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EE212 Passive AC Circuits Lecture Notes 2a 2010-2011 EE 212 1 Circuit Analysis When a circuit has more than one element, a circuit analysis is required to determine circuit parameters (v, i, power, etc.) in different parts of the circuit. Circuit Theories: •Ohm’s Law •Superposition Theorem •Kirchhoff’s Voltage and Current Laws •Mesh/Nodal Analysis •Source Transformation •Thevenin/Norton Theorem •Wye/Delta Transformation 2010-2011 EE 212 2 Mesh (Loop) Analysis • Kirchhoff’s voltage law applies to a closed path in an electric circuit. The close path is referred to as a loop. • A mesh is a simple loop. That is, there are no other loops inside it. • Mesh analysis applies to planar circuits. That is, a circuit that can be drawn on a plane with no crossed wires. 2010-2011 EE 212 3 Mesh (Loop) Analysis (cont) ~ I1 I2 I3 Steps: -assume mesh current I1 for Mesh 1, I2 for Mesh 2, etc. -apply KVL for each loop -obtain ‘n’ equations for ‘n’ meshes -i.e., the mesh currents are the unknown variables -solve equations to determine mesh currents -usually done using matrices -obtain currents through each circuit element of interest -apply Ohm’s Law to calculate voltages of interest 2010-2011 EE 212 4 Mesh (Loop) Analysis (continued) • • • • • Single voltage source Multiple voltage sources Voltage and current sources Supermesh Dependent sources 2010-2011 EE 212 5 Mesh (Loop) Analysis (continued) • Dependent sources: – Current controlled voltage source (CCVS) – Voltage controlled voltage source (VCVS) – Current controlled current source (CCCS) – Voltage controlled voltage source (VCCS) 2010-2011 EE 212 6 Example 1: Find the power dissipated in the 50 resistor using Mesh analysis. j5 50 -j100 -j5 + 50/300 V ~ - 2010-2011 + -j50 j50 ~ 50/300 V - EE 212 7 Mesh Analysis Using Matrix Method [V]=[Z][I] For a circuit with ‘n’ loops, the matrix equation is: V1 V2 .. .. Vn = Z11 -Z12 .. .. -Z1n -Z21 Z22 .. .. -Z2n … … .. … .. ... .. .. .. .. -Zn1 -Zn2 .. .. Znn I1 I2 .. .. In [ V ] is voltage vector Vi is the source voltage in Mesh i Sign convention: +ve voltage if going –ve to +ve (since voltage vector has been moved to the left hand side of the equation) [ Z ] is impedance matrix (square matrix) Diagonal element Zii = sum of all impedances in Mesh i Zik = impedance between Mesh i and Mesh k 2010-2011 [ I ] is the unknown current vector Ii = current in Mesh i EE 212 8 Example 1: Mesh (Loop) Analysis (cont) j5 + 50/300 V ~ - 2010-2011 I1 -j50 50 -j100 -j5 + I2 j50 EE 212 I3 ~ 50/300 V - 9 Nodal Analysis • A node is a point in an electric circuit where 2 or more components are connected. (Strictly speaking, it is the whole conductive surface connecting those components.) • Nodal analysis applies to planar and nonplanar circuits. • Nodal analysis is used to solve for node voltages. • Sign convention: current leaving node is +ve 2010-2011 EE 212 10 Nodal Analysis (continued) V1 V2 ~ ~ ref. node Steps: - identify all the nodes - select a reference node (usually the node with the most branches connected to it or the ground node of a power source) - assume voltage Vi (w.r.t . reference node) for Node i - assume current direction in each branch - apply KCL at each node - obtain ‘n-1’ equations for ‘n’ nodes (since one node is the reference node) -solve the equations to determine node voltages - apply Ohm’s Law to calculate the currents 2010-2011 EE 212 11 Nodal Analysis (continued) • • • • • Single current source Multiple current sources Voltage and current sources Supernode Dependent sources 2010-2011 EE 212 12 Example 1 (revisited): Find the power dissipated in the 50 resistor using Nodal analysis. j5 50 -j100 -j5 + 50/300 V ~ ~ - 2010-2011 + -j50 j50 EE 212 50/300 V - 13 Solving by Nodal Analysis: V3 j5 V1 50 -j100 V2 -j5 V4 + 50/300 V + ~ ~ - -j50 j50 50/300 V - ref. node KCL at Node 1: get Equation 1 KCL at Node 2: get Equation 2 Why no node equations at Nodes 3 and 4? 2010-2011 Solve Equations 1 and 2 toEEobtain V1 and V2 212 14 Nodal Analysis Using Matrix Method [I]=[Y][V] For a circuit with ‘n+1’ nodes, the matrix equation is: I1 I2 .. .. In = Y11 -Y21 … ... -Yn1 -Y12 .. .. Y22 .. .. … .. … .. .. .. -Yn2 .. .. -Y1n -Y2n .. .. Ynn V1 V2 .. .. Vn [ I ] is the known current vector Ii is the source current in Node i Sign convention: current entering node, +ve (since current vector has been moved to the left hand side of the equation) [ Y ] is admittance matrix (square matrix) Diagonal element Yii = sum of all admittances connected to Node i Yik = admittance between Node i and Node k [ V ] is the unknown voltage vector Vi = voltage at Node i w.r.t. the reference 2010-2011 EE 212 node 15 Source Transformation Voltage Source to Current Source Z ~ A V B Current Source to Voltage Source A 2010-2011 I Z B EE 212 16 Example 1 (revisited): Nodal Analysis and source transformation V1 V2 1. Current source equivalent for V1 2. Current source equivalent for V2 3. Admittances for all other components All admittance values in siemens 2010-2011 EE 212 17 Example 1 (revisited): Nodal Analysis and source transformation V1 = V2 V1= 2010-2011 = 55.56/300 Volts EE 212 Similarly, V2 = 55.56/300 Volts 18 Thevenin’s Theorem Any linear two terminal network with sources can be replaced by an equivalent voltage source in series with an equivalent impedance. Zth A Linear Circuit ~ B Eth A B Thevenin Voltage, Eth voltage measured at the terminals A & B with nothing connected to the external circuit Thevenin Impedance, Zth impedance at the terminals A & B with all the sources reduced to zero i.e. voltage sources short circuited (0 volts) current sources open circuited (0 amperes) 2010-2011 EE 212 19 Example: Use a Thevenin equivalent circuit at bus “A-B” to calculate the short circuit current at A-B. j5 -j2 A 5 + 3 5 110/300 V ~ - 2010-2011 B EE 212 20 Maximum Power Transfer Theorem Maximum power is transferred to a load when the load impedance is equal to the conjugate of the Thevenin Impedance. i.e. ZLoad = Zth* 2010-2011 EE 212 21