1. Single Phase System

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

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

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

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

6. From the phasor diagramV IR IC IT 
/2
From the phasor diagram
Admittance triangle V G BL Y 
-/2
 = 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)

13. 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. fA fB fC f

14. The current is 7.1o lags the voltage

15. Example 4 Lags the voltage
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

16. Leads the voltage

17. (b)
More capacitive

18. Resonance circuits

19. 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

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

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

22. 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

23. 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

24. (b)
(c)
(d)
(e)

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

26. Not at resonance V IC IL I VC q

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

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