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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 balanced Y-connected load with an impedance of 6 + j8 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 208 V. Find the total real power P absorbed by the load in kW.
VL = 208 VZ = 6 + j8 Ω per phase
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 60 kW at a power factor of 0.72 lagging from a 480 V (line-to-line), 60 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.92 lagging. The real power stays the same. The capacitors are connected in Δ. Find the capacitance Cph of each capacitor in µF.
P = 60 kWpf1 = 0.72pf2 = 0.92VL = 480 Vf = 60 Hz
Cph = µF
3.TransformersMedium
A 50 Ω resistor is connected across the secondary winding of an ideal transformer with a turns ratio N₁:N₂ = 12:1. Find the resistance R1′ seen looking into the primary winding.
N1:N2 = 12:1RL = 50 Ω
R1′ = Ω
4.TransformersHard
A single-phase 100 kVA transformer has a core loss of 600 W and a copper loss of 2400 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
S = 100 kVAPcore = 600 WPcu,fl = 2400 W
Sη,max = kVA
5.AC power & power factorMedium
A load with impedance Z = 12 + j16 Ω is connected to a 230 V (rms) single-phase supply. Find the apparent power S.
V = 230 VZ = 12 + j16 Ω
S = VA
6.AC power & power factorHard
A single-phase load takes 5 kW at a power factor of 0.65 lagging from a 240 V, 60 Hz supply. A capacitor connected in parallel raises the power factor to 0.95. Find the capacitance C needed.
P = 5 kWpf1 = 0.65pf2 = 0.95V = 240 Vf = 60 Hz
C = µF
7.AC impedance & phasorsMedium
A series RLC circuit is connected to a 230 V (rms), 50 Hz supply. Find the power factor cos φ of the circuit.
V = 230 V (rms)f = 50 HzR = 30 ΩL = 20 mHC = 100 µF
cos φ =
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 12 V (rms), 60 Hz supply. Find the rms voltage VC across the capacitor.
V = 12 V (rms)f = 60 HzR = 15 ΩL = 100 mHC = 100 µF
VC = V
9.DC motorsMedium
A DC motor (k = 1.5 V·s/rad, Ra = 0.4 Ω) runs from 240 V. At full load it draws 40 A. Neglecting the no-load current, find its speed regulation SR = (n₀ − nFL)/nFL in percent.
V = 240 Vk = 1.5 V·s/radRa = 0.4 ΩIFL = 40 A
SR = %
10.DC motorsHard
A small DC motor has an armature resistance of 3 Ω and runs from 110 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 110 VRa = 3 Ω
Pmax = W
3Solutions
1.Three-phase power
Answer: P = 2.596 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.
Phase voltage
Vph = 208/√3 = 120.1 V
Phase current
|Z| = √(6² + 8²) = 10 Ω
Iph = 120.1/10 = 12.01 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·12.01²·6 = 2.596 kW
Answer
P = 2.596 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
2.Three-phase power
Answer: Cph = 123.8 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 = 60·tan(acos 0.72) = 60·0.9639 = 57.83 kvar
Q2 = 60·tan(acos 0.92) = 60·0.426 = 25.56 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 32.27 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.
V = 480 V, ω = 2π·60 = 377 rad/s
Cph = (QC/3)/(ω·V²) = 10757/(377·480²) = 123.8 µF
Answer
Cph = 123.8 µF
3.Transformers
Answer: R1′ = 7200 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 = 12²·50 = 7200 Ω
Answer
R1′ = 7200 Ω
4.Transformers
Answer: Sη,max = 50 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
x²·2400 = 600 → x = √(600/2400) = 0.5
Load at maximum efficiency
Sη,max = x·S = 0.5·100 = 50 kVA
Answer
Sη,max = 50 kVA
5.AC power & power factor
Answer: S = 2645 VA
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| = √(12² + 16²) = 20 Ω
I = V/|Z| = 230/20 = 11.5 A
Apparent power
S = V·I = 230·11.5 = 2645 VA
Answer
S = 2645 VA
6.AC power & power factor
Answer: C = 193.5 µF
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 = P·tan φ1 = 5·tan(49.46°) = 5.846 kvar
Q2 = P·tan φ2 = 5·tan(18.19°) = 1.643 kvar
QC = Q1 − Q2 = 4.202 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 4202/(2π·60·240²) = 193.5 µF
Answer
C = 193.5 µF
7.AC impedance & phasors
Answer: cos φ = 0.7613
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 = 6.283 Ω, XC = 31.83 Ω
Impedance
|Z| = √(30² + (−25.55)²) = 39.4 Ω
Power factor
cos φ = R/|Z| = 30/39.4 = 0.7613
The circuit is capacitive, so the power factor is leading.
Answer
cos φ = 0.7613
8.AC impedance & phasors
Answer: VC = 17.02 V
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 = 377·0.1 = 37.7 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| = 12/18.7 = 0.6416 A
Voltage across the capacitor
VC = I·XC = 0.6416·26.53 = 17.02 V
Answer
VC = 17.02 V
9.DC motors
Answer: SR = 7.143 %
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.
No-load and full-load speed
n₀ = 60·240/(2π·1.5) = 1528 rpm
nFL = 60·(240 − 40·0.4)/(2π·1.5) = 1426 rpm
Speed regulation
SR = (1528 − 1426)/1426·100 = 7.143 %
Answer
SR = 7.143 %
10.DC motors
Answer: Pmax = 1008 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/(4·3) = 1008 W
Answer
Pmax = 1008 W
At this point the efficiency is only 50 %, so real motors run well above half speed.