Transcript Mesh Analysis

Objective of Lecture
 Provide 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 Analysis
 Technique to find voltage drops around a loop using
the currents that flow within the loop, Kirchoff’s
Voltage Law, and Ohm’s Law
 First result is the calculation of the mesh currents

Which can be used to calculate the current flowing through
each component
 Second result is a calculation of the voltages across the
components

Which can be used to calculate the voltage at the nodes.
Definition of a Mesh
 Mesh – the smallest loop around a subset of
components in a circuit
 Multiple meshes are defined so that every component in
the circuit belongs to one or more meshes
Steps in Mesh Analysis
Vin
Step 1
 Identify all of the meshes in the circuit
Vin
Step 2
 Label the currents flowing in each mesh
i2
Vin
i1
Step 3
 Label the voltage across each component in the circuit
+
V1
_
Vin
+ V2
-
+
V3
_
+ V4
i2
i1
+
V6
_
-
+
V5
_
Step 4
 Use Kirchoff’s Voltage Law
+
V1
_
Vin
+ V2
-
+
V3
_
+ V4
i2
i1
+
V6
_
-
+
V5
_
 Vin  V1  V2  V3  V6  0
 V3  V4  V5  0
Step 5
 Use 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.
VR  I a  I b R
Step 5
V1  i1 R1
V2  i1 R2
+
V1
_
Vin
+ V2
-
+
V3
_
+ V4
i2
i1
+
V6
_
-
+
V5
_
V3  i1  i2 R3
V4  i2 R4
V5  i2 R5
V6  i1 R6
Step 6
 Solve for the mesh currents, i1 and i2
 These currents are related to the currents found during
the nodal analysis.
i1  I 7  I1  I 2  I 6
i2  I 4  I 5
I 3  i1  i2
Step 7
 Once the mesh currents are known, calculate the
voltage across all of the components.
12V
From Previous Slides
 Vin  V1  V2  V3  V6  0
 V3  V4  V5  0
V1  i1 R1
V2  i1 R2
V3  i1  i2 R3
V4  i2 R4
V5  i2 R5
V6  i1 R6
Substituting in Numbers
 12V  V1  V2  V3  V6  0
 V3  V4  V5  0
V1  i1 4k 
V2  i1 8k 
V3  i1  i2 5k 
V4  i2 6k 
V5  i2 3k 
V6  i1 1k 
Substituting the results from Ohm’s
Law into the KVL equations
 12V  i1 4k  i1 8k  i1  i2 5k  i1 1k  0
 i1  i2 5k  i2 6k  i2 3k  0
Chugging through the Math
Mesh Currents
(mA)
i1
740
i2
264
 One or more of the mesh currents may have a negative
sign.
Chugging through the Math
Voltage across
resistors
(V)
VR1 = -i1R2
VR2 = i2 R2
VR3 =(i1 – i2) R3
VR4 = i2 R4
VR5 = (V4 – V5)
VR6 = (V5 – 0V)
-2.96
5.92
2.39
1.59
0.804
0.740
 The magnitude of any
voltage across a resistor
must be less than the
sum of all of the voltage
sources in the circuit
 In this case, no voltage
across a resistor can be
greater than 12V.
Chugging through More Math
The currents through each component in the
circuit.
Currents
(mA)
IR1 = i1
IR2 = i1
740
740
IR3 = i1- i2
IR4 = i2
476
264
IR5 = i2
IR6 = i1
264
740
I Vin = i1
740
Check
 None of the mesh currents should be larger than the
current that flows through the equivalent resistor in
series with the 12V supply.
Req  4k  8k  5k 6k  3k   1k
Req  16.2k
I eq  12V Req  740mA
Summary
Steps in Mesh Analysis
1. Identify all of the meshes in the circuit
2. Label the currents flowing in each mesh
3. Label the voltage across each component in the circuit
4. 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 currents
7. Once the mesh currents are known, calculate the
voltage across all of the components.