An ideal boost converter turns 15 V into 60 V for a 10 Ω load at 40 kHz. The output capacitor is 220 µF. Find the peak-to-peak output voltage ripple ΔVo in mV.
Vin = 15 VVo = 60 Vf = 40 kHzC = 220 µFR = 10 Ω
ΔVo = mV
3.DC motorsMedium
A DC motor with k = 0.8 V·s/rad runs from 220 V. Neglecting all losses at no load, find its no-load speed n0 in rpm.
V = 220 Vk = 0.8 V·s/rad
n0 = rpm
4.DC motorsHard
A separately excited DC motor on 120 V (Ra = 1.2 Ω) runs at 1440 rpm with an armature current of 6 A. The load changes and the current becomes 3.6 A, with the field unchanged. Find the new speed n2.
V = 120 VRa = 1.2 Ωn1 = 1440 rpmIa1 = 6 AIa2 = 3.6 A
n2 = rpm
5.AC power & power factorMedium
Two loads are connected in parallel to a 230 V single-phase supply: an 18 kW load at a power factor of 0.7 lagging and a 20 kW heater (unity power factor). Find the overall power factor cos φ seen by the supply.
A single-phase load takes 17 kW at a power factor of 0.65 lagging from a 400 V, 60 Hz supply. A 33 µF capacitor is connected in parallel with the load. Find the new line current I.
P = 17 kWpf1 = 0.65V = 400 Vf = 60 HzC = 33 µF
I = A
7.Three-phase powerMedium
A balanced three-phase load draws a line current of 20 A from a supply with a line-to-line voltage of 415 V. The power factor is 0.9 lagging. Find the total real power P drawn by the load in kW.
VL = 415 VIL = 20 Acos φ = 0.9
P = kW
8.Three-phase powerHard
A balanced three-phase load takes 300 kW at a power factor of 0.7 lagging from a 690 V (line-to-line), 50 Hz supply. A three-phase capacitor bank in parallel with the load must raise the overall power factor to 0.95 lagging. The real power stays the same. The capacitors are connected in Y. Find the capacitance Cph of each capacitor in µF.
P = 300 kWpf1 = 0.7pf2 = 0.95VL = 690 Vf = 50 Hz
Cph = µF
3Solutions
1.DC-DC converters
Answer: Imax = 0.767 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 = 7.5/15 = 0.5
Average and ripple current
Io = 7.5/12 = 0.625 A
ΔIL = VL·D/(L·f) = 7.5·0.5/(330·10⁻⁶·40·10³) = 0.2841 A
Peak current
Imax = 0.625 + 0.2841/2 = 0.767 A
Answer
Imax = 0.767 A
2.DC-DC converters
Answer: ΔVo = 511.4 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 = 1 − 15/60 = 0.75
Load current
Io = 60/10 = 6 A
Output ripple
ΔVo = 6·0.75/(40·10³·220·10⁻⁶) = 511.4 mV
Answer
ΔVo = 511.4 mV
3.DC motors
Answer: n0 = 2626 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 = 220/0.8 = 275 rad/s
n0 = 60·275/(2π) = 2626 rpm
Answer
n0 = 2626 rpm
4.DC motors
Answer: n2 = 1477 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.
Back-EMF before and after
E1 = 120 − 6·1.2 = 112.8 V
E2 = 120 − 3.6·1.2 = 115.7 V
New speed
n2 = 1440·115.7/112.8 = 1477 rpm
Answer
n2 = 1477 rpm
5.AC power & power factor
Answer: cos φ = 0.9004
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 = 18·tan(arccos 0.7) = 18.36 kvar
Q2 = 0 = 0 kvar
Totals
P = 18 + 20 = 38 kW
Q = 18.36 + 0 = 18.36 kvar
S = √(38² + 18.36²) = 42.2 kVA
Overall power factor
cos φ = P/S = 38/42.2 = 0.9004 lagging
Answer
cos φ = 0.9004
6.AC power & power factor
Answer: I = 61.69 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.
Reactive power before and after
The real power P does not change; only the reactive power drawn from the supply does.
Q1 = 17·tan(arccos 0.65) = 19.88 kvar
QC = V²·2πf·C = 400²·2π·60·33·10⁻⁶ = 1991 var = 1.991 kvar
Q2 = 19.88 − 1.991 = 17.88 kvar
New apparent power
S2 = √(17² + 17.88²) = 24.68 kVA
New line current
I = S2/V = 24675/400 = 61.69 A
Answer
I = 61.69 A
7.Three-phase power
Answer: P = 12.94 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.
Apparent power
S = √3·VL·IL = √3·415·20 = 14.38 kVA
Real power
P = S·cos φ = 14.38·0.9 = 12.94 kW
Answer
P = 12.94 kW
8.Three-phase power
Answer: Cph = 1387 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 0.7) = 300·1.02 = 306.1 kvar
Q2 = 300·tan(acos 0.95) = 300·0.3287 = 98.61 kvar
Reactive power from the capacitors
QC = Q1 − Q2 = 207.5 kvar
Capacitance per phase
In a Y bank each capacitor is connected between a line and the star point.