Page 20 - Basics of Electrical, Electronic and Communication

30 Chapter 2. Kirchoff’s laws

Example 5
Find the loop currents in the electric circuit shown in Figure 2.13.

5 Ω 2 Ω

I2
+ 1 2
5 V − 1 Ω

I1
2 Ω

+
10 V − 3 3 Ω

I3

Figure 2.13:

Solution
From the loop 1 : 5I 1 + (I 1 − I 2 ) = 5

5I 1 + (I 1 − I 2 ) = 5 (2.8)

From loop 2 : 2I 2 + (I 2 − I 1 ) + 2(I 2 − I 3 ) = 0

5I 2 − I 1 − 2I 3 ) = 0 (2.9)

From loop 3: (I 3 − I 2 )2 + 3I 3 = 10

5I 3 − 2I 2 = 10 (2.10)

Solving 2.8, 2.9 and 2.10, we get,
I 1 = 1.03 A
I 2 = 1.198 A

I 3 = 2.48 A

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