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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 18.5 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 415 V. Find the line current IL.
Pout = 18.5 kWη = 95%cos φ = 0.82VL = 415 V
IL = A
2.Three-phase powerHard
A balanced three-phase load takes 250 kW at a power factor of 0.7 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.9 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 250 kWpf1 = 0.7pf2 = 0.9VL = 690 Vf = 50 Hz
Cph = µF
3.TransformersMedium
An ideal transformer with a turns ratio N₁:N₂ = 5:1 has 1150 V applied to its primary winding. A 12 Ω resistor is connected across the secondary winding. Find the primary current I1.
N1:N2 = 5:1V1 = 1150 VRL = 12 Ω
I1 = A
4.TransformersHard
A single-phase 100 kVA transformer has a core loss of 800 W. At rated load and a power factor of 0.9 lagging its efficiency is 97.5%. Find the full-load copper loss Pcu,fl in watts.
S = 100 kVAPcore = 800 Wpf = 0.9η = 97.5%
Pcu,fl = W
5.AC power & power factorMedium
A single-phase motor takes 39 kW at a power factor of 0.75 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 39 kWpf = 0.75V = 230 V
Q = kvar
6.AC power & power factorHard
A single-phase load takes 13 kW at a power factor of 0.75 lagging from a 230 V, 60 Hz supply. A 470 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 13 kWpf1 = 0.75V = 230 Vf = 60 HzC = 470 µF
I = A
7.AC impedance & phasorsMedium
A series RLC circuit with C = 0.47 µF must resonate at 500 Hz. Find the inductance L needed.
f0 = 500 HzC = 0.47 µF
L = mH
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 24 V (rms), 60 Hz supply. Find the rms current I.
V = 24 V (rms)f = 60 HzR = 80 ΩL = 200 mHC = 100 µF
I = A
9.DC motorsMedium
A DC motor draws 10 A from a 220 V supply. Its armature resistance is 0.8 Ω and its rotational (friction, windage and iron) losses are 127 W. Find its efficiency η.
V = 220 VIa = 10 ARa = 0.8 ΩProt = 127 W
η = %
10.DC motorsHard
A 120 V DC motor has an armature resistance of 1 Ω. Find the series starting resistance Rst that limits the starting current to 40 A.
V = 120 VRa = 1 ΩImax = 40 A
Rst = Ω
3Solutions
1.Three-phase power
Answer: IL = 33.04 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 = 18.5/0.95 = 19.47 kW
Line current
IL = Pin/(√3·VL·cos φ) = 19474/(√3·415·0.82) = 33.04 A
Answer
IL = 33.04 A
2.Three-phase power
Answer: Cph = 298.6 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 = 250·tan(acos 0.7) = 250·1.02 = 255.1 kvar
Q2 = 250·tan(acos 0.9) = 250·0.4843 = 121.1 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 134 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.
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 = 1150/5 = 230 V
I2 = 230/12 = 19.17 A
Primary current
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
I1 = I2/5 = 3.833 A
Answer
I1 = 3.833 A
Check: the load seen from the primary is a²·RL = 300 Ω, and 1150/300 gives the same current.
4.Transformers
Answer: Pcu,fl = 1508 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 = 100·0.9 = 90 kW
Pin = 90000/0.975 = 92308 W
Total and copper losses
Ploss = 92308 − 90000 = 2308 W
Pcu,fl = 2308 − 800 = 1508 W
Answer
Pcu,fl = 1508 W
5.AC power & power factor
Answer: Q = 34.39 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(0.75) = 41.41°, tan φ = 0.8819
Reactive power
Q = P·tan φ = 39·0.8819 = 34.39 kvar
Answer
Q = 34.39 kvar
6.AC power & power factor
Answer: I = 57.25 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 = 13·tan(arccos 0.75) = 11.46 kvar
QC = V²·2πf·C = 230²·2π·60·470·10⁻⁶ = 9373 var = 9.373 kvar
Q2 = 11.46 − 9.373 = 2.092 kvar
New apparent power
S2 = √(13² + 2.092²) = 13.17 kVA
New line current
I = S2/V = 13167/230 = 57.25 A
Answer
I = 57.25 A
7.AC impedance & phasors
Answer: L = 215.6 mH
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·500 = 3142 rad/s
L = 1/(ω₀²·C) = 1/(3142²·0.47·10⁻⁶) = 215.6 mH
Answer
L = 215.6 mH
8.AC impedance & phasors
Answer: I = 0.256 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|.
Reactances
ω = 2πf = 2π·60 = 377 rad/s
XL = ωL = 377·0.2 = 75.4 Ω
XC = 1/(ωC) = 1/(377·100·10⁻⁶) = 26.53 Ω
Impedance
In series the impedances add as complex numbers: Z = R + j(XL − XC).
The same current flows through every element of a series circuit.
I = V/|Z| = 24/93.75 = 0.256 A
Answer
I = 0.256 A
9.DC motors
Answer: η = 90.59 %
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.
Powers
Pin = 220·10 = 2200 W
Pdev = 212·10 = 2120 W
Pout = 2120 − 127 = 1993 W
Efficiency
η = 1993/2200·100 = 90.59 %
Answer
η = 90.59 %
10.DC motors
Answer: Rst = 2 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 = 120/40 = 3 Ω
Starting resistor
Rst = 3 − 1 = 2 Ω
Answer
Rst = 2 Ω
The resistor is cut out step by step as the motor speeds up and the back-EMF grows.