objective of lecture - virginia techliab/lectures/ch_3/slides... · · 2011-05-31objective of...
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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
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Mesh Analysis Technique to find voltage drops around a loop using
the currents that flow within the loop, Kirchoff’sVoltage 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.
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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
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Steps in Mesh Analysis
Vin
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Step 1 Identify all of the meshes in the circuit
Vin
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Step 2 Label the currents flowing in each mesh
i1
i2
Vin
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Step 3 Label the voltage across each component in the circuit
i1
i2
+ V1
_
Vin
+ V3
_
+ V5
_
+ V6
_
+ V2 - + V4 -
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Step 4 Use Kirchoff’s Voltage Law
i1
i2
+ V1
_
Vin
+ V3
_
+ V5
_
+ V6
_
+ V2 - + V4 -
0
0
543
6321
VVV
VVVVVin
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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.
RIIV baR
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Step 5
i1
i2
+ V1
_
Vin
+ V3
_
+ V5
_
+ V6
_
+ V2 - + V4 -
616
525
424
3213
212
111
RiV
RiV
RiV
RiiV
RiV
RiV
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Step 6 Solve 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
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Step 7 Once the mesh currents are known, calculate the
voltage across all of the components.
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12V
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From Previous Slides
616
525
424
3213
212
111
RiV
RiV
RiV
RiiV
RiV
RiV
0
0
543
6321
VVV
VVVVVin
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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
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Substituting the results from Ohm’s Law into the KVL equations
0365
0158412
2221
12111
kikikii
kikiikikiV
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Chugging through the Math
One or more of the mesh currents may have a negative sign.
Mesh Currents (mA)
i1 740
i2 264
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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 circuit
In this case, no voltage across a resistor can be greater than 12V.
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Chugging through More Math
Currents (mA)
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.
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Check None 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
m74012
2.16
136584
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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.