Page 129 - Basic Electrical Engineering

For loop BEFCB, the voltage equation is

− 5 I − 6 − (I − I )8 = 0
2
2
1
or, 8 I − 13 I = 6 (ii)
2
1

solving eqs. (i) and (ii), we get

I = 1.17 A, I = 0.26A
1
2
and current flowing through R is (I − I ) = 0.91 A
2
1
3
I is flowing through R , I is flowing through R and (I − I ) is flowing
1
2
1
1
2
2
through R .
3
Example 2.8 Using the mesh current method calculate the current flowing
through the resistors in the circuit shown in Fig. 2.27.

Figure 2.27

Solution:

Applying KVL in mesh I,

10 − 2 I − 6(I − I ) − 6 − 4 I = 0
1
1
1
2
or, 6 I + 3 I = 2 (i)
1
2