Basic Electric Circuits Session 9B

Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2
Session 9B: Focus
 Circuits with current sources
◦ On the perimeter of meshes
 Summary of Steps
◦ For Mesh Analysis
◦ In between meshes
 Supermesh
◦ Example with Supermesh
 Home Work Problems

Mesh Analysis:
Circuits with Current Sources

Mesh Analysis: With Current sources
 There are two possible types of current sources that
could be found in a circuit
◦ Dependent and independent current sources
 There are two possible positions that they could be
found
a) On the perimeter of the circuit
b) As a common element between two meshes

Current Source on Perimeter
 Where is the independent current source in each circuit?
i4
On the Perimeter of the circuit As a Common element between meshes

Current Source on Perimeter
 Where is the dependent current source in each circuit?
On the Perimeter of the circuit As a Common element between meshes

Mesh Analysis: Perimeter Current sources
 When the current source is appearing at the perimeter
of a circuit
 The mesh in which the current source is found, is
ignored, by making
◦ The mesh current of that mesh in which the current source is
found, is made equal to the value of the current source
 We thus reduce the number of mesh currents to be
evaluated by 1 for each current source present on the
perimeter

Example 1: Mesh Analysis with
Independent Current source on Perimeter
ix = i3 – i4
How many meshes are there? 4
What is ix in terms of mesh currents?
How much is the mesh current i4 ? 10 A
- 3+ 8i1 + 4(i1 -i2) = 0
4(i2 - i1) + 12i2 + 8(i2 - i3) = 0
KVL for the mesh with i1 :
Note: This is problem 36 on page 116, Figure 4.61 in the Ref 1 book (by Hayt)
12i1 - 4i2 = 3
-4i1 + 24i2 - 8i3 = 0
8(i3 - i2) + 20(i3 - i4) + 5i3 = 0
-8i2 + 33i3 - 20i4 = 0
-8i2 + 33i3 – (20* 10) = 0
-8i2 + 33i3 = 200
-i1 + 6i2 - 2i3 = 0
i3 = 6.639 A ix = i3 - 10 = -3.36 A
Solving these equations:
i4
How many KVL equs to be written ? 3

Example 2: Mesh Analysis with
Dependent Current source on Perimeter
i3= -0.5vx
How many meshes are there? 3
What is the value of i3 ?
- 2+ (2+9)i1 + 1 + 3i1 = 0 -1 + 10(i2 - i3) – 5 = 0
KVL for the mesh with i1 : KVL for the mesh with i2 :
14i1 = 1 10i2 - 10i3 = 6
i1 = 71.43 mA
vx = 9i1 = 9*71.43*10-3 = 0.643V
i3= -0.5*0.643
i3= -0.5*0.643
i3= -321.4 mA
10i2 = 6 + 10i3 10i2 = 6 - 10 * 321.4 * 10-3
i2 = 278.6 mA

Summary of Steps: Mesh Analysis

Summary of Basic Mesh Analysis Procedure
1. Determine if the circuit is a planar circuit. If not, perform
nodal analysis instead
2. Count the number of meshes (M). Redraw the circuit if
necessary.
3. Label each of the M mesh currents. Generally, defining all
mesh currents to flow clockwise results in a simpler analysis.
4. Write a KVL equation around each mesh. Begin with a
convenient node and proceed in the direction of the mesh
current.
a. Pay close attention to “−” signs. If a current source lies on the
periphery of a mesh, no KVL equation is needed and the mesh current
is determined by inspection.

Summary of Basic Nodal Analysis Procedure
5. Express any additional unknowns such as voltages or currents
other than mesh currents in terms of appropriate mesh
currents. This situation can occur if current sources or
dependent sources appear in our circuit.
6. Organize the equations. Group terms according to mesh
currents.
7. Solve the system of equations for the mesh currents (there will
be M of them).

Mesh Analysis: Supermesh

Mesh Analysis: When Current source is in
between two meshes
 Create a kind of “supermesh” from two meshes that have a current
source as a common element
 The current source is in the interior of the supermesh
 We thus reduce the number of meshes by 1 for each dependent
current source present

Mesh Analysis: Supermesh
 Two meshes are merged which are separated by a current source
 Mesh current I1 is given as:
 The KVL equation of the Supermesh is given below:

Example 3: Supermesh
 Find i1 :
Note: This is Practice problem 4.9 on page 99, Figure 4.25 in the Ref 1 book (by Hayt)
Supermesh is created
by merging meshes 1 and 3

Example 3: Supermesh
i3 - i1 = 3 A
Express 3A current in terms of mesh currents
- 10+ 4(i1 - i2) + 10(i3 - i2) + (1+7)i3 = 0
4(i2 - i1) +(5+9) i2 +10(i2 - i3) = 0
KVL for the supermesh:
i1 = -1.93A
4i1 - 14i2 + 18i3 = 10
4i1 – 14i2 + 18*(3 + i1) = 10
22i1 - 14i2 = -44
-4i1 + 28i2 - 10i3 = 0
-2i1 + 14i2 – 5i3 = 0
-2i1 + 14i2 – 5*(3 + i1) = 0
-7i1 + 14i2 = 15
i3 = 3 + i1
4i1 – 14i2 + 54 + 18i1 = 10 Solving these two equations:

Home Work Problems

S9B_Problem 1: Mesh Analysis
 Find the mesh currents:
i1 = 5 A
12i2 +11(i2 - i3) + 13(i2 - i1) = 0
i3 - i1 = vx/3
The mesh current i1 is:
-13i1+36i2 - 11i3 = 0
i2 = 1.35 A
vx = 13i3
i3 - i1 = 13i3/3
3i3 - 3i1 = 13i3
- 3i1 = 10i3 i3 = -1.5 A

S9B_Problem 2: Mesh Analysis
 Find i1 and Power supplied by 1V: i1 = 19 A P1V = (i1 * 1) = 19 W

Session 9B: Summary
◦ On the perimeter of meshes
 Summary of Steps
◦ For Mesh Analysis
◦ In between meshes
 Supermesh
◦ Example with Supermesh
 Home Work Problems

References
Ref 1 Ref 2

Basic Electric Circuits Session 9B

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