AC Circuit Theorems

Wednesday, 20 February 2019

6:04 PM

<<ELEC2134_Circuits3_ACCircuitTheorems.pdf>>
Machine generated alternative text: ELEC2134: Circuits and Signals AC Circuit Theorems Impedance (Z), Admittance (Y), Conductance (G) and Susceptance (B) Delta ( -Wye (Y) Transformation in AC networks Independent and dependent current and voltage sources Nodal Analysis Mesh Analysis Superposition Theorem Thévenin s Professor Eliathamby Ambikairajah
Machine generated alternative text: Impedance and Admittance The reciprocal of resistance is known as conductance ( G ) and is expressed in siemens (S) or óï ò ¥²±¬» ¬¸·­ ·­ ±²´§ ¬®«» ·º ¬¸» ®»¿½¬¿²½» ·­ ¦»®±ô ·ò»ò ®»­·­¬·ª» ½·®½«·¬­ ±²´§£ G = The conductance ( G ) of a resistor with a resistance of 5 is equal to 200 mS(i.e. G = ) At times it is convenient to work with the reciprocal of impedance ( Z ) known as admittance ( Y ) Y = Y is the admittance, measured insiemens(S) or óï Z is the impedance, measured in ohms ( ÷ Admittance is used in parallel circuit analysis 2 Prof E Ambikairajah, UNSW, Australia
Machine generated alternative text: Susceptance The admittance Y can be written as Y = = G + jB The imaginary part of admittance is called susceptance and is denoted by B and measured in siemens (S) or óï . The two components of the susceptance are capacitive susceptance B C and the inductive susceptance B L . G is theconductance measured in siemens (S) or óï . Even though Y and Z are reciprocal to each other, in general, R and G are not reciprocal to each other and the same holds for X and B . 3 Prof E Ambikairajah, UNSW, Australia
Machine generated alternative text: Relationships between R , X , G and B [5] G = and B = { G = 1/ R only if X =0} Conversely, if G and B are known, R and X can be found as R = and X = 4 Prof E Ambikairajah, UNSW, Australia jB G jX R Z Y Y = = = = - j = G + jB Conductance (G) Susceptance (B) A simple series and parallel equivalents Resistance (R) Reactance (X)
Machine generated alternative text: Exercise 3.1 5 Prof E Ambikairajah, UNSW, Australia Assuming rad/s, Find G and B first and hence find R p and C p jB G C s R s Z Y 20 1 F = =Rp C P = and = Basic Elements : Impedance and Admittance R = R L = L C = =
Machine generated alternative text: 6 Prof E Ambikairajah, UNSW, Australia Consider the following parallel circuit and compute the impedance Z: Since this is a parallel circuit, it is easier to compute the admittance (Y) first: Y = = + + = + + = + } The reciprocal of admittance yields impedance. i.e. Z = Alternatively Z = R // jX L // (1/ jX C ) = 0.25// j 0.5 //(1/ j 5) R z C L Example 3.1 Alternatively Z = R// // (1/ ) Compute the impedance (Z) and Y (admittance) for the circuit shown below: R=0.25 Y X C =5 X L =0.5 Compute the admittance first: Y = + + = + + j 5 = 4 -2 j + j 5 = 4 + j 3 = 5 ± óï Since the impedance is the reciprocal of admittance Z = = = 0.2 óíêòç ± Z = 0.16 j 0.12
Machine generated alternative text: Example 3.2 Fortheparallelcircuitshownbelow,Findthebranchcurrents( and ),andalsothe supplyvoltage.Alsodrawthephasordiagram.Youmayassumethatthephaseangleof thesupplycurrent iszerodegrees. 7 Prof E Ambikairajah, UNSW, Australia = 10 ± Z 2 = 6 j 8 = 10 óëíòïí ± Y 2 = = =0.1 ± ã Åêð õ èðà ïð óí óï Z 1 = 8 + j 15 = 17 ± Y 1 = = =0.05882 óêïòçí ± ã Åîéòêè ëïòçà ïð óí óï ̱¬¿´ ¿¼³·¬¬¿²½» Ç ã Ç ï õ Ç î ã Åèéòêè õ îèòïà ïð óí ã çîòðé ïð óí ± ̱¬¿´ ׳°»¼¿²½» Z = = = ïðòèê óïéòéé ± Í«°°´§ ª±´¬¿¹» Ê ã × Æ ã øïð ± ÷øïðòèê óïéòéé ± óïéòéé ± Ê Ì¸» ¾®¿²½¸ ½«®®»²¬ 1 = = =6.39 óéçòé ± ß Ì¸» ¾®¿²½¸ ½«®®»²¬ 2 = = =10.86 ± ß Phasordiagram
Machine generated alternative text: Example 3.3 Calculate the total impedance of the circuit given below: 8 Prof E Ambikairajah, UNSW, Australia Method 1: = R = 250 = = 0.004 = j j 245.044 = = - j 0.0041 = = - j 1768.388 = = j (5.655) Total admittance (Y) = + + Y = 0.004 -j0.0035 = óìïòî ± ̱¬¿´ ׳°»¼¿²½» ø¦÷ ã = ± Method 2: = = 0 = - j 0.490A = = j 0.068A Total current (I) = + + = 0.48 - j 0.422 = óìïòí ± ̱¬¿´ ׳°»¼¿²½» ø¦÷ ã Z = = ±
