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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 7.5 kW of mechanical output power at full load. At this load its efficiency is 93% and its power factor is 0.88 lagging. The supply has a line-to-line voltage of 415 V. Find the line current IL.
Pout = 7.5 kWη = 93%cos φ = 0.88VL = 415 V
IL = A
2.Three-phase powerHard
A balanced three-phase load takes 80 kW at a power factor of 0.8 lagging from a 400 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. Find the total reactive power QC the capacitor bank must supply, in kvar.
P = 80 kWpf1 = 0.8pf2 = 0.9VL = 400 Vf = 60 Hz
QC = kvar
3.TransformersMedium
A 50 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 3:1. Find the resistance R1′ seen looking into the primary winding.
N1:N2 = 3:1RL = 50 Ω
R1′ = Ω
4.TransformersHard
A single-phase 50 kVA transformer has a core loss of 300 W and a copper loss of 600 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
S = 50 kVAPcore = 300 WPcu,fl = 600 W
Sη,max = kVA
5.AC power & power factorMedium
A single-phase motor takes 28 kW at a power factor of 0.6 lagging from a 400 V supply. Find the line current I.
P = 28 kWpf = 0.6V = 400 V
I = A
6.AC power & power factorHard
A 50 kVA single-phase transformer supplies a load of 26 kW at a power factor of 0.75 lagging. How much extra real power ΔP can be added as a resistive (unity power factor) load before the transformer reaches its kVA rating?
Smax = 50 kVAP1 = 26 kWpf1 = 0.75
ΔP = kW
7.AC impedance & phasorsMedium
A series RLC circuit is connected to a 230 V (rms), 100 Hz supply. Find the real power P taken from the supply.
V = 230 V (rms)f = 100 HzR = 80 ΩL = 150 mHC = 2.2 µF
P = W
8.AC impedance & phasorsHard
A series RLC circuit has R = 50 Ω, L = 200 mH and C = 0.022 µF. Find its quality factor Q at resonance.
R = 50 ΩL = 200 mHC = 0.022 µF
Q =
9.DC motorsMedium
A DC motor delivers 0.5 kW of mechanical power at 1800 rpm. Find the shaft torque T.
P = 0.5 kWn = 1800 rpm
T = N·m
10.DC motorsHard
A small DC motor has an armature resistance of 2.5 Ω and runs from 110 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 110 VRa = 2.5 Ω
Pmax = W
3Solutions
1.Three-phase power
Answer: IL = 12.75 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 = 7.5/0.93 = 8.065 kW
Line current
IL = Pin/(√3·VL·cos φ) = 8065/(√3·415·0.88) = 12.75 A
Answer
IL = 12.75 A
2.Three-phase power
Answer: QC = 21.25 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 = 80·tan(acos 0.8) = 80·0.75 = 60 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 21.25 kvar
Answer
QC = 21.25 kvar
3.Transformers
Answer: R1′ = 450 Ω
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.
Referred resistance
The voltage goes up by a and the current down by a, so the ratio V/I goes up by a².
R1′ = a²·RL = 3²·50 = 450 Ω
Answer
R1′ = 450 Ω
4.Transformers
Answer: Sη,max = 35.36 kVA
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 equals core loss
x²·600 = 300 → x = √(300/600) = 0.7071
Load at maximum efficiency
Sη,max = x·S = 0.7071·50 = 35.36 kVA
Answer
Sη,max = 35.36 kVA
5.AC power & power factor
Answer: I = 116.7 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) = 28000/(400·0.6) = 116.7 A
Answer
I = 116.7 A
6.AC power & power factor
Answer: ΔP = 18.43 kW
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 the existing load
Q1 = 26·tan(arccos 0.75) = 22.93 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(50² − 22.93²) = 44.43 kW
Extra real power
ΔP = 44.43 − 26 = 18.43 kW
Answer
ΔP = 18.43 kW
7.AC impedance & phasors
Answer: P = 10.52 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 = 94.25 Ω, XC = 723.4 Ω
Impedance
|Z| = √(80² + (−629.2)²) = 634.2 Ω
Real power
Inductors and capacitors only store and return energy; the resistor absorbs all of the real power.
I = 230/634.2 = 0.3626 A
P = I²·R = 0.3626²·80 = 10.52 W
Answer
P = 10.52 W
8.AC impedance & phasors
Answer: Q = 60.3
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·√(0.2·0.022·10⁻⁶)) = 2399 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·1800/60 = 188.5 rad/s
Torque
T = P/ω = 500/188.5 = 2.653 N·m
Answer
T = 2.653 N·m
10.DC motors
Answer: Pmax = 1210 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 = 55 V
Maximum power
Pmax = (V/2)²/Ra = V²/(4Ra) = 110²/(4·2.5) = 1210 W
Answer
Pmax = 1210 W
At this point the efficiency is only 50 %, so real motors run well above half speed.