1. Find out the number of mesh loops:
From the circuit, we can see two independent mesh loops:
Mesh 1 (Left Loop): (V1, R1, R2, R3)
Mesh 2 (Right Loop): (V2, R3, R4, R5)
So, the number of mesh loops = 2.
2. Find out the current flow direction in every mesh loop:
We assume clockwise (CW) current flow in each loop unless otherwise specified:
Mesh 1: Current ( I1 ) flows clockwise (CW). Mesh 2: Current ( I2 ) flows clockwise (CW).
Both currents ( I1 ) and ( I2 ) interact at ( R3 ).
3. Find out all resistors’ polarity in every mesh loop:
The polarity across each resistor is determined by the current direction:
For Mesh 1:
- (R1): Left side (+), Right side (-)
- (R2): Left side (-), Right side (+)
- (R3): Top side (+), Bottom side (-)
- (V1)
For Mesh 2:
- (R4): Left side (+), Right side (-)
- (R5): Top side (+), Bottom side (-)
- (R3): Top side (-), Bottom side (+)
- (V2)
At (R3), both (I1) and (I2) flow in opposite directions, so the net voltage drop depends on their magnitudes.
4. Use KVL law in every mesh loop:
Applying Kirchhoff’s Voltage Law (KVL):
For Mesh 1:
V1 – (I1 R1) – (I1 R2) – (I1 – I2) R3 = 0
or, V1 – I1 (R1 + R2 + R3) + I2 R3 = 0
For Mesh 2:
V2 – (I2 R4) – (I2 R5) – (I2 – I1) R3 = 0
or, V2 – I2 (R3 + R4 + R5) + I1 R3 = 0
These equations can be solved to find (I1) and (I2).
Mesh Analysis Solution
Step 1: Identify Mesh Loops
The circuit has two independent mesh loops:
- Mesh 1 (Left Loop): Contains V1, R1, R2, R3.
- Mesh 2 (Right Loop): Contains V2, R3, R4, R5.
Step 2: Assign Mesh Currents
- Let I1 be the mesh current for Mesh 1 (Clockwise).
- Let I2 be the mesh current for Mesh 2 (Clockwise).
- R3 is a shared resistor between both meshes.
Step 3: Apply Kirchhoff’s Voltage Law (KVL)
Using KVL for each loop:
For Mesh 1 (Left Loop):
Applying KVL,
V1 – (I1 * R1) – (I1 * R2) – (I1 – I2) * R3 = 0
Substituting values V1 = 10V, R1 = 1Ω, R2 = 2Ω, R3 = 3Ω:
10 – (I1 * 1) – (I1 * 2) – (I1 – I2) * 3 = 0
Simplifying,
10 – I1 – 2I1 – 3I1 + 3I2 = 0
10 – 6I1 + 3I2 = 0
6I1 – 3I2 = 10 —-(Equation 1)
For Mesh 2 (Right Loop):
Applying KVL,
V2 – (I2 * R4) – (I2 * R5) – (I2 – I1) * R3 = 0
Substituting values V2 = 10V, R4 = 4Ω, R5 = 5Ω, R3 = 3Ω:
10 – (I2 * 4) – (I2 * 5) – (I2 – I1) * 3 = 0
Simplifying,
10 – 4I2 – 5I2 – 3I2 + 3I1 = 0
10 + 3I1 – 12I2 = 0
3I1 – 12I2 = -10 —-(Equation 2)
Step 4: Solve for I1 and I2
We now have two simultaneous equations:
6I1 – 3I2 = 10 —-(1)
3I1 – 12I2 = -10 —-(2)
Multiply Equation (2) by 2:
6I1 – 24I2 = -20
Now subtract Equation (1):
(6I1 – 24I2) – (6I1 – 3I2) = -20 – 10
-24I2 + 3I2 = -30
-21I2 = -30
I2 = 30 / 21
I2 = 1.428 A
Substituting I2 = 1.428 into Equation (1):
6I1 – 3(1.428) = 10
6I1 – 4.284 = 10
6I1 = 14.284
I1 = 14.284 / 6
I1 = 2.381 A
Final Answer
- I1 = 2.381 A (Clockwise in Mesh 1)
- I2 = 1.428 A (Clockwise in Mesh 2)
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