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Single Phase System

Single Phase System. I L. I R. Simple parallel circuits. R-L parallel circuit. V L = V R = V. I L lags V by p /2. I R and V are in phase. I R. V. .  /2. I T. I L. From the phasor diagram. or. G. V. . - /2. B L. Y. Admittance triangle.

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Single Phase System

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  1. Single Phase System

  2. IL IR Simple parallel circuits R-L parallel circuit VL = VR = V IL lags V by p/2 IR and V are in phase

  3. IR V  /2 IT IL From the phasor diagram or

  4. G V  -/2 BL Y Admittance triangle For parallel circuit we look at admittance = tan-1 (BL/G) = tan-1 (1/LG) cos = G/Y = Z/R Power factor

  5. I IR IC R-L parallel circuit VC = VR = V IC leads V by p/2 IR and V are in phase

  6. G V  -/2 IC IT /2 BL Y  V IR From the phasor diagram Admittance triangle = tan-1 (BC/G) = tan-1 (C/G) cos = G/Y = Z/R power factor

  7. Example 1 A circuit consists of a 115W resistor in parallel with a 41.5 mF capacitor and is connected to a 230 V, 50Hz supply. Calculate: (a) The branch currents and the supply current; (b) The circuit phase angle; (c) The circuit impedance

  8. continue

  9. Example 2 Three branches, possessing a resistance of 50W, an inductance of 0.15H and a capacitor of 100mF respectively, are connected in parallel across a 100V, 50Hz supply. Calculate: (a) The current in each branch; (b) The supply current; (c) The phase angle between the supply current and the supply voltage

  10. solution

  11. Parallel impedance circuits Impedance sometime has the magnitude and phase . For example we combine the resistance and inductance such in inductor. In practical inductor has resistance and inductance. If we have two impedances in parallel the current passing through impedance 1 will be I1 and in impedance 2 will be I2. To solve this, these current components can be resolve into two components, i.e active and reactive , thus (active) (reactive)

  12. fC f fB fA Example 3 A parallel network consists of branches A,B and C. If IA=10/-60oA, IB=5/-30oA and IC=10/90oA, all phase angles, being relative to the supply voltage, determine the total supply current.

  13. The current is 7.1o lags the voltage

  14. Example 4 • A coil of resistance 50W and inductance 0.318H is connected in parallel with a circuit comprising a 75W resistor in series with a 159mF capacitor. The resulting circuit is connected to a 230V, 50Hz a.c supply. Calculate: • the supply current; • the circuit impedance; resistance and reactance. Lags the voltage

  15. Leads the voltage

  16. (b) More capacitive

  17. Resonance circuits

  18. RLC serial circuit From phasor diagram, since the voltage VL (BO) and VC (CO) are in line, thus the resultants for these two component is DO (BO-CO) which is not involved in phase. AO is the voltage across R. Thus EO2= AO2 + DO2

  19. Therefore the impedance is When resonance or f =0 and Z=R I=V/R ; therefore

  20. Z [] I [A] Z R I 0 f [Hz] fr Current (I) and impedance (Z) vs frequensi (f) in R, L & C serial circuit

  21. At f=fr increasing of voltage occured where VL > V or VL = QV VC > V or VC = QV Q = VL/V = VC/V In serial circuit: VL = IXL; VC = IXC and V = IR Q = IXL/IR = IXC/IR Q = XL/R = XC/R Substitute XL and XC Q = rL/R = 1/rCR where r = 2fr Q is a circuit tuning quality

  22. Example 5 • A circuit having a resistance of 12 W, an inductance of 0.15H and a capacitance of 100 mF in series, is connected across a 100V, 50Hz supply. Calculate: • The impedance; • The current; • The voltage across R, L and C; • The phase difference between the current and the supply voltage • Resonance frequency (a) note XL=47.1W XC=31.85W

  23. (b) (c) (d) (e)

  24. ZL IL IC V VL  VR IL Parallel resonance circuit ZL=ZL/j ZL XL VC IC R Red notation for impedance triangle

  25. VC q IC I V IL Not at resonance

  26. IC I V IL At resonance At resonance q=0 , thus sin q =0

  27. When R is so small we can put R=0 then Previously we have Q factor of the circuit

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