EE212 Passive AC Circuits Lecture Notes 2a EE 212

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 EE 212

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 EE 2123

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 EE 212

Mesh (Loop) Analysis (continued) Single voltage source Multiple voltage sources Voltage and current sources Supermesh Dependent sources EE 2125

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) EE 2126

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

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 Z 1n -Z 21 Z 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 EE 212

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

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 EE 21210

~ 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 EE 212

Nodal Analysis (continued) Single current source Multiple current sources Voltage and current sources Supernode Dependent sources EE 21212

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

Solving by Nodal Analysis: 50/30 0 V ~ -j50  50  j5  -j100  -j5  j50  ~ /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 V4V EE 212

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 Y 1n -Y 21 Y 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 EE 212

Source Transformation ~ A B Z V Voltage Source to Current Source  A B Z I Current Source to Voltage Source EE 212

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 EE 212

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 EE 212

~ 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) EE 212

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 EE 212

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 * EE 212