EE212 Passive AC Circuits

Lecture Notes 2a

2010-2011 1EE 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

2010-2011 2EE 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.

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 4EE 212

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.

50/300 V ~-j50

50 j5 -j100 -j5

j50 ~

+

-

50/300 V

+

-

2010-2011 7EE 212

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 vectorVi is the source voltage in Mesh iSign 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 iZik = impedance between Mesh i and Mesh k

[ I ] is the unknown current vectorIi = current in Mesh i2010-2011 8EE 212

50/300 V ~

-j50

50 j5 -j100 -j5

j50

~

+

-

50/300 V

+

-

I1I2

I3

Example 1: Mesh (Loop) Analysis (cont)

2010-2011 9EE 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 +ve2010-2011 EE 212 10

~

V1 V2

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 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 11EE 212

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.

50/300 V ~

-j50

50 j5 -j100 -j5

j50 ~

+

-

50/300 V

+

-

2010-2011 13EE 212

Solving by Nodal Analysis:

50/300 V ~-j50

50 j5 -j100 -j5

j50

~

+

-

50/300 V

+

-

V1 V2

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 V1 and V2

V3 V4

2010-2011 14EE 212

Nodal Analysis Using Matrix Method

[ I ] = [ Y ] [ V ]

For a circuit with ‘n+1’ nodes, the matrix equation is:

I1

I2

..

..In

Y11 -Y12 .. .. -Y1n

-Y21 Y22 .. .. -Y2n … … .. … .. ... .. .. .. ..-Yn1 -Yn2 .. .. Ynn

=

V1

V2

..

..Vn

[ I ] is the known current vectorIi is the source current in Node iSign 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 iYik = admittance between Node i and Node k

[ V ] is the unknown voltage vectorVi = voltage at Node i w.r.t. the reference node2010-2011 15EE 212

Source Transformation

~

A

B

Z

V

Voltage Source to Current Source

A

BZI

Current Source to Voltage Source

2010-2011 16EE 212

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 17EE 212

Example 1 (revisited): Nodal Analysis and source transformation

=V 1

V 2

V 1 = = 55.56/300

Volts

Similarly, V2 = 55.56/300 Volts

2010-2011 18EE 212

~

A

B

Zth

Eth

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, 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 19EE 212

Example: Use a Thevenin equivalent circuit at bus “A-B” to calculate the short circuit current at A-B.

110/300 V

j5 -j2 A

5

~

+

-

53

B

2010-2011 20EE 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. ZLoad = Zth*

2010-2011 21EE 212