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​ = ?M37 kW, η = 91%cos φ = 0.8

A three-phase induction motor delivers 37 kW of mechanical output power at full load. At this load its efficiency is 91% and its power factor is 0.8 lagging. The supply has a line-to-line voltage of 690 V. Find the line current IL.

Pout = 37 kWη = 91%cos φ = 0.8VL = 690 V
IL = A
2.Three-phase powerHard
P = 100 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 100 kW at a power factor of 0.75 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 0.9 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 100 kWpf1 = 0.75pf2 = 0.9VL = 415 Vf = 60 Hz
Cph = µF
3.TransformersMedium
4800 V240 VIHV​ = ?ILV​500 kVA

A single-phase 500 kVA, 4800/240 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 500 kVAV1 = 4800 VV2 = 240 V
IHV = A
4.TransformersHard
pf 0.8550 kVA, 50% load

A single-phase 50 kVA transformer has a core loss of 400 W and a copper loss of 1200 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 = 50 kVAPcore = 400 WPcu,fl = 1200 Wx = 0.5pf = 0.85
Ploss = W
5.AC power & power factorMedium
400 V19 kW, 0.8525 kW, 1I

Two loads are connected in parallel to a 400 V single-phase supply: a 9 kW load at a power factor of 0.85 lagging and a 5 kW heater (unity power factor). Find the overall power factor cos φ seen by the supply.

V = 400 VP1 = 9 kW, pf1 = 0.85 laggingP2 = 5 kW, pf2 = 1
cos φ =
6.AC power & power factorHard
P = 40 kWQ1​S1​φ1​QC​

A single-phase load takes 40 kW at a power factor of 0.8 lagging from a 230 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the reactive power QC the capacitor must supply.

P = 40 kWpf1 = 0.8pf2 = 0.95V = 230 Vf = 60 Hz
QC = kvar
7.AC impedance & phasorsMedium
230 V500 HzR = 15 ΩL = 20 mHC = 2.2 µF

A series RLC circuit is connected to a 230 V (rms), 500 Hz supply. Find the phase angle φ of the impedance, in degrees (positive when the current lags the voltage).

V = 230 V (rms)f = 500 HzR = 15 ΩL = 20 mHC = 2.2 µF
φ = °
8.AC impedance & phasorsHard
VfR = 50 ΩL = 20 mHC = 1 µF

A series RLC circuit has R = 50 Ω, L = 20 mH and C = 1 µF. Find its quality factor Q at resonance.

R = 50 ΩL = 20 mHC = 1 µF
Q =
9.DC motorsMedium
T (N·m)n (rpm)n0​ = ?

A DC motor with k = 0.8 V·s/rad runs from 110 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.

V = 110 Vk = 0.8 V·s/rad
n0 = rpm
10.DC motorsHard
+−120 VRa​ = 2 ΩIa​: 8 → 4.8 AM+−En = ?T

A separately excited DC motor on 120 V (Ra = 2 Ω) runs at 1800 rpm with an armature current of 8 A. The load changes and the current becomes 4.8 A, with the field unchanged. Find the new speed n2.

V = 120 VRa = 2 Ωn1 = 1800 rpmIa1 = 8 AIa2 = 4.8 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: IL = 42.53 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 = 37/0.91 = 40.66 kW

Line current

IL = Pin/(√3·VL·cos φ) = 40659/(√3·690·0.8) = 42.53 A

Answer

IL = 42.53 A
2.Three-phase power

Answer: Cph = 612.4 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 = 100·tan(acos 0.75) = 100·0.8819 = 88.19 kvar
Q2 = 100·tan(acos 0.9) = 100·0.4843 = 48.43 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 39.76 kvar

Capacitance per phase

In a Y bank each capacitor is connected between a line and the star point.

V = 415/√3 = 239.6 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 13253/(377·239.6²) = 612.4 µF

Answer

Cph = 612.4 µF
3.Transformers

Answer: IHV = 104.2 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.

Rated current

Both windings carry the same apparent power.

IHV = S/V = 500000/4800 = 104.2 A

Answer

IHV = 104.2 A
4.Transformers

Answer: Ploss = 700 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 = 0.25·1200 = 300 W

Total losses

The core loss is the same at any load.

Ploss = 400 + 300 = 700 W

Answer

Ploss = 700 W

The efficiency at this load is η = 96.81%.

5.AC power & power factor

Answer: cos φ = 0.929

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 = 9·tan(arccos 0.85) = 5.578 kvar
Q2 = 0 = 0 kvar

Totals

P = 9 + 5 = 14 kW
Q = 5.578 + 0 = 5.578 kvar
S = √(14² + 5.578²) = 15.07 kVA

Overall power factor

cos φ = P/S = 14/15.07 = 0.929 lagging

Answer

cos φ = 0.929
6.AC power & power factor

Answer: QC = 16.85 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.

Reactive power before and after

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

Q1 = P·tan φ1 = 40·tan(36.87°) = 30 kvar
Q2 = P·tan φ2 = 40·tan(18.19°) = 13.15 kvar
QC = Q1 − Q2 = 16.85 kvar

Answer

QC = 16.85 kvar
7.AC impedance & phasors

Answer: φ = −79.62 °

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·500 = 3142 rad/s
XL = ωL = 3142·0.02 = 62.83 Ω
XC = 1/(ωC) = 1/(3142·2.2·10⁻⁶) = 144.7 Ω

Impedance

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

|Z| = √(R² + (XL − XC)²) = √(15² + (−81.85)²) = 83.22 Ω

Phase angle

φ = arctan((XL − XC)/R) = arctan(−81.85/15) = −79.62°

Answer

φ = −79.62°
8.AC impedance & phasors

Answer: Q = 2.828

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|.

Resonant frequency

f0 = 1/(2π·√(0.02·1·10⁻⁶)) = 1125 Hz

Quality factor

Q compares the reactance at resonance with the resistance: Q = ω₀L/R.

Q = (1/R)·√(L/C) = (1/50)·√(0.02/(1·10⁻⁶)) = 2.828

Answer

Q = 2.828
9.DC motors

Answer: n0 = 1313 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.

No-load speed

ω₀ = V/k = 110/0.8 = 137.5 rad/s
n0 = 60·137.5/(2π) = 1313 rpm

Answer

n0 = 1313 rpm
10.DC motors

Answer: n2 = 1911 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 − 8·2 = 104 V
E2 = 120 − 4.8·2 = 110.4 V

New speed

n2 = 1800·110.4/104 = 1911 rpm

Answer

n2 = 1911 rpm

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