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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
Convert the frequency: ω = 2πf.
Find each reactance: XL = ωL and XC = 1/(ωC).
Add impedances like resistors, with Z = R + jX.
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
Draw the armature circuit: supply V, resistance Ra and the back-EMF E in series.
Use V = E + Ia·Ra to link voltage, current and back-EMF.
Use E = k·ω for speed and T = k·Ia for torque, with ω = 2π·n/60.
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
A three-phase induction motor delivers 30 kW of mechanical output power at full load. At this load its efficiency is 95% and its power factor is 0.82 lagging. The supply has a line-to-line voltage of 480 V. Find the line current IL.
Pout = 30 kWη = 95%cos φ = 0.82VL = 480 V
IL = A
2.Three-phase powerHard
A balanced three-phase load takes 300 kW at a power factor of 0.72 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 = 300 kWpf1 = 0.72pf2 = 0.95VL = 480 Vf = 50 Hz
QC = kvar
3.TransformersMedium
An amplifier with an output resistance of 288 Ω drives an 8 Ω loudspeaker through an ideal transformer. For maximum power transfer the loudspeaker must look like 288 Ω from the amplifier. Find the turns ratio a = N₁/N₂.
Rs = 288 ΩRL = 8 Ω
a =
4.TransformersHard
A single-phase 75 kVA transformer has a core loss of 450 W. At rated load and a power factor of 0.85 lagging its efficiency is 96%. Find the full-load copper loss Pcu,fl in watts.
S = 75 kVAPcore = 450 Wpf = 0.85η = 96%
Pcu,fl = W
5.AC power & power factorMedium
A single-phase motor takes 17 kW at a power factor of 0.75 lagging from a 240 V supply. Find the line current I.
P = 17 kWpf = 0.75V = 240 V
I = A
6.AC power & power factorHard
A single-phase load takes 5 kW at a power factor of 0.6 lagging from a 230 V, 60 Hz supply. A 47 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 5 kWpf1 = 0.6V = 230 Vf = 60 HzC = 47 µF
I = A
7.AC impedance & phasorsMedium
A resistor and a capacitor are connected in parallel across a 12 V (rms), 100 Hz supply. Find the magnitude of the impedance |Z|.
V = 12 V (rms)f = 100 HzR = 100 ΩC = 22 µF
|Z| = Ω
8.AC impedance & phasorsHard
A series RLC circuit has R = 5 Ω, L = 100 mH and C = 2.2 µF. Find its quality factor Q at resonance.
R = 5 ΩL = 100 mHC = 2.2 µF
Q =
9.DC motorsMedium
A DC motor delivers 0.5 kW of mechanical power at 1450 rpm. Find the shaft torque T.
P = 0.5 kWn = 1450 rpm
T = N·m
10.DC motorsHard
A DC motor with k = 0.3 V·s/rad and Ra = 2 Ω must drive a load of 0.6 N·m at 400 rpm. Find the armature voltage V it needs.
k = 0.3 V·s/radRa = 2 ΩT = 0.6 N·mn = 400 rpm
V = V
3Solutions
1.Three-phase power
Answer: IL = 46.32 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 = 30/0.95 = 31.58 kW
Line current
IL = Pin/(√3·VL·cos φ) = 31579/(√3·480·0.82) = 46.32 A
Answer
IL = 46.32 A
2.Three-phase power
Answer: QC = 190.6 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 = 300·tan(acos 0.72) = 300·0.9639 = 289.2 kvar
Q2 = 300·tan(acos 0.95) = 300·0.3287 = 98.61 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 190.6 kvar
Answer
QC = 190.6 kvar
3.Transformers
Answer: a = 6
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.
Match the referred load
a²·8 = 288
a = √(288/8) = √36 = 6
Answer
a = 6
4.Transformers
Answer: Pcu,fl = 2206 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 = 75·0.85 = 63.75 kW
Pin = 63750/0.96 = 66406 W
Total and copper losses
Ploss = 66406 − 63750 = 2656 W
Pcu,fl = 2656 − 450 = 2206 W
Answer
Pcu,fl = 2206 W
5.AC power & power factor
Answer: I = 94.44 A
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.
Line current
Only the real part of the power does work, so divide by the power factor.
I = P/(V·cos φ) = 17000/(240·0.75) = 94.44 A
Answer
I = 94.44 A
6.AC power & power factor
Answer: I = 33.06 A
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 = 5·tan(arccos 0.6) = 6.667 kvar
QC = V²·2πf·C = 230²·2π·60·47·10⁻⁶ = 937.3 var = 0.9373 kvar
Q2 = 6.667 − 0.9373 = 5.729 kvar
New apparent power
S2 = √(5² + 5.729²) = 7.604 kVA
New line current
I = S2/V = 7604/230 = 33.06 A
Answer
I = 33.06 A
7.AC impedance & phasors
Answer: |Z| = 58.61 Ω
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·100·22·10⁻⁶) = 72.34 Ω
Branch currents
Both branches are connected straight across the supply, so each sees the full voltage V.
IR = V/R = 12/100 = 0.12 A
IX = V/XC = 12/72.34 = 0.1659 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) = √(0.12² + 0.1659²) = 0.2047 A
Impedance
|Z| = V/I = 12/0.2047 = 58.61 Ω
Answer
|Z| = 58.61 Ω
8.AC impedance & phasors
Answer: Q = 42.64
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·√(0.1·2.2·10⁻⁶)) = 339.3 Hz
Quality factor
Q compares the reactance at resonance with the resistance: Q = ω₀L/R.
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·1450/60 = 151.8 rad/s
Torque
T = P/ω = 500/151.8 = 3.293 N·m
Answer
T = 3.293 N·m
10.DC motors
Answer: V = 16.57 V
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.