An ideal buck-boost converter turns 48 V into −112 V for an 8 Ω load at 20 kHz. The output capacitor is 100 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
A DC motor draws 50 A from a 110 V supply. Its armature resistance is 0.2 Ω and its rotational (friction, windage and iron) losses are 250 W. Find its efficiency η.
V = 110 VIa = 50 ARa = 0.2 ΩProt = 250 W
η = %
4.DC motorsHard
A separately excited DC motor (k = 2 V·s/rad at full field, Ra = 1 Ω, V = 240 V) drives a constant load torque of 50 N·m. The field flux is reduced to 85 % of its full value. Find the new speed n.
V = 240 Vk = 2 V·s/radRa = 1 ΩT = 50 N·mΦ = 85 %
n = rpm
5.AC power & power factorMedium
Two loads are connected in parallel to a 240 V single-phase supply: a 5 kW load at a power factor of 0.85 lagging and a 29 kW load at a power factor of 0.8 leading. Find the overall power factor cos φ seen by the supply.
V = 240 VP1 = 5 kW, pf1 = 0.85 laggingP2 = 29 kW, pf2 = 0.8 leading
cos φ =
6.AC power & power factorHard
A single-phase load takes 50 kW at a power factor of 0.8 lagging from a 400 V, 50 Hz supply. A capacitor connected in parallel raises the power factor to 0.98. Find the capacitance C needed.
P = 50 kWpf1 = 0.8pf2 = 0.98V = 400 Vf = 50 Hz
C = µF
7.Three-phase powerMedium
A balanced Δ-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 80 kW at a power factor of 0.65 lagging from a 480 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 = 80 kWpf1 = 0.65pf2 = 0.9VL = 480 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 12.37 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 = 28.8/48 = 0.6
Average and ripple current
Io = 28.8/2.5 = 11.52 A
ΔIL = VL·D/(L·f) = 19.2·0.6/(68·10⁻⁶·100·10³) = 1.694 A
Peak current
Imax = 11.52 + 1.694/2 = 12.37 A
Answer
Imax = 12.37 A
2.DC-DC converters
Answer: ΔVo = 4900 mV
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 = 112/(48 + 112) = 0.7
Load current
Io = 112/8 = 14 A
Output ripple
ΔVo = 14·0.7/(20·10³·100·10⁻⁶) = 4900 mV
Answer
ΔVo = 4900 mV
3.DC motors
Answer: η = 86.36 %
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 = 110·50 = 5500 W
Pdev = 100·50 = 5000 W
Pout = 5000 − 250 = 4750 W
Efficiency
η = 4750/5500·100 = 86.36 %
Answer
η = 86.36 %
4.DC motors
Answer: n = 1183 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.
Field weakening is how a DC drive runs above its base speed.
5.AC power & power factor
Answer: cos φ = 0.8767
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 each load
A lagging load takes reactive power (Q > 0); a leading load supplies it (Q < 0).
Q1 = 5·tan(arccos 0.85) = 3.099 kvar
Q2 = −29·tan(arccos 0.8) = −21.75 kvar
Totals
P = 5 + 29 = 34 kW
Q = 3.099 + (−21.75) = −18.65 kvar
S = √(34² + (−18.65)²) = 38.78 kVA
Overall power factor
cos φ = P/S = 34/38.78 = 0.8767 leading
Answer
cos φ = 0.8767
6.AC power & power factor
Answer: C = 544.1 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 = 50·tan(36.87°) = 37.5 kvar
Q2 = P·tan φ2 = 50·tan(11.48°) = 10.15 kvar
QC = Q1 − Q2 = 27.35 kvar
Capacitance
A capacitor across the supply delivers Q = V²·ωC.
C = QC/(2πf·V²) = 27347/(2π·50·400²) = 544.1 µF
Answer
C = 544.1 µF
7.Three-phase power
Answer: P = 20.77 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 = 208 V
Phase current
|Z| = √(4² + 3²) = 5 Ω
Iph = 208/5 = 41.6 A
Total real power
Only the resistive part of each impedance absorbs real power.
P = 3·Iph²·R = 3·41.6²·4 = 20.77 kW
Answer
P = 20.77 kW
Check: cos φ = R/|Z| = 0.8, and √3·VL·IL·cos φ gives the same power.
8.Three-phase power
Answer: Cph = 252.3 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 = 80·tan(acos 0.65) = 80·1.169 = 93.53 kvar
Q2 = 80·tan(acos 0.9) = 80·0.4843 = 38.75 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 54.78 kvar
Capacitance per phase
In a Δ bank each capacitor is connected between two lines.