Machine generated alternative text: Exercise 3.2 9 Prof E Ambikairajah, UNSW, Australia Q1: Compute the total impedance (Z T ) and the admittance (Y T ) for the circuit shown below: Q2: Determine the input impedance (Z) of the circuit shown below at =10 rad/s. 50 Z 4mF 2H 20 2mF Ans: Z = 32.38 j73.76 R2=20 Z T ,Y T Z C2 =-j16 R1=10 Z L2 =j5 Z C1 =-j8 Z L1 =j13 Ans: Z T = ± Y T = óíëòî ±
Machine generated alternative text: -Wye (Y) Transformation in AC networks A delta network can be converted easily to a wye network (or Star network) or vice-versa. The concept of this conversion is normally used in three-phase AC circuits. 10 Prof E Ambikairajah, UNSW, Australia Delta ( to Wye (Y) Transformation Each impedance in the Wye network is the product of the impedance in the two adjacent branches, divided by the sum of the three impedances Wye (Y) to Delta ( Transformation Each impedance in the network is thesum of all possible products of Y impedances taken two at a time, divided by the opposite Y impedance.
Machine generated alternative text: Balanced Delta circuit 11 Prof E Ambikairajah, UNSW, Australia A Wye or Delta circuit is balanced if it has equal impedances (load) in all three branches. That is, For a balanced load For a balanced load For a balanced load the equivalent delta connected impedance is 3 times that of the Wye or star connected impedance. For a balanced load the equivalent star or Wye connected impedance is 1/3 times that of the delta connected impedance.
Machine generated alternative text: Example 3.4 12 Prof E Ambikairajah, UNSW, Australia Convert the network of resistors shown below to a Y network of resistors: network Ynetwork Delta ( ) to Wye (Y) Transformation
Machine generated alternative text: Example 3.5 13 Prof E Ambikairajah, UNSW, Australia Find the current I for the circuit shown:
Machine generated alternative text: Alternative forms of Y and networks 14 Prof E Ambikairajah, UNSW, Australia T-network Pi -network -network Y -network PI and T networks are commonly used in power transmission line models.
Machine generated alternative text: Exercise 3.3 15 Prof E Ambikairajah, UNSW, Australia Q1: Calculate the impedance ( Z ab ) for the network shown below: ß²­æ éòëêé õ ¶ðòëçìê Z ab Q2: Use a -to -Y transformation to find the current in the circuit shown below: 10 -j15 j40 40 ïíê ð ± Ê ó õ 50 14 I ß²­æ × ± ß
Machine generated alternative text: Independent and dependent current and voltage sources 16 Prof E Ambikairajah, UNSW, Australia Independent Voltage source Dependent Voltage source Independent current source Dependent current source Dependent source are drawn with a diamond outline Dependent source is controlled by another voltage or current. The dependent current source supplies 0.5 times the current flowing through the 4 ohm resistor in the above example Example
Machine generated alternative text: Current and Voltage sources 17 Prof E Ambikairajah, UNSW, Australia Voltage source An ideal voltage source has zero internal resistance ( R S = 0) and it cannot exist in nature A practical voltage source has a very small internal resistance. This resistance ( R S ) is typically less than 1 ohm. (A car battery has an internal resistance of less the 0.1 ohm) The internal resistance of a voltage source appears in series with an ideal voltage source(V s ) R S I S Current source An ideal current source has an infinite internal resistance ( R S = ) and it cannot exist in nature A practical current source has a very high internal resistance( R S ) . This resistance is typically in the order of a Mega ohm. The internal resistance ( R S ) of a practical current source is always shown in parallel with an ideal current source ( I S ) Note: The current sources are rarely built with a battery and a resistor combination. Instead, transistors are connected in such a way as to produce current sources. Ideal voltage source and its I-V characteristic 0 - õ ó Ê ­ + ª Load × × ª 0 - × ­ + ª Load × Ê × Ideal current source and its I-V characteristic
Machine generated alternative text: Example 3.6: Node Voltage Analysis 18 Prof E Ambikairajah, UNSW, Australia Use node voltage analysis to find the voltages and in the circuit shown below: Convert to Phasor Domain Nodal Analysis at Node 1 provides: (1) Nodal Analysis at Node 2 provides: Node 1 Node 2 (2) Nodal analysis applies KCL to find node voltages in a given circuit.
