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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 Δ-connected load with an impedance of 18 + j24 Ω per phase is connected to a balanced three-phase supply with a line-to-line voltage of 400 V. Find the total real power P absorbed by the load in kW.
VL = 400 VZ = 18 + j24 Ω per phase
P = kW
2.Three-phase powerHard
A balanced three-phase load takes 300 kW at a power factor of 0.75 lagging from a 415 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 = 300 kWpf1 = 0.75pf2 = 0.9VL = 415 Vf = 50 Hz
Cph = µF
3.TransformersMedium
An ideal transformer with a turns ratio N₁:N₂ = 20:1 has 4600 V applied to its primary winding. A 20 Ω resistor is connected across the secondary winding. Find the primary current I1.
N1:N2 = 20:1V1 = 4600 VRL = 20 Ω
I1 = A
4.TransformersHard
A single-phase 500 kVA transformer has a core loss of 2000 W and a copper loss of 4000 W at rated load. At what load Sη,max, in kVA, is its efficiency highest?
S = 500 kVAPcore = 2000 WPcu,fl = 4000 W
Sη,max = kVA
5.AC power & power factorMedium
A single-phase motor takes 22 kW at a power factor of 0.6 lagging from a 230 V supply. Find the reactive power Q it draws.
P = 22 kWpf = 0.6V = 230 V
Q = kvar
6.AC power & power factorHard
A 150 kVA single-phase transformer supplies a load of 52 kW at a power factor of 0.65 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 = 150 kVAP1 = 52 kWpf1 = 0.65
ΔP = kW
7.AC impedance & phasorsMedium
A series RLC circuit with L = 10 mH must resonate at 50 Hz. Find the capacitance C needed.
f0 = 50 HzL = 10 mH
C = µF
8.AC impedance & phasorsHard
A series RLC circuit is connected to a 100 V (rms), 100 Hz supply. Find the rms current I.
V = 100 V (rms)f = 100 HzR = 50 ΩL = 50 mHC = 2.2 µF
I = A
9.DC motorsMedium
A DC motor with k = 0.8 V·s/rad runs from 120 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 120 Vk = 0.8 V·s/rad
n0 = rpm
10.DC motorsHard
A small DC motor has an armature resistance of 1.5 Ω and runs from 120 V. Neglecting rotational losses, find the largest mechanical power Pmax it can develop at any speed.
V = 120 VRa = 1.5 Ω
Pmax = W
3Solutions
1.Three-phase power
Answer: P = 9.6 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 = VL = 400 V
Phase current
|Z| = √(18² + 24²) = 30 Ω
Iph = 400/30 = 13.33 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·13.33²·18 = 9.6 kW
Answer
P = 9.6 kW
Check: cos φ = R/|Z| = 0.6, and √3·VL·IL·cos φ gives the same power.
2.Three-phase power
Answer: Cph = 734.8 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 = 300·tan(acos 0.75) = 300·0.8819 = 264.6 kvar
Q2 = 300·tan(acos 0.9) = 300·0.4843 = 145.3 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 119.3 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 = 4600/20 = 230 V
I2 = 230/20 = 11.5 A
Primary current
An ideal transformer has no losses: V₁·I₁ = V₂·I₂.
I1 = I2/20 = 0.575 A
Answer
I1 = 0.575 A
Check: the load seen from the primary is a²·RL = 8000 Ω, and 4600/8000 gives the same current.
4.Transformers
Answer: Sη,max = 353.6 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
x²·4000 = 2000 → x = √(2000/4000) = 0.7071
Load at maximum efficiency
Sη,max = x·S = 0.7071·500 = 353.6 kVA
Answer
Sη,max = 353.6 kVA
5.AC power & power factor
Answer: Q = 29.33 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(0.6) = 53.13°, tan φ = 1.333
Reactive power
Q = P·tan φ = 22·1.333 = 29.33 kvar
Answer
Q = 29.33 kvar
6.AC power & power factor
Answer: ΔP = 85.13 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 = 52·tan(arccos 0.65) = 60.79 kvar
Largest total real power
At full rating S = Smax, and Q is unchanged by the resistive load.
P = √(Smax² − Q1²) = √(150² − 60.79²) = 137.1 kW
Extra real power
ΔP = 137.1 − 52 = 85.13 kW
Answer
ΔP = 85.13 kW
7.AC impedance & phasors
Answer: C = 1013 µF
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·50 = 314.2 rad/s
C = 1/(ω₀²·L) = 1/(314.2²·0.01) = 1013 µF
Answer
C = 1013 µF
8.AC impedance & phasors
Answer: I = 0.1441 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·100 = 628.3 rad/s
XL = ωL = 628.3·0.05 = 31.42 Ω
XC = 1/(ωC) = 1/(628.3·2.2·10⁻⁶) = 723.4 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| = 100/693.8 = 0.1441 A
Answer
I = 0.1441 A
9.DC motors
Answer: n0 = 1432 rpm
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 speed
ω₀ = V/k = 120/0.8 = 150 rad/s
n0 = 60·150/(2π) = 1432 rpm
Answer
n0 = 1432 rpm
10.DC motors
Answer: Pmax = 2400 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/(4·1.5) = 2400 W
Answer
Pmax = 2400 W
At this point the efficiency is only 50 %, so real motors run well above half speed.