC = 0.47 µF
VL = V
9.DC motorsMedium
+−VRa​M+−En = 1000 rpmT = ?

A DC motor delivers 0.5 kW of mechanical power at 1000 rpm. Find the shaft torque T.

P = 0.5 kWn = 1000 rpm
T = N·m
10.DC motorsHard
+−120 VRa​ = 1.5 ΩIa​: 6 → 12 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 1.5 Ω) runs at 1440 rpm with an armature current of 6 A. The load changes and the current becomes 12 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 1.5 Ωn1 = 1440 rpmIa1 = 6 AIa2 = 12 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: IL = 82.09 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 = 45/0.92 = 48.91 kW

Line current

IL = Pin/(√3·VL·cos φ) = 48913/(√3·400·0.86) = 82.09 A

Answer

IL = 82.09 A
2.Three-phase power

Answer: QC = 67.24 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 = 80·tan(acos 0.65) = 80·1.169 = 93.53 kvar
Q2 = 80·tan(acos 0.95) = 80·0.3287 = 26.29 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 67.24 kvar

Answer

QC = 67.24 kvar
3.Transformers

Answer: I1 = 20 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.

Secondary voltage and current

V2 = 480/2 = 240 V
I2 = 240/6 = 40 A

Primary current

An ideal transformer has no losses: V₁·I₁ = V₂·I₂.

I1 = I2/2 = 20 A

Answer

I1 = 20 A

Check: the load seen from the primary is a²·RL = 24 Ω, and 480/24 gives the same current.

4.Transformers

Answer: Sη,max = 163.3 kVA

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 equals core loss

x²·1500 = 1000 → x = √(1000/1500) = 0.8165

Load at maximum efficiency

Sη,max = x·S = 0.8165·200 = 163.3 kVA

Answer

Sη,max = 163.3 kVA
5.AC power & power factor

Answer: Q = 2.906 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(0.9) = 25.84°, tan φ = 0.4843

Reactive power

Q = P·tan φ = 6·0.4843 = 2.906 kvar

Answer

Q = 2.906 kvar
6.AC power & power factor

Answer: C = 66.31 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* = 8 − j40 Ω

Capacitance

XC = 40 Ω → C = 1/(2π·60·40) = 66.31 µF

Answer

C = 66.31 µF
7.AC impedance & phasors

Answer: L = 21.56 mH

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
L = 1/(ω₀²·C) = 1/(3142²·4.7·10⁻⁶) = 21.56 mH

Answer

L = 21.56 mH
8.AC impedance & phasors

Answer: VL = 145.9 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/√(0.01·0.47·10⁻⁶) = 14586 rad/s
I = V/R = 10/10 = 1 A

Voltage across the inductor

XL = ω₀L = 14586·0.01 = 145.9 Ω
VL = I·XL = 1·145.9 = 145.9 V

Answer

VL = 145.9 V

This is Q = 14.59 times the supply voltage.

9.DC motors

Answer: T = 4.775 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·1000/60 = 104.7 rad/s

Torque

T = P/ω = 500/104.7 = 4.775 N·m

Answer

T = 4.775 N·m
10.DC motors

Answer: n2 = 1323 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 − 6·1.5 = 111 V
E2 = 120 − 12·1.5 = 102 V

New speed

n2 = 1440·102/111 = 1323 rpm

Answer

n2 = 1323 rpm

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