Machine generated alternative text: 19 Prof E Ambikairajah, UNSW, Australia We have to solve following two simultaneous equations to find V 1 and V 2 Determinant of the matrix is: : In time domain, cos cos
Machine generated alternative text: Exercise 3.4 20 Prof E Ambikairajah, UNSW, Australia Q1: Find V 1 and V 2 via nodal analysis for the circuit shown below: Q2: Find and via nodal analysis for the circuit shown below: {Combine the parallel elements to obtain a simpler circuit and then apply nodal analysis} ß²­æ V 1 ± Ê V 2 ± Ê ß²­æ ã ½±­øïð¬óìíòìç ± ÷ Ê ã ½±­øïð¬óìéòêé ± ÷ Ê v 1 0.2H î½±­ïð¬ ß 1 I 2 0.1F 0.05F 0.1H õ ë½±­ïð¬ Ê ó v 2 î íð ± ß 2 -j1 j2 I V 2 2V 1 V 1 V 2 Ü»°»²¼»²¬ ½«®®»²¬ ­±«®½»æ îÊ ï ¿³°­
Machine generated alternative text: Example 3.7:Mesh current Analysis 21 Prof E Ambikairajah, UNSW, Australia Find I o for the following circuit using mesh analysis. For mesh 1, KVL provides: For mesh 2 , KVL provides: For mesh 3 : Note: By solving the above three equations we obtain: 1 2 3 Mesh analysis applies KVL to find mesh or loop currents in a given circuit. be the mesh currents in the circuit
Machine generated alternative text: Exercise 3.5 22 Prof E Ambikairajah, UNSW, Australia Q2: Use the mesh-current method to find the phasorcurrent in the circuit shown below: ß²­æ ± ß 2 j2 -j5 1 3 õ ó ííòè ð ± Ê V x õ ó 0.75V x × Ü»°»²¼»²¬ ½«®®»²¬ ­±«®½»æ ðòéëÊ ¨ ¿³°­ Q1: Solve for the mesh currents shown in the circuit below: ß²­æ ø¬÷ã ïòìïì ½±­øïððð¬óìë ± ÷ ß ø¬÷ã ½±­øïððð¬÷ ß 100 i 1 (t) 5 F ïðð ½±­øïððð¬÷ Ê 0.1H õ ó 0.1H i 2 (t) Q3: Find and via mesh analysis: -j3 ê ð ± Ê 1 j2 õ ó õ ó × ï × î ì× ï ß²­æ óêëòîí ± ß óïéíòêê ± ß
Machine generated alternative text: The Principle of superposition 23 Prof E Ambikairajah, UNSW, Australia Step1:Retainoneindependentsourceatatimeinthe circuitandreplaceallothersourceswiththeirinternal impedances. Step2:Determinetheoutput(currentorvoltage)dueto thesinglesourceactingaloneusinganycircuitanalysis methods. Step3:Repeatsteps1and2foreachoftheother independentsources. Step4:Findthetotalcontributionbyaddingalgebraically allthecontributionsduetotheindependentsources. Voltage source short circuit Current source open circuit
Machine generated alternative text: Example 3.8: Superposition Theorem 24 Prof E Ambikairajah, UNSW, Australia Calculate for the circuit shown below using the superposition theorem. Let is due to the voltage source is due to the current source. Note that frequencies of voltage and current sources are not the same. To calculate , we remove the current source (open circuit) By voltage division
Machine generated alternative text: 25 Prof E Ambikairajah, UNSW, Australia To calculate , we short circuit the voltage source Add all contributions algebraically
Machine generated alternative text: Exercise 3.6 26 Prof E Ambikairajah, UNSW, Australia Q1: Solvefor forthecircuitshownbelowusingthe superpositionprinciple. = 10 + 21.45 sin(2t + ) + 10.73 cos(3t - ) V Q2: Determinethevoltage (t)across8 resistorinthecircuit shownbelow: (t)= 15.46 cos (50t + ) + 20.24 cos(10t - ) V 6 v o ïî ½±­øí¬÷ Ê 2H õ ó õ ó ïð Ê ì ­·²øî¬÷ ß - + 2mF 8 v o (t) îð ½±­øëð¬÷ Ê 150mH õ ó õ ó - + îð ½±­øïð¬÷ Ê
Machine generated alternative text: Transformation between voltage and current Sources Voltage Source Current Source Note: The Internal resistance of the voltage source has the same value as the internal resistance of the current source. Thevenin's theorems are applied to ac circuits in the same way as they are to dc circuits. Always disconnect the load impedance (Z L ) from the circuit (between a and b ) before finding the or
