BFF1303: ELECTRICAL / ELECTRONICS ENGINEERING Alternating Current Circuits : Basic Law Ismail Mohd Khairuddin, Zulkifil Md Yusof Faculty of Manufacturing Engineering Universiti Malaysia Pahang
Alternating Current Circuit (AC)-Basic Laws & Circuit Techniques BFF1303 ELECTRICAL/ELECTRONICS ENGINEERING Faculty of Manufacturing Universiti Malaysia Pahang Kampus Pekan, Pahang Darul Makmur Tel: +609-424 5800 Fax: +609-4245888 Contents: Outcome Kirchhoff s Law Series Impedances and Voltage Division Parallel Impedances and Current Division Nodal Analysis Mesh Analysis Superposition Source Transformation Thevenin and Norton Equivalent Circuit BFF1303 Electrical/Electronic Engineering 2 2
Draw the power triangle, and compute the capacitor size required to perform power factor correction on a load. Convert time-domain sinusoidal voltages and currents to phasor notation, and vice versa; represent circuit using impedance Learn complex power notation; compute apparent power, real, and reactive power for complex load. Solve steady-state ac circuits, using phasors and complex impedances. BFF1303 Electrical/Electronic Engineering 3 3
This principles used in dc analysis, are also applicable in the phasor domain. The difference is simply the voltages, currents and resistance/inductance/capacitance are converted to phasor and impedance. Kirchhoff s current law (KCL) that the algebraic sum of phasor currents entering a node (or a closed boundary) is zero. Kirchhoff s voltage law (KVL) the algebraic sum of all phasor voltages around a closed path (or loop) is zero. N I n1 n 0 M V m1 m 0 BFF1303 Electrical/Electronic Engineering 4 4
Consider the following figure, the same current I flow through the impedance Applying KVL around the loop BFF1303 Electrical/Electronic Engineering 5 5
The equivalent impedance at the input terminals V Z eq Z1 Z2 Z I Any number of impedances connected in series is the sum of the individual impedances. N If N = 2 The current through impedances I Z V Z 1 2 BFF1303 Electrical/Electronic Engineering 6 6
Since V Z I 1 1 V Z I 2 2 Then Z 1 2 1, Z V V V 2 V Z1 Z2 Z1 Z2 Principle of voltage division. BFF1303 Electrical/Electronic Engineering 7 7
The voltage across each impedance is the same. Applying KCL at top node 1 1 1 I I1 I2 I N V Z 1 Z 2 Z N BFF1303 Electrical/Electronic Engineering 8 8
The equivalent impedance 1 1 1 1 1 Z V Z Z Z eq 1 2 N When N = 2 1 1 1 2 eq 1 1 ZZ Z Z Z Z Z 1 2 1 2 BFF1303 Electrical/Electronic Engineering 9 9
Since V IZ I Z I Z eq 1 1 2 2 The current in the impedances are Z 2 1 1, Z I I I 2 I Z1 Z2 Z1 Z2 BFF1303 Electrical/Electronic Engineering 10 10
Find the input impedance of the circuit shown. Assume that the circuit operates at ω = 50 rad/s Solution To get Z in, we combine resistors, resistor-capacitor and resistorinductor in series and in parallel. BFF1303 Electrical/Electronic Engineering 11 11
Let Z Z Z 1 2 3 Impedance of the 2 mf capacitor Impedance of the 3 Ω resistor in series with the 10 mf capacitor Impedance of the 0.2 H inductor in series with the 8 Ω resistor Then Z Z Z 1 2 3 1 j10 jc 1 3 3 j2 jc 8 jl 8 j10 BFF1303 Electrical/Electronic Engineering 12 12
in in in The input impedance Z Z Z Z Z Z 3 j28 j10 1 2 3 j10 11 j8 3.22 j11.07 11.52 73.78 BFF1303 Electrical/Electronic Engineering 13 13
Determine the input impedance of the circuit in figure shown at ω = 10 rad/s. Answer Z in 149.52 j195 245.7352.52 BFF1303 Electrical/Electronic Engineering 14 14
