ELE 205
Lecture # 12 – Network Analysis using Mesh Equations
(Floyd page 347)
A loop = a complete current path within a circuit.
• In loop analysis there are not constraints on how we assign loops.
To perform loop analysis:
– create a loop until all components are included in one loop.
- Do not make additional loops once all components have
been included
– A loop can include other loops within it.
• Mesh Analysis is a special case of Loop Analysis
– A Mesh is a loop that contains no other loops within it.
We use Mesh Current analysis when we have circuits:
– that contain multiple sources
– we wish to solve for currents
Note: Since Mesh Currents is a special case of Loop Analysis, we
will not perform any examples of Loop Analysis as the procedure
for writing KVL equations is the same for both.
This method is
- based on Ohm’s and Kirchhoff’s laws.
- useful in the analysis of multiple-loop circuits having two or
more voltage or current sources.
205 Unit 1
W2007 IC
1
ELE 205
Procedure:
1.
Convert all current sources to voltage sources and redraw
circuit.
2.
Draw all loop currents in a clockwise direction and identify
them by number.(although direction of an assigned loop
current is arbitrary)
3.
Indicate the voltage drop polarities in each loop (as + to –)
based on the assigned loop current direction.
4.
Identify all voltage sources according to their current polarity.
5.
Apply KVL around each closed loop - Write the equations for
the voltage drops around each loop in turn, by equating the
sum of the voltage drops to zero.
6.
Solve equations to find the unknown currents.
In the mesh (loop) current method, you can solve for the currents
in a circuit using simultaneous equations.
Simultaneous equations:
- A set of n equations containing n unknowns, where n is a
number with a value of 2 or more.
Three methods for solving simultaneous equations are:
Algebraic substitution
The determinant method
Using a calculator
205 Unit 1
W2007 IC
2
ELE 205
To simplify solving simultaneous equations, they are usually set up
in standard form. Standard form for two equations with two
unknowns is: Simultaneous Equations
Example
A circuit has the following equations.
Set up the equations in standard form.
−10 + 270IA +1000(IA − IB ) = 0
1000 ( IB − IA ) + 680IB + 6 = 0
Solution:
Rearrange so that variables and their coefficients are in order
and put constants on the right.
1270 I A − 1000 I B = 10
−1000 I A + 1680 I B = −6
Using a calculator:
a1 x + b1 y = c1
a2 x + b2 y = c2
a1 = 1270 , b1 = -1000 , c1 = 10
a2 = -1000 , b2 = 1680 , c2 = -6
x = 9.53 mA = IA
y = 2.10 mA = IB
205 Unit 1
W2007 IC
3
ELE 205
Example 1 (Floyd page 348-353)
Solve the resulting equations for the loop currents
Solution:
VS1 - R1IA - R2IA + R2IB = 0
- VS2 - R2IB + R2IA – R3IB = 0
or
-VS1 + R1IA + R2IA - R2IB = 0
VS2 + R2IB - R2IA + R3IB = 0
− 10 + 0.270 I A + 1.0 ( I A − I B ) = 0
1.0 ( I B − I A ) + 0.68 I B + 6.0 = 0
Rearranging the loop equations into standard form:
1.270 I A − 1.0 I B = 10
− 1.0 I A + 1.68 I B = − 6.0
I1=IA = 9.53 mA (x)
I2= IA-IB = 7.43 mA
I3=IB = 2.10 mA (y)
I A = 9 .5 3 m A
I B = 2 .1 0 m A
205 Unit 1
W2007 IC
4