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Mesh Analysis
Mesh analysis is a strong method that computes the currents flowing in every mesh or loop in a network. The principles behind the mesh analysis approach are KVLs stated as the sum of the potentials around every closed loop sums to zero.
1.0Key Concepts
- Mesh: A mesh is an enclosed path in a circuit with no other meshes within it. That is to say, the circuit that contains a loop of many branches, then in such cases, the enclosed path of branches will form a mesh.
- Mesh Current: The current flowing in a mesh is called a mesh current. Mesh analysis assumes a current in each mesh. This current can be flowing in any direction, either clockwise or counterclockwise. If the calculated value is negative, then the current is flowing in the opposite direction.
2.0Kirchhoff’s Voltage Law (KVL)
Kirchhoff’s voltage law states that the sum of all the voltages around any mesh loop equals zero:
The Sign Convention for Kirchhoff’s Voltage Law:
- Voltage Rise: When the direction of movement is from the negative terminal to the positive terminal of a component, such as a battery or power source, the voltage is considered a rise. The potential difference is considered positive.
- Voltage Drop: When passing through a component from positive to negative terminals, say through resistors or wires, the voltage is said to be a drop. The potential difference is considered a negative value.
3.0Steps for Mesh Analysis
- Recognize Meshes:
Identify the meshes in a given circuit and name a mesh current to each. In a circuit with 3 loops, you can label the mesh currents as I1, I2, I3.
- Apply Kirchhoff’s Voltage Law (KVL):
In a given circuit, for each mesh apply KVL. The Voltage in a resistor V = IR, where I is the current and R is the resistance. The direction of voltage sources is treated according to the direction of the mesh current.
- Forming Mesh current equations:
Write down the KVL equations for each mesh. For instance, if mesh 1 has resistors R1 and R2 and a voltage of V1, the KVL equation for mesh 1 is:
- Solve and Calculate
After you are done with forming equations, solve them as well to find the values of the mesh currents. Once the mesh current is found, use Ohm’s law to find voltage, resistance, or other circuit quantities.
4.0Mesh Analysis in Special Cases
- Mesh Analysis with 3 loops (3 Meshes):
For a three-loop circuit, you assign three mesh currents, say I1, I2, and I3, to each loop. Use KVL for each loop and write three equations. Solving this system will give you the three mesh currents.
- Mesh Analysis and Nodal Analysis:
Mesh analysis deals with currents flowing through loops, while nodal analysis deals with voltages that can be determined at all the nodes in a circuit. Both methods are used to analyze electrical circuits, but it is applied if you have more loops with fewer voltage sources, whereas nodal analysis is appropriate for circuits that have too many nodes and fewer resistors.
- Mesh Analysis Current Sources:
When the circuit is contained with current sources, the mesh analysis can be modified. The current sources affect the mesh currents directly by setting the current in certain branches. You will have to revise your KVL equations to take into account known currents from these sources.
5.0Mesh Analysis Solved Problems
Example 1: Simple circuit (2 Loops)
Consider a circuit with two loops, a 10V battery, and resistors R1=2Ω and R2=3Ω.
Solution: Let the mesh currents be I1 and I2, with a clockwise direction.
Apply KVL for Mesh 1:
For mesh 1:
…….(1)
Apply KVL for Mesh 2:
For mesh 2:
Put in equation 1:
(Ans)
Example 2: Mesh Analysis with 3 Loops
Consider a circuit with three meshes, resistors R1=2Ω, R2=4 Ω, R3 = 3Ω, and two voltage sources of V1=5V and V2 = 6V in one loop.
Solution: Let the mesh currents be I1, I2, and I3.
Apply KVL for mesh 1 for V1 = 5V
5 + 2I1 + (I1 - I2)4 = 0
5 + 2I1 + 4I1 - 4I2 = 0
–6I1 + 4I2 = –5 …… (1)
Apply KVL for Mesh 2 for V2 = 6V
–(I2 - I1)R2 + R3I2 – V2= 0
–(I2 - I1)4 + 3I2 –6 = 0
4I1 – 4I2 + 3I2 –6 = 0
4I1 – I2 = 6 ……(2)
For Loop 3 (with only R3)
–I3R3 + (I3 – I2)R2 = 0
–I3(3) + (I3 – I2)(4) = 0
–3I3 + 4I3 – 4I2 = 0
I3 – 4I2 = 0 ……(3)
Now solve these three equations are:
6I1 – 4I2 = –5 …… (1)
4I1 – I2 = 6 ……(2)
I3 – 4I2 = 0 ……(3)
From equation 2
4I1 – 6 = I2
Putting the value of I2 in equation (1)
–6I1 + 4(4I1 – 6) = –5
–6I1 + 16I1 – 24 = –5
10I1 = 19
I1 = 1.9A
4(1.9) – 6 = I2
I2 = 1.6A
From equation 3
I3 = 4I2 = 4(1.6)
I3 = 6.4 A