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​ = 690 VIL​ = 60 Acos φ = 0.85lagging

A balanced three-phase load draws a line current of 60 A from a supply with a line-to-line voltage of 690 V. The power factor is 0.85 lagging. Find the total real power P drawn by the load in kW.

VL = 690 VIL = 60 Acos φ = 0.85
P = kW
2.Three-phase powerHard
P = 75 kWφ1​S1​Q2​QC​ = ?

A balanced three-phase load takes 75 kW at a power factor of 0.72 lagging from a 400 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. Find the total reactive power QC the capacitor bank must supply, in kvar.

P = 75 kWpf1 = 0.72pf2 = 0.92VL = 400 Vf = 60 Hz
QC = kvar
3.TransformersMedium
RL​ = 8 Ω21460 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 2:1 has 460 V applied to its primary winding. An 8 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 2:1V1 = 460 VRL = 8 Ω
I1 = A
4.TransformersHard
pf 0.8500 kVA, 50% load

A single-phase 500 kVA transformer has a core loss of 1500 W and a copper loss of 4500 W at rated load. It supplies 50% of its rated kVA at a power factor of 0.8 lagging. Find the total losses Ploss in watts at this load.

S = 500 kVAPcore = 1500 WPcu,fl = 4500 Wx = 0.5pf = 0.8
Ploss = W
5.AC power & power factorMedium
P = 6 kWQ = ?Sφ

A single-phase motor takes 6 kW at a power factor of 0.8 lagging from a 230 V supply. Find the reactive power Q it draws.

P = 6 kWpf = 0.8V = 230 V
Q = kvar
6.AC power & power factorHard
10 V1000 HzRs​ = 25 ΩXs​ = j3 ΩZL​ = ?

A 10 V (rms), 1000 Hz source has an internal impedance Zs = 25 + j3 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 = 10 V (rms)f = 1000 HzZs = 25 + j3 Ω
C = µF
7.AC impedance & phasorsMedium
24 V60 HzR = 30 ΩC = 100 µFI

A resistor and a capacitor are connected in parallel across a 24 V (rms), 60 Hz supply. Find the rms current I.

V = 24 V (rms)f = 60 HzR = 30 ΩC = 100 µF
I = A
8.AC impedance & phasorsHard
50 V200 HzR1​ = 30 ΩR2​ = 60 ΩC = 47 µFI

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 rms current I.

V = 50 V (rms)f = 200 HzR1 = 30 ΩR2 = 60 ΩC = 47 µF
I = A
9.DC motorsMedium
T (N·m)n (rpm)22.5 N·mn0​nFL​

A DC motor (k = 0.75 V·s/rad, Ra = 0.5 Ω) runs from 220 V. At full load it draws 30 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.

V = 220 Vk = 0.75 V·s/radRa = 0.5 ΩIFL = 30 A
SR = %
10.DC motorsHard
+−48 VRst​ = ?Ra​ = 0.5 Ω50 AM+−E = 0n = 0T

A 48 V DC motor has an armature resistance of 0.5 Ω. Find the series starting resistance Rst that limits the starting current to 50 A.

V = 48 VRa = 0.5 ΩImax = 50 A
Rst = Ω

3Solutions

1.Three-phase power

Answer: P = 60.95 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 = √3·690·60 = 71.71 kVA

Real power

P = S·cos φ = 71.71·0.85 = 60.95 kW

Answer

P = 60.95 kW
2.Three-phase power

Answer: QC = 40.34 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 = 75·tan(acos 0.72) = 75·0.9639 = 72.29 kvar
Q2 = 75·tan(acos 0.92) = 75·0.426 = 31.95 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 40.34 kvar

Answer

QC = 40.34 kvar
3.Transformers

Answer: I1 = 14.38 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 = 460/2 = 230 V
I2 = 230/8 = 28.75 A

Primary current

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

I1 = I2/2 = 14.38 A

Answer

I1 = 14.38 A

Check: the load seen from the primary is a²·RL = 32 Ω, and 460/32 gives the same current.

4.Transformers

Answer: Ploss = 2625 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 = 0.25·4500 = 1125 W

Total losses

The core loss is the same at any load.

Ploss = 1500 + 1125 = 2625 W

Answer

Ploss = 2625 W

The efficiency at this load is η = 98.7%.

5.AC power & power factor

Answer: Q = 4.5 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(0.8) = 36.87°, tan φ = 0.75

Reactive power

Q = P·tan φ = 6·0.75 = 4.5 kvar

Answer

Q = 4.5 kvar
6.AC power & power factor

Answer: C = 53.05 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* = 25 − j3 Ω

Capacitance

XC = 3 Ω → C = 1/(2π·1000·3) = 53.05 µF

Answer

C = 53.05 µF
7.AC impedance & phasors

Answer: I = 1.208 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|.

Reactance

XC = 1/(ωC) = 1/(2π·60·100·10⁻⁶) = 26.53 Ω

Branch currents

Both branches are connected straight across the supply, so each sees the full voltage V.

IR = V/R = 24/30 = 0.8 A
IX = V/XC = 24/26.53 = 0.9048 A

Total current

The resistor current is in phase with V and the reactive current is 90° out of phase, so they add like the sides of a right triangle.

I = √(IR² + IX²) = √(0.8² + 0.9048²) = 1.208 A

Answer

I = 1.208 A
8.AC impedance & phasors

Answer: I = 1.322 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|.

Reactance

XC = 1/(2π·200·47·10⁻⁶) = 16.93 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 = 4.425 − j15.68 Ω

Total impedance

Z = 30 + 4.425 − j15.68 = 34.43 − j15.68 Ω
|Z| = √(34.43² + 15.68²) = 37.83 Ω

Current

I = V/|Z| = 50/37.83 = 1.322 A

Answer

I = 1.322 A
9.DC motors

Answer: SR = 7.317 %

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 = 60·220/(2π·0.75) = 2801 rpm
nFL = 60·(220 − 30·0.5)/(2π·0.75) = 2610 rpm

Speed regulation

SR = (2801 − 2610)/2610·100 = 7.317 %

Answer

SR = 7.317 %
10.DC motors

Answer: Rst = 0.46 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.

Total resistance

Ra + Rst = 48/50 = 0.96 Ω

Starting resistor

Rst = 0.96 − 0.5 = 0.46 Ω

Answer

Rst = 0.46 Ω

The resistor is cut out step by step as the motor speeds up and the back-EMF grows.

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