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AC Circuits 2 1 For Capacitor Current For Inductance Leads Voltage Current Voltage Leads 2 Real in phase quantities along the real axis Quadrature Components Along the j – j axis 3 Draw voltage phasor diagram for the above circuit 4 VS VL-VC θ VR VS Find VS VS 6 2 4 2 7.2 Volts VL VC 4 arctan arctan arctan0.66 6 33.690 VR 6 5 VS VL-VC θ VR VS VS 6 4 7.2 Volts 2 arctan 2 VL VC 4 arctan arctan0.66 6 33.690 VR 6 VS 6 j 4Volts in Cartesian form or 7.2V33.690 in Polar form 6 Power in A.C. Circuits Consider the purely resistive circuit shown. The voltage and current are in phase as shown (on the next slide) 7 Power in A.C. Circuits Voltage and current in phase Power V̂sint Iˆ sin ωt power V̂Î sin t 2 8 Power in A.C. Circuits Power V̂sint Iˆ sin ωt power V̂Î sin t 2 using cos2 A 1-2 sin 2 A 1 sin 2 A (1 cos 2 A) 2 1 thus sin t 1 cos 2t 2 2 1 ˆˆ sub into above..... .. power VI 1 cos 2ωt 2 1 ˆˆ 1 ˆˆ power VI VI cos 2ωt 2 2 Fixed component Time varying component 9 Resultant power curve for resistor in a.c. circuit 10 Resultant power curve for resistor in a.c. circuit ÎV̂ average power watts 2 11 Resultant power curve for resistor in a.c. circuit ÎV̂ average power watts 2 Iˆ I rms 2, Vˆ Vrms 2 thus average power I rms 2 Vrms 2 I rmsVrms watts 2 12 Voltage and current 900 out of phase 13 Voltage and current 300 out of phase 14 Voltage and current 600 out of phase 15 Apparent power (S) = V x I (volt-amperes, VA) True power (active power) (P) = Apparent power x Cos θ (W) Reactive power Q = VI sin θ reactive (voltamperes var) 16 It can be shown that the power in an AC circuit is equal to power VI cos θ Where is the phase angle between th e current and voltage cos is known as the power factor p.f. R p.f. also equals Z 17 Series Circuits 18 In a series circuit consisting of L,C and R if XC = XL the voltage across each will be equal but in anti-phase. As can be seen from the phasor diagram they will cancel so that the series is effectively a pure resistance. Adding VL and VC VL Leaves just VR VR This condition is known as Series resonance VC 19 What happens if the frequency rises or falls ? The frequency at which this condition occurs is known as the Resonant Frequency XC X L 1 2fL 2fC 1 4 f LC 2 f 2 1 2 LC Hertz 20 Remember that the impedance Z R 2 X L X C 2 It should be clear that at resonance when X C X L that opposition to current flow will be minimum and only limited by the ohmic resistance in the circuit Also, as tan XC X L R the phase difference between voltage and current will be 0 0 What is the p.f. of a series resonant circuit ? 21 What is the resonant frequency of the circuit ? f 1 2 LC Hertz 10 Volts 1 f 159.15 Hertz 2 1H 1F Plot the impedance of the above series circuit from 140 Hertz to 180 Hertz Plot a graph of current flowing for the same frequency range 22 And that the current flowing is Maximum . f 0 159 Hz IMAX Current IMIN Note that the Impedance is purely resistive and minimum at the resonant frequency 23 Q factor Q factor is a measure of merit for a resonant circuit It is defined as the ratio of reactive power of either the inductor or the capacitor to the average power of the resistor at resonance reactive power QS average power I2XL XL 2 R I R Note that the quantity is dimensionless 24 25 Admittance (Y) is defined as 1/Z a.c. equivalent of the d.c. conductance G = 1/R Unit is siemens (S) Product or for two use Sum 26 Admittance (Y) is defined as 1/Z e.g. the Admittance of a circuit of impedance (4 j6) IS 1 4 6 j 4 6 j 4 6 j S 4 6 j 4 6 j 4 6 j 16 36 52 Note use of complex conjugate Note also it would have been easy to convert to polar first 4 j 6 7.256.3 0 the reciprocal of which os 0.139S 56.30 27 1 Susceptanc e B is the reciprocal of reactance i.e. X for inductance for capacitanc e 1 X L 90 0 1 X C 900 or 1 2fL90 0 siemens, S or 2fC900 siemens, S 28 Find admittance of each branch Find the total i/p admittance Calculate the i/p impedance 1 1 YR G0 00 0o R 20 0.05 S00 ( 0.05 j 0 )S 0 YL 1 1 0 90 X L900 10 0.1 S 900 ( 0 j 0.1 )S YT YR YL 0.05 j0 0 j0.1 S ( 0.5 j 0.1 )S 0.1118 63.430 S ZT 1 8.9463.430 ( 4 j 8 ) YT 29 Draw admittance phasor YT ( 0.5 j 0.1 ) S 0.1118 63.430 S j -j 30 Draw impedance vector ZT 1 8.9463.430 YT ( 4 j 8 ) Z=8.94Ω 63.430 It should be clear that the parallel circuit may be replaced by a series circuit of (4+j8)Ω 31 Determine the impedance of the circuit ZT Z1 // Z 2 Use product over sum ZT 6 j 64 j 5 6 j 6 4 j5 54 j 610 j 10 j 10 j 54 j 6 10 j 534 j114 10 1 2 Or in polar form 2 conjugate 5.287 j1.129 5.4120 32 impedance of the circuit 5.287 j1.128 5.4120 Determine the current from the source VS IS ZT 50V00 5.412 0 9.26 A 120 9.06 j1.63A Determine the current through Z1 and Z2 50V00 I Z1 4.166 j 4.166 A 6 j 6 4.878 6.097A 50V00 IZ 2 4 j5 33 I S 9.06 j1.63 A I Z 1 4.166 j 4.166 A I Z 2 4.878 6.097A If the calculations up to now had not been rounded at each stage then if the two branch currents were added they should give the same value as the source current 34