An inductive load takes 20 A and 4320 W from a 240 V (rms) single-phase supply. Model it as a resistance R in series with a reactance X. Find R.
V = 240 VI = 20 AP = 4320 W
R = Ω
6.AC power & power factorHard
A 100 V (rms), 1000 Hz source has an internal impedance Zs = 50 + j30 Ω. The load impedance ZL can be chosen freely. Find the maximum real power Pmax that can be delivered to the load.
Vs = 100 V (rms)f = 1000 HzZs = 50 + j30 Ω
Pmax = W
7.Three-phase powerMedium
A balanced Y-connected load with an impedance of 4 + j3 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 = 4 + j3 Ω per phase
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 120 kW at a power factor of 0.65 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 = 120 kWpf1 = 0.65pf2 = 0.9VL = 400 Vf = 60 Hz
QC = kvar
3Solutions
1.DC-DC converters
Answer: ΔIL = 0.8182 A
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 18/24 = 0.75
Inductor voltage during the on-time
While S is on, the inductor sits between Vin and Vo.
VL = 24 − 18 = 6 V
Ripple current
ΔIL = VL·D/(L·f) = 6·0.75/(220·10⁻⁶·25·10³) = 0.8182 A
Answer
ΔIL = 0.8182 A
2.DC-DC converters
Answer: Lcrit = 72 µH
DC-DC converters: a switch S turns on for the fraction D of each period T = 1/f. In steady state the average inductor voltage is zero (volt-second balance), which gives the ideal conversion ratios in continuous conduction: buck Vo = D·Vin, boost Vo = Vin/(1 − D), inverting buck-boost |Vo| = Vin·D/(1 − D). An ideal converter keeps the power: Vin·Iin = Vo·Io. While S is on the inductor current rises by ΔIL = VL·D·T/L, with VL = Vin − Vo for the buck and Vin for the boost and buck-boost. The output ripple is ΔVo = ΔIL/(8fC) for the buck and Io·D/(fC) for the boost and buck-boost. The inductor current just touches zero (the CCM boundary) when the average inductor current equals ΔIL/2.
Duty cycle
D = 6/(9 + 6) = 0.4
Boundary condition
Set the average inductor current equal to half the ripple current and solve for L.
Lcrit = (1 − D)²·R/(2f)
Critical inductance
Lcrit = (1 − 0.4)²·10/(225·10³) = 72 µH
Answer
Lcrit = 72 µH
3.DC motors
Answer: Tstall = 360 N·m
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.
Stall current
Ia = V/Ra = 240/1 = 240 A
Stall torque
Tstall = 1.5·240 = 360 N·m
Answer
Tstall = 360 N·m
4.DC motors
Answer: Rs = 1.387 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.
Current and back-EMF
Ia = 25/1 = 25 A
E = 1·62.83 = 62.83 V
Series resistance
Ra + Rs = (110 − 62.83)/25 = 1.887 Ω
Rs = 1.387 Ω
Answer
Rs = 1.387 Ω
The resistor wastes 866.7 W; a chopper controls the speed with far less loss.
5.AC power & power factor
Answer: R = 10.8 Ω
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.
Resistance from the real power
R = P/I² = 4320/20² = 10.8 Ω
Answer
R = 10.8 Ω
6.AC power & power factor
Answer: Pmax = 50 W
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.
Conjugate match
The load reactance cancels the source reactance, and the load resistance equals the source resistance.
ZL = Zs* = 50 − j30 Ω
Maximum power
I = Vs/(2Rs) = 100/100 = 1 A
Pmax = I²·RL = Vs²/(4Rs) = 100²/(4·50) = 50 W
Answer
Pmax = 50 W
7.Three-phase power
Answer: P = 6.922 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| = √(4² + 3²) = 5 Ω
Iph = 120.1/5 = 24.02 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·24.02²·4 = 6.922 kW
Answer
P = 6.922 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: QC = 82.18 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.