1. EE212 Passive AC Circuits Lecture Notes 2a 2010-20111EE 212

2. 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 2010-20112EE 212

3. 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. 2010-2011EE 2123

4. Mesh (Loop) Analysis (cont) ~ I1I1 I2I2 I3I3 Steps: -assume mesh current I 1 for Mesh 1, I 2 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 2010-20114EE 212

5. Mesh (Loop) Analysis (continued) Single voltage source Multiple voltage sources Voltage and current sources Supermesh Dependent sources 2010-2011EE 2125

6. 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) 2010-2011EE 2126

7. Example 1: Find the power dissipated in the 50  resistor using Mesh a nalysis. 50/30 0 V ~ -j50  50  j5  -j100  -j5  j50  ~ + - 50/30 0 V + - 2010-20117EE 212

8. Mesh Analysis Using Matrix Method [ V ] = [ Z ] [ I ] For a circuit with ‘n’ loops, the matrix equation is: V 1 V 2.. V n Z 11 -Z 12.... -Z 1n -Z 21 Z 22.... -Z 2n … ….. …............. -Z n1 -Z n2.... Z nn = I 1 I 2.. I n [ V ] is voltage vector V i 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 Z ii = sum of all impedances in Mesh i Z ik = impedance between Mesh i and Mesh k [ I ] is the unknown current vector I i = current in Mesh i 2010-20118EE 212

9. 50/30 0 V ~ -j50  50  j5  -j100  -j5  j50  ~ + - 50/30 0 V + - I1I1 I2I2 I3I3 Example 1: Mesh (Loop) Analysis (cont) 2010-20119EE 212

10. 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 non- planar circuits. Nodal analysis is used to solve for node voltages. Sign convention: current leaving node is +ve 2010-2011EE 21210

11. ~ V1V1 V2V2 ref. node ~ Nodal Analysis (continued) 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 V i (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 2010-201111EE 212

12. Nodal Analysis (continued) Single current source Multiple current sources Voltage and current sources Supernode Dependent sources 2010-2011EE 21212

13. Example 1 (revisited): Find the power dissipated in the 50  resistor using Nodal analysis. 50/30 0 V ~ -j50  50  j5  -j100  -j5  j50  ~ + - 50/30 0 V + - 2010-201113EE 212

14. Solving by Nodal Analysis: 50/30 0 V ~ -j50  50  j5  -j100  -j5  j50  ~ + - 50/30 0 V + - V1V1 V2V2 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? Solve Equations 1 and 2 to obtain V 1 and V 2 V3V3 V4V4 2010-201114EE 212

15. Nodal Analysis Using Matrix Method [ I ] = [ Y ] [ V ] For a circuit with ‘n+1’ nodes, the matrix equation is: I 1 I 2.. I n Y 11 -Y 12.... -Y 1n -Y 21 Y 22.... -Y 2n … ….. …............. -Y n1 -Y n2.... Y nn = V 1 V 2.. V n [ I ] is the known current vector I i 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 Y ii = sum of all admittances connected to Node i Y ik = admittance between Node i and Node k [ V ] is the unknown voltage vector V i = voltage at Node i w.r.t. the reference node 2010-201115EE 212

16. Source Transformation ~ A B Z V Voltage Source to Current Source  A B Z I Current Source to Voltage Source 2010-201116EE 212

17. Example 1 (revisited): Nodal Analysis and source transformation  V1V1 V2V2  1.Current source equivalent for V 1 2.Current source equivalent for V 2 3.Admittances for all other components All admittance values in siemens 2010-201117EE 212

18. Example 1 (revisited): Nodal Analysis and source transformation = V 1 V 2 V 1 = = 55.56/30 0 Volts Similarly, V 2 = 55.56/30 0 Volts 2010-201118EE 212

19. ~ A B Z th E th A B Linear Circuit Thevenin’s Theorem Any linear two terminal network with sources can be replaced by an equivalent voltage source in series with an equivalent impedance. Thevenin Voltage, E th voltage measured at the terminals A & B with nothing connected to the external circuit Thevenin Impedance, Z th 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) 2010-201119EE 212

20. Example: Use a Thevenin equivalent circuit at bus “A-B” to calculate the short circuit current at A-B. 110/30 0 V j5  -j2  A 55 ~ + - 55 33 B 2010-201120EE 212

21. 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. Z Load = Z th * 2010-201121EE 212