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​ = ?M22 kW, η = 93%cos φ = 0.86

A three-phase induction motor delivers 22 kW of mechanical output power at full load. At this load its efficiency is 93% 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 = 22 kWη = 93%cos φ = 0.86VL = 400 V
IL = A
2.Three-phase powerHard
P = 150 kWφ1​S1​Q2​QC​

A balanced three-phase load takes 150 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. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.

P = 150 kWpf1 = 0.65pf2 = 0.95VL = 400 Vf = 50 Hz
Cph = µF
3.TransformersMedium
6600 V230 VIHV​ = ?ILV​250 kVA

A single-phase 250 kVA, 6600/230 V transformer operates at its rated load. Treat it as ideal. Find the current IHV in the high-voltage winding.

S = 250 kVAV1 = 6600 VV2 = 230 V
IHV = A
4.TransformersHard
pf 0.825 kVA, η = 98.2%

A single-phase 25 kVA transformer has a core loss of 75 W. At rated load and a power factor of 0.8 lagging its efficiency is 98.2%. Find the full-load copper loss Pcu,fl in watts.

S = 25 kVAPcore = 75 Wpf = 0.8η = 98.2%
Pcu,fl = W
5.AC power & power factorMedium
230 VR + jXP = 897 W6.5 A

An inductive load takes 6.5 A and 897 W from a 230 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.

V = 230 VI = 6.5 AP = 897 W
R = Ω
6.AC power & power factorHard
230 V60 Hzload29 kW, 0.6C220 µFI

A single-phase load takes 29 kW at a power factor of 0.6 lagging from a 230 V, 60 Hz supply. A 220 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 29 kWpf1 = 0.6V = 230 Vf = 60 HzC = 220 µF
cos φ =
7.AC impedance & phasorsMedium
Vf0​ = 100 HzRL = 5 mHC = ?

A series RLC circuit with L = 5 mH must resonate at 100 Hz. Find the capacitance C needed.

f0 = 100 HzL = 5 mH
C = µF
8.AC impedance & phasorsHard
24 V500 HzR = 15 ΩL = 100 mHC = 1 µF

A series RLC circuit is connected to a 24 V (rms), 500 Hz supply. Find the resonant frequency f0 of the circuit.

V = 24 V (rms)f = 500 HzR = 15 ΩL = 100 mHC = 1 µF
f0 = Hz
9.DC motorsMedium
+−VRa​M+−En = 1800 rpmT = ?

A DC motor delivers 7.5 kW of mechanical power at 1800 rpm. Find the shaft torque T.

P = 7.5 kWn = 1800 rpm
T = N·m
10.DC motorsHard
+−220 VRa​ = 2 ΩIa​: 4 → 6 AM+−En = ?T

A separately excited DC motor on 220 V (Ra = 2 Ω) runs at 1200 rpm with an armature current of 4 A. The load changes and the current becomes 6 A, with the field unchanged. Find the new speed n2.

V = 220 VRa = 2 Ωn1 = 1200 rpmIa1 = 4 AIa2 = 6 A
n2 = rpm

3Solutions

1.Three-phase power

Answer: IL = 39.7 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 = 22/0.93 = 23.66 kW

Line current

IL = Pin/(√3·VL·cos φ) = 23656/(√3·400·0.86) = 39.7 A

Answer

IL = 39.7 A
2.Three-phase power

Answer: Cph = 2508 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 = 150·tan(acos 0.65) = 150·1.169 = 175.4 kvar
Q2 = 150·tan(acos 0.95) = 150·0.3287 = 49.3 kvar

Reactive power from the capacitors

QC = Q1 − Q2 = 126.1 kvar

Capacitance per phase

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

V = 400/√3 = 230.9 V, ω = 2π·50 = 314.2 rad/s
Cph = (QC/3)/(ω·V²) = 42022/(314.2·230.9²) = 2508 µF

Answer

Cph = 2508 µF
3.Transformers

Answer: IHV = 37.88 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 = 250000/6600 = 37.88 A

Answer

IHV = 37.88 A
4.Transformers

Answer: Pcu,fl = 291.6 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.

Output and input power

Pout = 25·0.8 = 20 kW
Pin = 20000/0.982 = 20367 W

Total and copper losses

Ploss = 20367 − 20000 = 366.6 W
Pcu,fl = 366.6 − 75 = 291.6 W

Answer

Pcu,fl = 291.6 W
5.AC power & power factor

Answer: R = 21.23 Ω

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.

Resistance from the real power

R = P/I² = 897/6.5² = 21.23 Ω

Answer

R = 21.23 Ω
6.AC power & power factor

Answer: cos φ = 0.6459

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 = 29·tan(arccos 0.6) = 38.67 kvar
QC = V²·2πf·C = 230²·2π·60·220·10⁻⁶ = 4387 var = 4.387 kvar
Q2 = 38.67 − 4.387 = 34.28 kvar

New apparent power

S2 = √(29² + 34.28²) = 44.9 kVA

New power factor

cos φ = P/S2 = 29/44.9 = 0.6459

Answer

cos φ = 0.6459
7.AC impedance & phasors

Answer: C = 506.6 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·100 = 628.3 rad/s
C = 1/(ω₀²·L) = 1/(628.3²·0.005) = 506.6 µF

Answer

C = 506.6 µF
8.AC impedance & phasors

Answer: f0 = 503.3 Hz

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 inductive and capacitive reactances cancel: ωL = 1/(ωC). The supply frequency does not matter.

f0 = 1/(2π√(LC)) = 1/(2π·√(0.1·1·10⁻⁶)) = 503.3 Hz

Answer

f0 = 503.3 Hz
9.DC motors

Answer: T = 39.79 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·1800/60 = 188.5 rad/s

Torque

T = P/ω = 7500/188.5 = 39.79 N·m

Answer

T = 39.79 N·m
10.DC motors

Answer: n2 = 1177 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 = 220 − 4·2 = 212 V
E2 = 220 − 6·2 = 208 V

New speed

n2 = 1200·208/212 = 1177 rpm

Answer

n2 = 1177 rpm

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