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Mesh Current Method
In the mesh current method, you will work with loop currents instead of branch currents. A
branch current is the actual current through a branch. An ammeter placed in a given branch
will measure the branch current Loop currents are different because they are mathematical
quantities that are used to make circuit analysis somewhat easier than with the branch
current method. The term mesh comes from the fact that a multiple-loop circuit, when drawn,
can be imagined to resemble a wire mesh. After completing this section, you should be able
to:
1.
2.
3.
Use mesh analysis to find unknown quantities in a circuit
Assign loop currents. Apply Kirchhoff's voltage law around each loop
Develop the loop (mesh) equations. Solve the loop equations
A systematic method of mesh analysis is listed in the following steps and is illustrated in
Figure 1, which is the same circuit configuration used in the branch current section. It
demonstrates the basic principles well.
Step 1:Assign a current in the clockwise (CW) direction around each closed loop. This may
not be the actual current direction, but it does not matter. The number of current assignments
must be sufficient to include current through all components in the circuit. No redundant
current assignments should be made. The direction does not have to be clockwise, but we will
use CW for consistency.
Step 2:Indicate the voltage drop polarities in each loop based on the assigned current
directions.
R1
R2
VS1
I1
R3
I2
VS2
Step 3:Apply Kirchhoff's voltage law around each closed loop. When more than one loop
current passes through a component, include its voltage drop. This results in one equation for
each loop.
Step 4: Using substitution or determinants solve the resulting equations for the loop currents.
First, the loop currents I1 and I2 are assigned in the CW direction as shown in figure. A loop
current could be assigned around the outer perimeter of the circuit, but information would be
redundant since I1 and I2 already pass through all of the components.
Second, the polarities of the voltage drops across R1, R2, and R3 are shown based on the loop
current directions. Notice that I1 and I2 flow in opposite directions through R2 because R2 is
common to both loops. Therefore, two voltage polarities are indicated. In reality, R2 currents
cannot be separated into two parts, but remember that the loop currents are basically
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mathematical quantities used for analysis purposes. The polarities of voltage sources are
fixed and are not affected by the current assignments.
Third, Kirchhoff’s voltage law applied to the two loops results in the following equations:
R1I1 + R2(I1 - I2) = VS1
for loop 1
R3I2 + R2(I2 – I1) = -VS2
for loop 2
Fourth, the like terms in the equations are combined and rearranged for convenient solution
so that they have the same position in each equation, that is, the 1 term is first the I2 term is
second. The equations are rearranged into the following form. Once the currents are
evaluated, all of the branch currents can be determined.
(R1 + R2) I1 - R2 I2 = VS1
for loop 1
-R2 I1 + (R2 + R3) I2 = -VS2
for loop 2
Notice that in the mesh current method only two equations are required for the same circuit
that required three equations in the branch current method. The last two equations (developed
in the fourth step) follow a certain format to make mesh analysis easier. Referring to these
last two equations, notice that for loop 1, the total resistance in the R1 + R2, is multiplied by I1
(its loop current). Also in the loop 1 equation, the resistance common to both loops, R2, is
multiplied by the other loop current, I2, and subtracted the first term. The same general form
is seen in the loop 2 equation except that the have been rearranged. From these observations,
the format for setting up the equation for a loop circuit can be stated as follows:
1.
2.
3.
4.
Sum the resistances around the loop, and multiply by the loop current.
Subtract the common resistance(s) times the adjacent loop current(s).
Set the terms in Steps 1 and 2 equal to the total source voltage in the loop. The sign of
the source voltage is positive if the assigned loop current flows out of its positive
terminal. The sign is negative if the loop current flows into its positive terminal.
Rearrange the terms so that like terms appear in the same position in each equation.
Example 1 illustrates the application of this format to the mesh current analysis.
Example1.
Using the mesh current method, find the branch currents in Figure 2.
47
10V
I1
82
22 I2
5V
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Solution Assign the loop currents as shown in Figure 2. Use the format described to set up
the two loop equations.
