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​ = 30 Acos φ = 0.88lagging

A balanced three-phase load draws a line current of 30 A from a supply with a line-to-line voltage of 400 V. The power factor is 0.88 lagging. Find the total real power P drawn by the load in kW.

VL = 400 VIL = 30 Acos φ = 0.88
P = kW
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 690 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.

P = 150 kWpf1 = 0.65pf2 = 0.95VL = 690 Vf = 60 Hz
Cph = µF
3.TransformersMedium
RL​ = 48 Ω2014600 VV2​I1​ = ?I2​

An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4600 V applied to its primary winding. A 48 Ω resistor is connected across the secondary winding. Find the primary current I1.

N1:N2 = 20:1V1 = 4600 VRL = 48 Ω
I1 = A
4.TransformersHard
pf 0.9150 kVA, 75% load

A single-phase 150 kVA transformer has a core loss of 900 W and a copper loss of 2700 W at rated load. It supplies 75% of its rated kVA at a power factor of 0.9 lagging. Find the total losses Ploss in watts at this load.

S = 150 kVAPcore = 900 WPcu,fl = 2700 Wx = 0.75pf = 0.9
Ploss = W
5.AC power & power factorMedium
120 VZ8 + j6 ΩI

A load with impedance Z = 8 + j6 Ω is connected to a 120 V (rms) single-phase supply. Find the reactive power Q it absorbs.

V = 120 VZ = 8 + j6 Ω
Q = var
6.AC power & power factorHard
230 V60 Hzload9 kW, 0.65C100 µFI

A single-phase load takes 9 kW at a power factor of 0.65 lagging from a 230 V, 60 Hz supply. A 100 µF capacitor is connected in parallel with the load. Find the new power factor cos φ seen by the supply.

P = 9 kWpf1 = 0.65V = 230 Vf = 60 HzC = 100 µF
cos φ =
7.AC impedance & phasorsMedium
12 V60 HzR = 20 ΩL = 200 mHC = 10 µFI

A series RLC circuit is connected to a 12 V (rms), 60 Hz supply. Find the real power P taken from the supply.

V = 12 V (rms)f = 60 HzR = 20 ΩL = 200 mHC = 10 µF
P = W
8.AC impedance & phasorsHard
120 V200 HzR = 25 ΩL = 250 mHC = 22 µF

A series RLC circuit is connected to a 120 V (rms), 200 Hz supply. Find the resonant frequency f0 of the circuit.

V = 120 V (rms)f = 200 HzR = 25 ΩL = 250 mHC = 22 µF
f0 = Hz
9.DC motorsMedium
+−VRa​M+−En = 3000 rpmT = ?

A DC motor delivers 2.2 kW of mechanical power at 3000 rpm. Find the shaft torque T.

P = 2.2 kWn = 3000 rpm
T = N·m
10.DC motorsHard
+−120 VRa​ = 1 ΩM+−EωT

A small DC motor has an armature resistance of 1 Ω and runs from 120 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.

V = 120 VRa = 1 Ω
Pmax = W

3Solutions

1.Three-phase power

Answer: P = 18.29 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 = √3·400·30 = 20.78 kVA

Real power

P = S·cos φ = 20.78·0.88 = 18.29 kW

Answer

P = 18.29 kW
2.Three-phase power

Answer: Cph = 234.1 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 Δ bank each capacitor is connected between two lines.

V = 690 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 42022/(377·690²) = 234.1 µF

Answer

Cph = 234.1 µF
3.Transformers

Answer: I1 = 0.2396 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 = 4600/20 = 230 V
I2 = 230/48 = 4.792 A

Primary current

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

I1 = I2/20 = 0.2396 A

Answer

I1 = 0.2396 A

Check: the load seen from the primary is a²·RL = 19200 Ω, and 4600/19200 gives the same current.

4.Transformers

Answer: Ploss = 2419 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 = 0.5625·2700 = 1519 W

Total losses

The core loss is the same at any load.

Ploss = 900 + 1519 = 2419 W

Answer

Ploss = 2419 W

The efficiency at this load is η = 97.67%.

5.AC power & power factor

Answer: Q = 864 var

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.

Current

|Z| = √(8² + 6²) = 10 Ω
I = V/|Z| = 120/10 = 12 A

Reactive power

Q = I²·X = 12²·6 = 864 var

Answer

Q = 864 var
6.AC power & power factor

Answer: cos φ = 0.7259

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 = 9·tan(arccos 0.65) = 10.52 kvar
QC = V²·2πf·C = 230²·2π·60·100·10⁻⁶ = 1994 var = 1.994 kvar
Q2 = 10.52 − 1.994 = 8.528 kvar

New apparent power

S2 = √(9² + 8.528²) = 12.4 kVA

New power factor

cos φ = P/S2 = 9/12.4 = 0.7259

Answer

cos φ = 0.7259
7.AC impedance & phasors

Answer: P = 0.07902 W

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

XL = 75.4 Ω, XC = 265.3 Ω

Impedance

|Z| = √(20² + (−189.9)²) = 190.9 Ω

Real power

Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.

I = 12/190.9 = 0.06286 A
P = I²·R = 0.06286²·20 = 0.07902 W

Answer

P = 0.07902 W
8.AC impedance & phasors

Answer: f0 = 67.86 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·√(0.25·22·10⁻⁶)) = 67.86 Hz

Answer

f0 = 67.86 Hz
9.DC motors

Answer: T = 7.003 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·3000/60 = 314.2 rad/s

Torque

T = P/ω = 2200/314.2 = 7.003 N·m

Answer

T = 7.003 N·m
10.DC motors

Answer: Pmax = 3600 W

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.

Power as a function of the back-EMF

P(E) = E·(V − E)/Ra
dP/dE = 0 → E = V/2 = 60 V

Maximum power

Pmax = (V/2)²/Ra = V²/(4Ra) = 120²/(4·1) = 3600 W

Answer

Pmax = 3600 W

At this point the efficiency is only 50 %, so real motors run well above half speed.

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