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)

Mesh Current Analysis