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ES
Elements of Electrical Engineering
Lecture
Jeffrey Miller, Ph.D.
Outline
• Chapter 4.5-4.
Mesh-Current Method
• Mesh
- A loop with no other loops inside of it
• Mesh Current
- Current that exists only in the perimeter of a mesh
NOTE: Mesh-Current Method is applicable
only to planar circuits
Mesh-Current Method
- Identify all of the meshes in the circuit by drawing a curved arrow to identify the direction the current is flowing in the mesh
- Assign a variable name to the current in each mesh
- Use Kirchhoff’s Voltage Law to write an equation around each of the meshes in the direction of the current arrow
- Solve the equations
- Calculate the desired output value
Mesh-Current/Kirchhoff Example
Assume v 1 , v 2 , R 1 , R 2 , R 3 are given Find i 1 , i 2 , i 3 First do this problem using Kirchhoff’s Laws, then using the Mesh-Current method
Mesh-Current Example –
Kirchhoff’s Approach
- Using Kirchhoff’s Current Law, we get: i 1 = i 2 + i 3
- Using Kirchhoff’s Voltage Law, we get: v 1 = i 1 R 1 + i 3 R 3 -v 2 = i 2 R 2 – i 3 R 3
Mesh-Current Example –
Kirchhoff’s Approach
Using i 3 = i 1 – i 2
v 1 = i 1 R 1 + i 3 R 3 v 1 = i 1 R 1 + (i 1 – i 2 )R 3 v 1 = i 1 (R 1 + R 3 ) – i 2 R 3
-v 2 = i 2 R 2 – i 3 R 3 -v 2 = i 2 R 2 – (i 1 – i 2 ) R 3 -v 2 = -i 1 R 3 + i 2 (R 2 + R 3 )
Mesh-Current Example –
Mesh-Current Approach
- Apply Kirchhoff’s voltage law around the two meshes v 1 = iaR 1 + (ia – ib) R 3 -v 2 = (ib – ia) R 3 + ibR 2
Mesh-Current Sample Problem
40V = 2Ω ia + 8Ω (ia – ib)
0 = 8Ω (ib – ia) + 6Ω ib + 6Ω (ib – ic)
-20V = 6Ω (ic – ib) + 4Ω ic
Mesh-Current Sample Problem
ia = 5.6A ib = 2.0A ic = -0.80A p40V = -40V ia = -40V * 5.6A = -224W p20V = 20V ic = 20V * -0.80A = -16W So both voltage sources are delivering power to the circuit
Mesh-Current Sample Problem
The current across the 8Ω resistor is ia – ib So the voltage drop across the resistor is v 0 = 8Ω (ia – ib) = 8Ω (5.6A – 2.0A) = 8Ω (3.6A) = 28.8V
Node-Voltage vs Mesh-Current
• What is the advantage to using the Node-
Voltage or the Mesh-Current method over
Kirchhoff’s and Ohm’s Laws?
- It reduces the number of simultaneous equations that need to be manipulated
• When does one use the Node-Voltage
method over the Mesh-Current method?
- There is no answer, but one method may result in fewer simultaneous equations
Mesh-Current Example
- Find i 0 in the following diagram using the mesh- current method.
Mesh-Current Example
-50V + 2Ω (ia – ib) + 6Ω (ia – ic) = 0 10Ω ib + 8Ω (ib – ic) + 2Ω (ib – ia) = 0 -10V + 6Ω (ic – ia) + 8Ω (ic – ib) = 0 ia = 16A ib = 6A ic = 11A i 0 = ia – ic = 5A
Homework
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