Mesh-Current Analysis

General circuit analysis method

Based on KVL and Ohm’s Law

Advantages: ALWAYS worksSimple to set up

Disadvantages: Leads to systems of equationsCan be tedious to solveCould be an easier method

Matrix methods (Cramer’s Rule) and computers can be very useful!

1

Mesh-Current AnalysisExample

20

5

10

25

15

35

30

Assume mesh currents

Direction of currents is arbitrary

Write KVL equations Vs = (I R)

2

Mesh 1: 0 = 20 i1 + 5 i1 + 35 (i1 – i3) + 10 (i1 – i2)

Mesh 2: 5 = 10 (i2 – i1) + 15 (i2 – i3) + 25 i2

Mesh 3: 0 = 15 (i3 – i2) + 35 (i3 – i1) + 30 i3

70 i1 – 10 i2 – 35 i3 = 0

– 10 i1 + 50 i2 – 15 i3 = 5

– 35 i1 – 15 i2 + 80 i3 = 0

Mesh-Current AnalysisExample

20

5

10

25

15

35

30

3

70 i1 – 10 i2 – 35 i3 = 0

– 10 i1 + 50 i2 – 15 i3 = 5

– 35 i1 – 15 i2 + 80 i3 = 0

Cramer’s Rule:

500184801535155010351070

,D

6256801501550535100

,A

87521800351551035070

,B

0007015355501001070

,C

mA.,

,

D

Ai 935

5001846256

1

mA.,

,

D

Bi 6118

50018487521

2

mA.,

,

D

Ci 937

5001840007

3

Mesh-Current AnalysisExample

20

5

10

25

15

35

30

4

mA.,

,

D

Ai 935

5001846256

1

mA.,

,

D

Bi 6118

50018487521

2

mA.,

,

D

Ci 937

5001840007

3 V1

V2V3

V4

V5Vref

Vref = 0

V1 = 5 V

V2 = V1 – 20 i1 = 5 – 20 (0.0359) = 4.282 V

V3 = V2 – 5 i1 = 4.282 – 5 (0.0359) = 4.103 V

V4 = V3 – 35 (i1 – i3) = 4.103 – 35 (0.0359 – 0.0379) = 4.173 V

V5 = V4 – 15 (i2 – i3) = 4.173 – 15 (0.1186 – 0.0379) = 2.963 V

Check: V25 = 25 i2 = 25 (0.1186) = 2.965 V 2.963 V

Also, V1 – V4 = 5 – 4.173 = 0.827 V V10 = 10 (0.1186 – 0.0359) = 0.827 V

Let’s verify our answers

Mesh-Current AnalysisWhat if there’s a current source?

5

Mesh 2: vs = R2 (i2 – i1) + R3 i2

Mesh 1: i1 = is

1 equation with 1 unknown

Mesh-Current AnalysisWhat if there’s a dependent source?

6

Supermesh: 7 = 2 i1 + 1 i2

2nd Equation: 2 i1 = i2 – i1

Dependent sources are no big deal!

Node-Voltage Analysis

General circuit analysis method

Based on KCL and Ohm’s Law

Advantages: ALWAYS worksSimple to set up

Disadvantages: Leads to systems of equationsCan be tedious to solveCould be an easier method

Matrix methods (Cramer’s Rule) and computers can be very useful!

7

Node-Voltage AnalysisExample

Assume node voltages

Choice of reference node is arbitrary, but there is often a “best choice”

Write KCL equations for each node using node voltages

8

79 v1 – 15 v2 = 0

– 7 v1 + 15 v2 – 8 v3 = – 224

– 6 v2 + 41 v3 = 0

Reference Node

Node 1: 08

VV

5

V

3

V 2111

Node 2: 047

VV

8

VV 3212

Node 3: 03

V

2

V

7

VV 3323

Node-Voltage AnalysisExample

9

79 v1 – 15 v2 = 0

– 7 v1 + 15 v2 – 8 v3 = – 224

– 6 v2 + 41 v3 = 0Reference Node (V=0)

Using Cramer’s Rule:

v1 = – 3.40 V

v2 = – 17.92 V

v3 = – 2.62 V

Check (iin = iout):

A13313

403.

.

A6805

403.

.

A81518

4039217.

..

A1862

7

6229217.

..

Don’t let the signs confuse you!Use absolute values and determine

current direction (Vhigh to Vlow).

A3112

622.

.

A87303

622.

.

Node 1: 1.133 + 0.68 = 1.813

Node 2: 1.815 + 2.186 = 4.001

Node 3: 1.31 + 0.873 = 2.183

Node-Voltage AnalysisWhat if there’s a voltage source?

10

Supernode

Supernode:

Second Equation: V1 – V2 = 9

02000

V

3000

V

5000

V

3000

3V 2211

Node-Voltage AnalysisWhat if there’s a dependent source?

11

Dependent sources are no big deal!

Supernode

Supernode:

Second Equation: V2 – V1 = 3 Ix

3220

V

50

V 21

50

VI 1

x 50

V3VV 1

12

Extra Tools for your ToolboxThevenin and Norton Equivalents

12

When “looking into” two ports of a circuit, you cannot tell exactly what components make up that circuit.

Vth = Vopen ckt

Ishort ckt

cktshort

cktopenth I

VR

In = Ishort ckt

cktshort

cktopenn I

VR

Thevenin Equivalent Norton Equivalent

Thevenin and Norton EquivalentsExample

13

Vopen ckt =

3ix

Open circuit case:

= 10 (i10) = 10 (3 ix)V10

72

310= = V7

60

Supermesh: 10 = 5 ix + 10 (3 ix) A72

=10 = 35 ix

3510

ix =

Short circuit case: Supermesh: 10 = 5 ix ix = 2

Ishort ckt = 3 ix = 6 A

3ix

Vth = 60/7 V, In = 6 A, and Rth = Rn = 10/7

7

10A6V

IV

RR 760

cktshort

cktopennth

The short-circuit case is a different circuit than the original problem!

Extra Tools for your ToolboxSource Transformations

14

mA51000

5R

VI

series

ss

Rshunt = Rseries = 1 k

1 k5 mA

Rseries

Vs RshuntIs Rshunt

Extra Tools for your ToolboxSuperposition Principle

15

In general, if a circuit has more than one source, we can determine the response of the circuit to ALL sources by analyzing the circuit considering one source at a time (ignoring the other sources), then combining all the partial responses to get the total response.

This is called the superposition principle.

It sounds like a great idea, but it has some caveats when applied to electric circuits…

Extra Tools for your ToolboxSuperposition Principle

16

Caveats when using the Superposition Principle

1. Only linear quantities (voltage, current) can be found using superposition – nonlinear quantities (power) cannot.

2. Dependent sources cannot be ignored. For this reason, superposition is of limited (questionable) use on circuits containing dependent sources.

Comment: The superposition principle is very useful in other areas (such as electromagnetics, and several non-EE fields), but it seldom (if ever) simplifies the process of analyzing a circuit.

Recommendation: Use another method.