Single Phase Parallel AC Circuits 1
Single Phase Parallel A.C. Circuits (Much of this material has come from Electrical & Electronic Principles & Technology by John Bird) n parallel a.c. circuits similar to those we will be considering, the voltage will be common to each branch, for that reason it will be taken as the reference phasor R-L Parallel A.C. Circuit From the phasor diagram, we see that R is in phase with the supply voltage V, and L lags V by 90 o. t will also be seen that the supply current is the phasor sum of R and L, which means that lags the supply voltage by an angle between 0 and 90 o. The relevant equations are: = R 2 + L 2 R = V R L = V X L tan ϕ = L R sin ϕ = L cos ϕ = R Z = V R-C Parallel A.C. Circuit As already stated in the previous section, R is in phase with the supply voltage, and we see that C leads V by 90 o. 2
The relevant equations are: = R 2 + C 2 R = V R C = V X C tan α = C R sin α = C cos α = R Z = V L-C Parallel A.C. Circuit n theory, there are three possible phase diagrams depending on the relative values of L and C. Two are shown above, for the third this would be when C = L and when this is the case = 0. n practice this third case is not possible because of the existence of circuit resistance. For the two cases shown: When L > C the current lags V by 90 o When C > L the current leads V by 90 o The relevant equations are: 3
L = V X L C = V X C = phasor difference between L and C Z = V Worked Example A pure inductor of 120mH is connected in parallel with a 25µF capacitor and the network is connected to a 100v, 50Hz supply. Determine (a) the branch currents, (b) the supply current and its phase angle, (c) the circuit impedance and (d) the power consumed. 1) The branch currents X L = 2πfL = 2π 50 120 10 3 = 37.7Ω X c = 1 = 1 = 127.3Ω 2πfC 2π 50 25 10 6 L = V X L = 100 37.7 = 2.653A C = V X C = 100 127.3 = 0.786A 2) Supply current = L C = 2.653 0.786 = 1.867A and lags the supply voltage by 90 o 3) Circuit impedance Z = V = 100 1.867 = 53.56Ω 4) Power consumed (the same power calculation equation apply as for series ac circuits) P = V cos ϕ = 100 1.867cos90 = 0W Note: n a parallel a.c. circuit with no resistor present, the power dissipated will always be 0. 4
R-L-C Parallel A.C. Circuit The phasor diagrams for the circuit is shown below C R V R ϕ L - C L From the current triangle, we see that = R 2 + ( L C ) 2 and tanϕ = ( L C ) Since R = V R, L = V X L and C = V X C we can write R = ( V R )2 + ( V X L V X C ) 2 = V Z Therefore 1 Z = ( 1 R )2 + ( 1 X L 1 X C ) 2 The above equation leads us to three new quantities which are found to be useful in the solving of parallel a.c. circuits. The three quantities are: Admittance Y This is the reciprocal of impedance and has units of siemens (S) Y = 1 Z Note: The total admittance of a number of admittances in parallel is found by summing the individual admittances. 5
Conductance G This is the reciprocal of resistance, and also has units of siemens G = 1 R Note: n an a.c. circuit the conductance is the reciprocal of resistance only when the circuit possesses no reactance. Susceptance B This is the reciprocal of reactance, it too has units of siemens B L = 1 X L and B C = 1 X C n series a.c. circuits the opposition to current flow is impedance, Z which has two components, resistance, R and reactance, X and from these we are able to construct a impedance triangle. n parallel a.c. circuits, admittance has two components, conductance, G and susceptance, B. These make it possible to construct an admittance triangle. See below. Z X ϕ R mpedance Triangle ϕ Y G B Admittance Triangle From the admittance triangle, we see that Y = G 2 + (B L B C ) 2 Where Y = 1 Z G = 1 R B L = 1 X L = 1 ωl 6
B C = 1 X C = ωc Therefore Y = ( 1 R )2 + ( 1 ωl ωc)2 and Z = 1 ( 1 R )2 +( 1 ωl ωc)2 Also tan ϕ = B G Admittance in a parallel a,c, circuit is a complex quantity and may be written as Y = G jb = G 2 + B 2 tan 1 B G Worked Example A 50Ω resistor, a 20mH coil and a 5uF capacitor are all connected in parallel across a 50V, 100Hz supply. Calculate the total current drawn from the supply, the current for each branch, the total impedance of the circuit and the phase angle. Also, construct the current and admittance triangles representing the circuit. X L = ωl = 2π 100 20 10 3 = 12.56Ω X C = Z = 1 2π 100 5 10 1 ( 1 R )2 +( 1 1 2 ) X L X C 6 = 318.3Ω = = V Z = 50 12.65 = 3.95A R = V R = 50 50 = 1A L = V X L = 50 12.56 = 3.98A 1 = 12.65Ω ( 1 50 )2 +( 1 12.56 1 )2 318.3 7
