The method of loop or mesh currents is generally used in solving networks having some degree of complexity. Such a degree of complexity already begins for a network of three meshes. It might even be convenient at times to use the method of loop or mesh currents for solving a two-mesh circuit.

The mesh-current method is preferred to the general or branch-current method because the unknowns in the initial stage of solving a network are equal to the number of meshes, i.e., the mesh currents. The necessity of writing the node-current equations, as done in the general or branchcurrent method where branch currents are used, is obviated. There are as many mesh-voltage equations as these are independent loop or mesh, currents. Hence, the M-mesh currents are obtained by solving the M-mesh voltages or loop equations for M unknowns. After solving for the mesh currents, only a matter of resolving the confluent mesh currents into the respective branch currents by very simple algebraic manipulations is required.

This method eliminates a great deal of tedious work involved in branch-current method and is best suited when energy sources are voltage sources rather than current sources. This method can be used only for planar circuits.

The procedure for writing the equations is as follows :

1. Assume the smallest number of mesh currents so that at least one mesh current links every element. As a matter of convenience, all mesh currents are assumed to have a clockwise direction.

The number of mesh currents is equal to the number of meshes in the circuit.

2. For each mesh write down the Kirchhoffs voltage law equation. Where more than one mesh current flows through an element, the algebraic sum of currents should be used. The algebraic sum of mesh currents may be sum or the difference of the currents flowing through the element depending on the direction of mesh currents.

3. Solve the above equations and from the mesh currents find the branch currents.

Fig. 36 shows two batteries E_{1} and E_{2} connected in a network consisting of three resistors.

Let the loop currents for two meshes be I_{1} and I_{2} (both clockwise-assumed). It is obvious that current through R3 (when considered as a part of first loop) is (I_{1}– 1_{2}). How ever, when R3 is considered part of the second loop, current through it is (I_{2} – I _{1}).

Applying Kirchhoffs voltage law to the two loops, we get

E_{1} – R_{1} R_{1} – R_{3}(I_{1} – I_{2}) = 0

E_{1} – I_{1} ( R_{1} + R_{3}) + I_{2} R_{3} = 0

– I_{2} R_{2} – E_{2} – R_{3} ( I_{2} – I_{1} ) = 0

– I_{2} R_{2} – E_{2} –I_{2 }R_{3} + I_{2} R_{3} = 0

I_{1} R_{3} – I_{2} ( R_{2} – R_{3})- E_{3} = 0

The above two equations can be solved not only to find loop currents but branch currents as

**Example 7**. Determine the currents through various resistors of the circuit shown in Fig. 37 using the concept of mesh

**Solution. **Refer Fig. 37.

Since there are two meshes, let the loop currents be as shown.

Applying Kirchhoffs law to loop 1, we get

24 – 4 I_{1} – 2(I_{1} – I_{2}) = 0

– 6 I_{1} + 2 I_{1} + 24 = 0

3 I_{1 }– I_{2 }= 12

For loop 2, we have

-2 (I_{2 }– I_{1}) – 6 I_{2} – 12 = 0

2 I_{1} – 8 I_{2} – 1_{2} = 0

I_{1} – 4 I_{1} = 6

Solving (i) and (ii), we get, I_{2} = 42 / 11 A

and I_{2} = – 6 /11

Hence Current through 4 Q resistor = 42 / 11 A (from L to M). (Ans.)

Current through 6 Q resistor = 6 / 11 A (from N to M). (Ans.)

Current through 2 Q resistor = 42 / 11 – ( – 6 /11) = 48 / 11 A ( From M to P)

**Example 8**. Determine the current supplied by each battery in the circuit shown in Fig. 38.

**Solution.** Refer Fig. 38.

As there are three meshes, let the three loop currents be as shown.

Applying Kirchhoff’s law to loop 1, we get

20 – 5 I1 – 3 ( I_{1 }– I_{2}) – 5 = 0

8 I_{1} – 3 I_{2} = 15

For loop 2, we have

– 4 I_{2} + 5 – 2 (I_{2} – I_{3}) + 5 + 5 – 3 (I_{2} – I_{1} ) = 0

3 I_{1} – 9 I_{2} + 2 I_{3} = – 15

For loop 3, we have

– 8 I_{3 }– 30 – 5 – 2( I_{3} – I_{2}) = 0

2 I_{2} – 10 I_{3} = 35

Eliminating I_{1} from (i) and (ii ), we get

63 I_{2} – 16 I_{3} = 165

Solving (iii) and (iv),we get

I_{2} = 1.82 A and I_{3} = – 3.15 A

(- ve sign means direction of current is counter-clockwise)

Substituting the value of 12 in (i), we get

I_{1} = 2.56 A

Current through battery B 1 (discharging current) = I_{1 }= 2.56 A. (Ans.)

Current through battery B2 (charging current) = I_{1}– I_{2} = 2.56- 1.82 = 0.74 A. (Ans.)

Current through battery B s (discharging current) = I_{2} + I_{3} = 1.82 + 3.15 = 4.97 A. (Ans.)

Current through battery B4 (discharging current)= I_{2} = 1.82 A. (Ans.)

Current through battery B5 (discharging current) = I_{3} = 3.15 A. (Ans.)

**Example 9. **Determine the currents through the different branches of the bridge circuit shown in Fig. 39.

**Solution.** Refer Fig. 39.

The three mesh currents are assumed as shown

The equations for the three meshes are :

For loop 1: 240 – 20 (I_{1} – I_{2}) – 500 (I_{1} – I_{2}) =0

– 70 I_{1} + 20 I_{2} + 50 I_{3} = – 240

70 I_{1} – 20 I_{2} – 50 I_{3} = 240

For loop 2: – 30 i2 – 40 ( I_{2} – I_{2}) – 200(I_{2} – I_{1}) = 0

20 I_{1} – 90 I_{2} + 40 I_{3} = 0

2 I1 – 9 I_{2} + 40 I_{3 }= 0

For loop 3: 60 i_{3} – 500 ( I_{3} – I_{1}) – 40 (I_{3} – I_{2}) = 0

50 I_{1} + 40 I_{2} – 150 I_{3 }= 0

5 I_{1} + 4 I_{2} – 15 I_{3} = 0

Solving these equations, we get

I_{1} = 6.10 A, I_{2} = 2.56 A, I_{3} = 2.72 A

Current through 30 Q resistor = I3 = 2.73 A (A to B)

Current through 60 Q resistor= I_{3} = 2.72 A (B to C). (Ans.)

Current through 20 Q resistor= I_{1}– I _{2} = 6.10-2.56 = 3.54 A (A to D).

Current through 50 Q resistor= I_{1}– I _{3} = 6.10-2.72 = 3.38 A (D to C).

Current through 40 Q resistor= I _{3} – I _{2} = 2. 72- 2.56 = 0.16 A (D to B).