Determine v o t for the given circuit. Solution Transform the time domain equivalent to phasor form V s 20 15 1 V Z C j25 jc ZL jl j20 BFF1303 Electrical/Electronic Engineering 15 15
Let Z Z 1 2 Impedance of the 60 Ω resistor Impedance of the parallel combination of the 10 mf capacitor and the 5 H inductor Z1 60 Z2 ZC ZL j100 BFF1303 Electrical/Electronic Engineering 16 16
By using voltage divider V V V o o o Z Z2 Z 1 2 V j100 20 15 60 j100 17.1515.96V In time domain v t 17.15cos 4t 15.96 V o BFF1303 Electrical/Electronic Engineering 17 17
Find v o t in the given circuit Answer v t 35.36 cos 10t 105 V o BFF1303 Electrical/Electronic Engineering 18 18
The basis of nodal analysis is Kirchhoff s Current Law. Since KCL is valid for phasors, we can analyze ac circuit by nodal analysis. BFF1303 Electrical/Electronic Engineering 19 19
Find i x in the given circuit using nodal analysis Solution Convert the circuit to phasor form BFF1303 Electrical/Electronic Engineering 20 20
20 cos 4t 200 1H jl j4 0.5H jl j2 0.1F 1 j2.5 jc BFF1303 Electrical/Electronic Engineering 21 21
Applying KCL at node V 1 20 V V V V 10 j2.5 j4 1 1 1 2 1 j1.5 V j2.5v 20 1 2 And I x = V 1 j2.5 2V V V V j2.5 j4 j 1 1 2 2 11V 15V 0 1 2 Then in matrix form Applying KCL at node V 2 2I x V V V j4 j 1 2 2 1 j1.5 j2.5 V1 20 11 15 2 0 V BFF1303 Electrical/Electronic Engineering 22 22
Then by using Cramer s Rule V 1 20 j2.5 0 15 300 18.97 18.43 V 1 j1.5 j2.5 15 j5 11 15 V 2 1 j1.5 20 11 0 220 13.91198.3V 1 j1.5 j2.5 15 j5 11 15 BFF1303 Electrical/Electronic Engineering 23 23
Then I x I x V1 18.9718.43 7.59108.4A j2.5 2.590 In time domain i t 7.59 cos 4t 108.4 A x BFF1303 Electrical/Electronic Engineering 24 24
Find v 1 and v 2 in the given circuit using nodal analysis Answer v t 11.325cos 2t 60.01 V 1 2 v t 33.02 cos 2t 57.12 V BFF1303 Electrical/Electronic Engineering 25 25
Compute V 1 and V 2 in the following circuit Answer V 1 V 2 25.78 70.48V 31.41 87.18V BFF1303 Electrical/Electronic Engineering 26 26
Kirchhoff s Voltage Law form the basis of mesh analysis. Since KVL is valid for phasors, we can analyze ac circuit by mesh analysis. BFF1303 Electrical/Electronic Engineering 27 27
Determine I o for the given circuit by using mesh analysis. Solution Apply KVL to mesh 1 BFF1303 Electrical/Electronic Engineering 28 28
8 j10 j2 I j2 I j10 I 0 Apply KVL to mesh 2 1 2 3 4 j2 j2 I j2 I j2 I 290 0 2 1 3 For mesh 3 I3 5 Then 8 j8 I j2i j50 1 2 j2i 4 j4 I j30 1 2 BFF1303 Electrical/Electronic Engineering 29 29
I In matrix form By using Cramer s Rule 2 8 j8 j2 I1 j50 j2 4 j4 2 j30 I 8 j8 j50 j2 j30 416.17 35.22 6.12 35.22 A 8 j8 j2 68 j2 4 j4 The desired current I I o 2 6.12 144.78 A BFF1303 Electrical/Electronic Engineering 30 30
Solve for V o in the following circuit using mesh analysis Answer V 9.756 o 222.32 V BFF1303 Electrical/Electronic Engineering 31 31
Thevenin and Norton theorem are applied to AC circuit in the same way as they are to DC circuits. The only additional effort arises from the need to manipulate complex number. If the circuit has sources operating at different frequencies the Thevenin or Norton equivalent circuit must be determined at each frequency. BFF1303 Electrical/Electronic Engineering 32 32
V Z I Z Z Th N N Th N BFF1303 Electrical/Electronic Engineering 33 33