C = V X C = 50 318.3 = 0.157A For the total current, could also use = R 2 + ( L C ) 2 = 1 2 + (3.98 0.157) 2 = 3.95A Admittance Y = 1 = 1 = 0.0796S = 79.6mS z 12.56 Conductance G = 1 = 1 = 0.02S = 20mS R 50 nductive susceptance B L = 1 X L = 1 12.56 = 0.0796S = 79.6mS Capacitive susceptance B C = 1 = 1 = 0.00314S = 3.14mS X C 318.3 Phase angle ϕ = tan 1 B G = tan 1 79.6 3.13 20 = 75.34 lagging Note: On the above diagram, figures have been rounded up and down. Power in Parallel a.c. Circuits The equations that applied to series a.c. circuits also apply here Tutorial Problems 1) A 30Ω resistor is connected in parallel with a pure inductance of 3mH across a 110v, 2kHz supply. Calculate (a) the current in each branch, (b) the circuit current, (c) the circuit phase angle, (d) the circuit impedance, (e) the power consumed and (f) the circuit power factor. (3.67A, 2.92A, 4.69A, 38.5 o lagging, 23.45Ω, 404W, 0.783 lagging) 2) A 40Ω resistor is connected in parallel with a coil of inductance L and negligible resistance across a 200v, 50Hz supply and the supply current is found to be 8A. Sketch the phasor diagram and determine the inductance of the coil. (102mH) 3) A 1500nF capacitor is connected in parallel with a 16Ω resistor across a 10v, 10kHz supply. Calculate (a) the current in each branch, (b) the supply current, (c) the circuit 8
phase angle, (d) the circuit impedance, (e) the power consumed, (f) the apparent power and (g) the circuit power factor. (0.625A, 0.943A, 1.131A, 56.46 o leading, 8.84Ω, 6.25W, 11.31vA, 0.552 leading) 4) A capacitor C is connected in parallel with a resistance R across a 60v, 100Hz supply. The supply current is 0.6A at a power factor of 0.8 leading. Calculate the value of R and C. (9.55µF, 125Ω) 5) An inductance of 80mH is connected in parallel with a capacitor of 10µF across a 60v, 100Hz supply. Determine (a) the branch currents, (b) the supply current, (c) the circuit phase angle, (d) the circuit impedance and (e) the power consumed. (0.377A, 1.194A, 0.817A, 90 o lagging, 73.44Ω, 0W) 6) Repeat Q5 but with a supply frequency of 200Hz. (0.597A, 0.754A, 0.157A, 90 o leading, 382.2Ω, 0W) 7) A 1kΩ resistor, a 142mH coil and a 160µF capacitator are all connected in parallel across a 240v, 60Hz supply. Calculate: (a) The total current drawn from the supply (b) The current for each branch (c) The total impedance, conductance, inductive and capacitive susceptance, and admittance for the circuit (d) The phase angle ( =10A, R = 0.24A, L = 4.48A, C = 14.47A, Z = 24Ω, G = 0.001S, BL = 0.0187S, BC = 0.0603S, Y = 0.0416S, ϕ = 88.62 o leading) 9
mpedances in Parallel The summation of the branch current is now quite difficult since they no longer remain wither in phase or quadrature with each other. For this reason, it becomes necessary to introduce a new analytical device current component. Consider the diagrams below in which the current is shown to lag or lead the voltage. n each case the current is made up of two components: 1. cos ϕ which is in phase with the voltage and is called the active or power component. 2. sin ϕ which is in quadrature with the voltage and called the quadrature or reactive component. t follows that 2 = ( cos ϕ) 2 + ( sin ϕ) 2 The Addition of Current Phasors Consider the diagrams below. = 1 + 2 (phasor sum) cos ϕ = 1 cos ϕ 1 + 2 cos ϕ 2 And sin ϕ = 1 sin ϕ 1 + + 2 sin ϕ 2 So that 2 = ( cos ϕ) 2 + ( sin ϕ) 2 t can also be shown that ϕ = cos 1 cos ϕ 1 + 2 cos ϕ 2 10
Consider the following Example: f V = 230 V, 50 Hz, Calculate: (a) The supply current (b) The circuit impedance, resistance and reactance. (a) X L = 2πfL = 100 Ω Z 1 = R 1 2 + X L 2 = 112 Ω 1 = V Z 1 = 230 112 = 2.05 A ϕ = tan 1 X l R = tan 1 100 50 Therefore 1 = 2.05 / -63.43 o X c = 1 2πfC = 20 Ω = 63,43 (lagging) Z 2 = R 2 2 + X C 2 = 77.7 Ω 2 = V Z 2 = 230 77.7 = 2.96 A ϕ = tan 1 X C 1 20 = tan = 15 (leading) R 2 75 Therefore 2 = 2.96 / 15 o So, we can sketch in the following 11
Now cos ϕ = 1 cos ϕ 1 + 2 cos ϕ 2 = 3.77 A And sin ϕ = 1 sin ϕ 1 + 2 sin ϕ 2 = -1.07 A So that = 3.77 2 + 1.07 2 = 3.92 A (b) Z = V = 230 3.92 = 58.7 Ω cos φ R = Z cos φ = Z = 56 Ω sin φ X = Z sin φ = Z = 16 Ω 12
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