2p ffiffiffiffiffi
pLC ð1:141Þ
Qs¼ 1
2pCR 2p ffiffiffiffiffiffi pLC
ð1:142Þ
Qs¼1 R
ffiffiffiffiL C r
1.18 Parallel Resonance
A RLC parallel circuit where the inductor contains a small resistance is shown in Fig.1.41. This circuit is said to be in resonance when the power factor is unity, i.e., the reactive component of the current in the inductive branch is equal to the reactive component of the current in the capacitive branch. The phasor diagrams for dif-ferent currents are shown in Fig.1.42. The reactive component of the current in the inductive branch is,
IRL¼ ILsin/ ð1:143Þ
The reactive component of the current for the capacitive branch is,
IRC ¼ IC ð1:144Þ
Fig. 1.41 A RLC parallel ac circuit
The current in the capacitive reactance is, IC ¼ V
XC ð1:145Þ
The current in the inductive branch is, IL¼ V
ZL ð1:146Þ
sin/ ¼XL
ZL ð1:147Þ
According to the definition of parallel resonance, the following equation can be written as,
IC¼ ILsin/ ð1:148Þ
Substituting Eqs. (1.145), (1.146) and (1.147) into Eq. (1.143) provides, V
XC¼ V ZLXL
ZL ð1:149Þ
ZL2¼ XLXC ð1:150Þ
ZL2¼L
C ð1:151Þ
Equation (1.150) can also be derived in an alternative way. The parallel circuit would be in resonance when the j-component of the total admittance is zero. The total admittance of the circuit as shown in the Fig.1.41is,
Yt¼ YLþ YC¼ 1 Rþ jXL
þ 1
jXC
ð1:152Þ Fig. 1.42 Phasor diagram for
currents
1.18 Parallel Resonance 39
Yt¼R jXL
Consider the j-component of Eq. (1.154) is equal to zero which becomes, 1
From Eq. (1.154), the impedance at resonance can be derived as, Yt¼ 1
Zr¼ R R2þ XL2
þ 0 ð1:161Þ
Equation (1.161) can be modified as, 1 Zr¼ R
ZL2 ð1:162Þ
Substituting Eq. (1.157) into Eq. (1.162) yields, 1
From Eq. (1.164), it is observed that the impedance at resonance condition is equal to the ratio of the inductance to the product resistance and capacitance.
Exercise Problems
1:1 The excitation voltage of the circuit as shown in Fig. P1.1 is vðtÞ ¼ 20 sinðxt 45Þ V and the value of the impedance is Z ¼ 3 30j X. Find the current in time domain, and the instantaneous power.
1:2 The current and the impedance of the circuit as shown in Fig. P1.2 are iðtÞ ¼ 2 sinð314t þ 35Þ A and Z ¼ 5 45j X respectively. Find the voltage vðtÞ and the instantaneous power.
1:3 The current of 5 38j A is flowing through the impedance of 5 34j X.
Determine the average power absorbed by the impedance.
1:4 The expression of current through the 4 X resistance is given by iðtÞ ¼ 2 sin 3t A. Calculate the average power absorbed by the resistance.
1:5 The expression of current in the 2 þ j3 X impedance is given by iðtÞ ¼ 2 sin 3t 5 sin 4t A. Find the average power absorbed by the impedance.
1:6 The current of iðtÞ ¼ 3 sin t 2 sin 2t þ 6 sin 3t A flows through the impe-dance of 4 j10 X. Calculate the average power absorbed by the impedance.
1:7 Determine the average power of each element of the circuit as shown in Fig. P1.3. The value of the voltage excitation is vðtÞ ¼ 5 sinð314t þ 36Þ V.
1:8 In the circuit as shown in Fig.P1.4,find the total average power supplied and the power absorbed by the 3X and 4 X resistors respectively.
1:9 Determine the power supplied by the voltage source and the power absorbed by the resistive elements of the circuit as shown in Fig.P1.5.
1:10 Determine the total average power supplied by the source and the average power absorbed by the 2X resistor of the circuit as shown in Fig.P1.6.
Fig. P1.1 A simple series ac circuit
Fig. P1.2 A simple series ac circuit
Exercise Problems 41
1:11 Figure P1.7 shows a parallel ac circuit. Calculate the (i) source current, (ii) total complex power, (iii) total real power, and (iv) total reactive power.
1:12 Find the source current, total complex power and total real power of the circuit as shown in Fig.P1.8.
1:13 Figure P1.9shows an induction motor that is connected across the 220 V rms, 50 Hz source. A capacitor is connected across the induction motor to raise the power factor to 0.95 lagging. Determine the value of the capacitor.
Also determine the value of the capacitor if it is used to convert the power factor to 0.85 leading.
