Page 131 - Basic Electrical Engineering

Solution:

Applying KVL in mesh I, II, and III respectively, we get

−4I , −2(I , − I )−3(I , −I )−V + 24 = 0
1
2
1
1
3
or, 9I − 2I − 3I = 24− V (i)
2
1
3
−4I − 6(I −I )−2(I −1 ) = 0
2
1
2
3
2
or, I − 61 + 3I = 0 (ii)
2
3
1
− 6 (I −I )− 2I + V − 3(I − I ) = 0
3
1
3
3
2
or, 3I + 6I − 11I = − V (iii)
3
2
1
From (i), (ii), and (iii) the determinants ∆ and ∆ or ∆ are
1
a

According to Cramer’s rule,

Here condition is that I must be zero.
1

So, Δ must be zero
1