Machine generated alternative text: Removealltheindependent sources. Thedependentsources cannotberemoved. Assumeavoltageorcurrent sourceconnectedacrossa-b terminalsasshown. Thevalueassumedis generally1 0 . Calculatethevoltageor current( v o or i o )andhence 28 Calculating Z Th for circuits with dependent sources
Machine generated alternative text: ðòïè ð ± ß -j125 j360 Example 3.9 Use source transformation and equivalent impedance to find Thevenin equivalent circuit of the ac circuit shown below 200 íê ½±­øïêð¬÷ Ê õ ó 2.25H 50 F It can be seen that the above circuit contains a series combination of a voltage source and impedance A source transformation replaces this series combination with a parallel combination of a current source and impedance as shown below: 200 õ ó íê ð ± Ê -j125 j360 200 õ ó íê ð ± Ê -j125 j360 The impedances of the resistor and inductor are connected in parallel and we can calculate the equivalent impedance 200 j360 = =
Machine generated alternative text: A source transformation replaces this parallel combination with a series combination of a voltage source and impedance. We can calculate the equivalent impedance: ðòïè ð ± ß -j125 ïéìòè îçòï ± íïòë îçòï ± Ê -j125 ïéìòè îçòï ± õ ó Theveninequivalent circuit is: Ê ÌØ ã íïòë îçòï ± Ê Æ ÌØ ãïëîòèíó¶ìðòðçì õ ó
Machine generated alternative text: Example 3.10 Find Theveninequivalent circuit of the following ac circuit: 200 íê ½±­øïêð¬÷ Ê õ ó 2.25H 50 F 200 õ ó íê ð ± Ê -j125 j360 The voltage across that open circuit is the open circuit voltage . Ê ±½ 200 õ ó íê ð ± Ê -j125 j360 õ ó = = = The Theveninimpedance is determined using the circuit below, where the voltage source is set to zero voltage (ie equivalent to a short circuit). = + = The Theveninequivalent circuit is : Ê ÌØ ã íïòë îçòï ± Ê Æ ÌØ ãïëîòèíó¶ìðòðçì õ ó 200 -j125 j360 Z TH
Machine generated alternative text: Example 3.11 Find I o for the circuit shown below using source transformation. Step1: source to a voltage source: 4,43 s jAZj S I Step 2: Now convert the voltage source ,with new impedance Z ,(shown by the dotted line) to a current source. Find equivalent impedance of (6-j2) & j5. and then apply current division to find Ans: I o = 3.288 99.46 o A
Machine generated alternative text: Example 3.12 Find the equivalent between terminals a and b for the circuit shown below. Step1: The voltage source is turned off by setting it to zero (i.eshort circuit). Step 2: V TH is open circuit voltage at terminals a and b with the voltage source turned on. V TH Z TH
Machine generated alternative text: Exercise 3.7 Q1: Between terminals and , obtain Theveninand Norton rad/s. ß²­æ ã ± = óíòê ± Ê , óïííòïè ± ß Q2: Find the Theveninand Norton equivalent circuits, for the circuit shown below, between terminals a and b : ß²­æ ã ã ë = ± Ê , ± ß î­·²ø ¬÷ ß 10 10mF 0.5H ó v o - + ïî½±­ø t ÷ Ê a b 2v o 100 õ ó ïðð ð ± Ê -j100 ï çð ± ß a b
Machine generated alternative text: Exercise 3.8: Use Supernode/Supermeshprinciple Q1: Obtain V 0 for the Figure shown below, using nodal analysis. ß²­æ ± Ê Q2: Using mesh analysis, obtain I 0 for the circuit shown below: ß²­æ ± ß ¶î ó ïî ð ± Ê 4 ðòîÊ ð 2 Ê ± õ ó ó¶ì ¶î 1 1 ì ð ± ß ó¶ì 2 î ð ± ß õ ó ïð çð ± Ê × ð
Machine generated alternative text: References [1] Alexander, C. K., & Sadiku th edition, McGraw Hill. th edition, Wiley & sons. School of Electrical engineering and Telecommunications, UNSW, Australia. [4] Hambley Publishing.

 

Created with Microsoft OneNote 2016.