1:14 A 15 kW, wye-connected, three-phase induction motor with a lagging power factor of 0.85 is connected to an 220 V rms, 50 Hz source. A capacitor is Fig. P1.3 A simple RC series
circuit with voltage source
Fig. P1.4 A simple parallel ac circuit with current source
Fig. P1.5 A simple parallel ac circuit
Fig. P1.6 A parallel ac circuit with voltage source
connected across the motor to raise the power factor to 0.95 lagging. Find the value of the capacitor.
1:15 A 12 kW, delta-connected, three-phase induction motor with a lagging power factor of 0.80 is connected to the 120 V rms, 50 Hz source. A capacitor is connected across the induction motor to improve the power factor to 0.95 leading. Calculate the value of the capacitor.
1:16 Figure P1.10 shows the circuit where the load is connected to the source through the transmission line. Find the line current, the source voltage and the source power factor.
1:17 In the circuit shown in Fig.P1.11, determine the value of the source current.
1:18 A three-phase Y-connected generator is shown in the Fig. P1.12. The magnitude of the phase voltage is 130 V rms. For abc phase sequence, write down the phase voltages with angles, and calculate the line voltage.
1:19 A balanced three-phase Y-connected load is shown in the Fig.P1.13. The magnitude of the line-to-line voltage is 380 V rms. The resistance of each phase of the load is 15X. For abc phase sequence, write down the phase voltages, andfind the phase currents in the load.
Fig. P1.7 A simple parallel ac circuit
Fig. P1.8 A simple series ac circuit
Fig. P1.9 A parallel load with capacitor
Exercise Problems 43
Fig. P1.10 A series load
Fig. P1.11 A parallel load
Fig. P1.12 A wye-connected generator
Fig. P1.13 A wye-connected load
1:20 A wye-connected load with three coils, each having a resistance of 6 X and an inductive reactance of 8X is shown in the Fig. P1.14. For abc phase sequence and delta-connected source, find (i) the line currents with phase angles, (ii) the neutral current, and (iii) power factor.
1:21 An unbalanced three-phase Y-connected load is shown in the Fig.P1.15. The magnitude of the line voltage is 200 V rms. For abc phase sequence and delta-connected source, calculate (i) the line currents with angles, and (ii) the neutral current.
1:22 A balanced three-phase Y-connected load is driven by Y-connected source as shown in Fig.P1.16. Each phase of the load is having a resistance of 3X and a capacitive reactance of 6X. For abc phase sequence, determine the (i) phase voltages, (ii) line currents with phase angles, and (ii) power factor.
1:23 A three-phase Y-connected load is driven by Y-connected source as shown in the Fig.P1.17. The magnitude of the line-to-line voltage is 200 V rms and the
Fig. P1.14 A wye-connected load
Fig. P1.15 A wye-connected load
Exercise Problems 45
line impedances are Za¼ Zb¼ Zc¼ 2 þ j3 X. For abc phase sequence, cal-culate the (i) line currents with phase angles, and (ii) power factor of the load.
1:24 A balanced three-phase, D-connected load is shown in the Fig. P1.18. The per phase impedance of the load is 4þ j6 X. If 415 V, three-phase supply is connected to this load,find the magnitude of (i) the phase current, and (ii) the line current.
Fig. P1.16 A wye-connected load
Fig. P1.17 A wye-connected generator
Fig. P1.18 A
delta-connected generator
References
1. C.K. Alexander, M.N.O. Sadiku, Fundamentals of Electric Circuits, 5th edn (McGraw-Hill Higher Education, 12 March 2014)
2. M.A. Salam, Electrical Circuits, 1st edn. (Alpha Science International Ltd., Oxford, 2007) 3. J.D. Irwin, R.M. Nelms, Basic Engineering Circuit Analysis, 10th edn (Wiley, 2011)
References 47
Transformer: Principles and Practices
2.1 Introduction
There are many devices such as three-phase ac generator, transformer etc., which are usually used in a power station to generate and supply electrical power to a power system network. In the power station, the three-phase ac generator generates a three-phase alternating voltage in the range between 11 and 20 kV. The magni-tude of the generated voltage is increased to 120 kV or more by means of a power transformer. This higher magnitude of voltage is then transmitted to the grid sub-station by a three-phase transmission lines. A lower line voltage of 415 V is achieved by stepping down either from the 11 or 33 kV lines by a distribution transformer. In these cases, a three-phase transformer is used in either to step-up or step-down the voltage. Since a transformer plays a vital role in feeding an electrical network with the required voltage, it becomes an important requirement of a power system engineer to understand the fundamental details about a transformer along with its analytical behavior in the circuit domain. This chapter is dedicated towards this goal. On the onset of this discussion it is worth mentioning that a transformer, irrespective of its type, contains the following characteristics (i) it has no moving parts, (ii) no electrical connection between the primary and secondary windings, (iii) windings are magnetically coupled, (iv) rugged and durable in construction, (v) efficiency is very high i.e., more than 95 %, and (vi) frequency is unchanged.