(47 + 22)I1 – 22I2 = 10
for loop 1
69 I1 - 22 I2 = 10
-22 I1 + (22 + 82) I2 = -5 for loop 2
-22 I1 + 104 I2 = -5
Use determinants to find I1.
10
-22
I1= -5
104 = (10)(104) - (-5)(-22) = 1040 – 110 = 139mA
69
-22 (69)(104) - (-22)(-22) 7176 - 484
-22
104
Solving for I2 yields
69 10
I2 = -22 -5 = (69)(-5) - (-22)(10) _ -345 - (-220) = -18.7 mA
6692
6692
6692
The negative sign on I2 means that its direction must be reversed.
Now find the actual branch currents. Since I1 is the only current through R1, it is also the
branch current IR1
IR1 = I1 = 139 mA
Since I2 is the only current through R3, it is also the branch current IR3.
IR3 = I2 = - 18.7 mA
(opposite direction of that originally assigned to I2)
Both loop currents I1and I2 flow through R2 in the same direction. Remember, the negative I2
value told us to reverse its assigned direction.
IR2 = I1 – I2 = 139 mA - (-18.7 mA) = 158 mA
Keep in mind that once we know the branch currents, we can find the voltages by using
Ohm's law.
Circuits with More than Two Loops
The mesh method also can be systematically applied to circuits with any number. Of course,
the more loops there are, the more difficult is the solution. However, the basic procedure still
applies. For example, for a three-loop circuit, three simultaneous equations are required.
Example 2 illustrates the analysis of a three-loop circuit.
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EXAMPLE 2 Find I3 in Figure 3.
47
12V
I1
22
I3
6V
33
10
8V
I2
Solution Assign three CW loop currents as shown in Figure 3. Then use procedure to set up
the loop equations. A concise restatement of this procedure as follows:
(Sum of resistors in loop) times (loop current) minus (each common resistor) times
(associated adjacent loop current) equals (source voltage loop). The polarity of a voltage
source is positive when the assigned current flows out of the positive terminal.
102I1 - 22 I2 - 33 I3 = 12
for loop I
-22 I1 + 32 I2 - 10 I3 = 6
for loop 2
-33 I1 - 10I2 + 43 I3 = 8
for loop 3
These three equations can be solved for the currents by substitution or, more easily with
third-order determinants or by Gausian elimination. I3 is found using determinants as follows.
The characteristic determinant is evaluated as follows:
102
-22
-33
=
=
-22
32
-10
-33
-10
43
32 -10
= 102
-22
-33
- (-22)
-10 43
-22 -33
+ (-33)
-10
43
32
-10
102[(32)(43) - (-10)(-10)] - (-22)[(-22)(43) - (-10)(-33)] + (-33)[(-22)(-10)-(32)(-33)]
102(1276) + 22(-1276) - 33(1276) = 130,152 - 28,072 - 42,108 = 59,972
The I3 determinant is evaluated as follows:
102
-22
12
-22
32
6 =102 32 6 - (-22) -22 12
-33
-10
8
-10 8
-10 8
+(-33) -22 12
32
6
= 102[(33)(8) - (- l0)(6)] + 22[(-22)(8) - (- l0)(12)] – 33[(-22)(6) - (32)(12)]
=102(316) + 22(-56) - 33(-516) = 32,232 - 1232 + 17,028 = 48,028
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I3 is determined by dividing the value of the I3 determinant by the value of the characteristic
determinant.
I3 = 48,028 = 80lmA
59,972
The other loop currents are found similarly.
REVIEW
1.
2.
3.
Do the loop currents necessarily represent the actual currents in the branches?
When you solve for a loop current and get a negative value, what does it mean
What circuit law is used in the mesh current method?
Exercises1. (Page 248 Introductory Circuit Analysis)
Write the mesh equations for the network of Fig 5 and find the current through the 7
resistor.
6
8
I1
4V
2
I2
7
9V
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Exercise 2 Write the mesh equations for the network of Fig 6
2
2V
1
2V
I1
4V
1
I2
I3
3
4
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Exercise 3. Find the current through the 10 resistor of the work of Fig. 7.
10
I3
8
15V
I1
5
3
I2
2
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