Objective of LectureProvide step-by-step instructions for mesh

analysis, which is a method to calculate voltage drops and mesh currents that flow around loops in a circuit.Chapter 3.4 and Chapter 3.5

Mesh AnalysisTechnique to find voltage drops around a

loop using the currents that flow within the loop, Kirchoff’s Voltage Law, and Ohm’s LawFirst result is the calculation of the mesh

currents Which can be used to calculate the current flowing

through each componentSecond result is a calculation of the voltages

across the components Which can be used to calculate the voltage at the

nodes.

Definition of a MeshMesh – the smallest loop around a subset of

components in a circuitMultiple meshes are defined so that every

component in the circuit belongs to one or more meshes

Steps in Mesh Analysis

Vin

Step 1Identify all of the meshes in the circuit

Vin

Step 2Label the currents flowing in each mesh

i1

i2

Vin

Step 3Label the voltage across each component in

the circuit

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

Step 4Use Kirchoff’s Voltage Law

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

0

0

543

6321

VVV

VVVVVin

Step 5Use Ohm’s Law to relate the voltage drops

across each component to the sum of the currents flowing through them.Follow the sign convention on the resistor’s

voltage.

RIIV baR

Step 5

i1

i2

+ V1

_

Vin

+ V3

_

+ V5

_

+ V6

_

+ V2 -

+ V4 -

616

525

424

3213

212

111

RiV

RiV

RiV

RiiV

RiV

RiV

Step 6Solve for the mesh currents, i1 and i2

These currents are related to the currents found during the nodal analysis.

213

542

62171

iiI

IIi

IIIIi

Step 7Once the mesh currents are known, calculate

the voltage across all of the components.

12V

From Previous Slides

616

525

424

3213

212

111

RiV

RiV

RiV

RiiV

RiV

RiV

0

0

543

6321

VVV

VVVVVin

Substituting in Numbers

kiV

kiV

kiV

kiiV

kiV

kiV

1

3

6

5

8

4

16

25

24

213

12

11

0

012

543

6321

VVV

VVVVV

Substituting the results from Ohm’s Law into the KVL equations

0365

0158412

2221

12111

kikikii

kikiikikiV

Chugging through the Math

One or more of the mesh currents may have a negative sign.

Mesh Currents (A)

i1 740

i2 264

Chugging through the MathVoltage across

resistors(V)

VR1 = -i1R2 -2.96

VR2 = i2 R2 5.92

VR3 =(i1 – i2) R3 2.39

VR4 = i2 R4 1.59

VR5 = (V4 – V5) 0.804

VR6 = (V5 – 0V) 0.740

The magnitude of any voltage across a resistor must be less than the sum of all of the voltage sources in the circuitIn this case, no

voltage across a resistor can be greater than 12V.

Chugging through More Math

Currents (A)

IR1 = i1 740

IR2 = i1 740

IR3 = i1- i2 476

IR4 = i2 264

IR5 = i2 264

IR6 = i1 740

I Vin = i1 740

The currents through each component in the circuit.

CheckNone of the mesh currents should be larger

than the current that flows through the equivalent resistor in series with the 12V supply.

ARVI

kR

kkkkkkR

eqeq

eq

eq

74012

2.16

136584

SummarySteps in Mesh Analysis

1. Identify all of the meshes in the circuit2. Label the currents flowing in each mesh3. Label the voltage across each component in

the circuit4. Write the voltage loop equations using

Kirchoff’s Voltage Law.5. Use Ohm’s Law to relate the voltage drops

across each component to the sum of the currents flowing through them.

6. Solve for the mesh currents7. Once the mesh currents are known, calculate

the voltage across all of the components.