Single Phase System

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

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

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

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

From the phasor diagram V 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

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

continue

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

solution

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)

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

The current is 7.1o lags the voltage

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

Leads the voltage

(b) More capacitive

Resonance circuits

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

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

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

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

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

(b) (c) (d) (e)

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

Not at resonance V IC IL I VC q

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

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