Page 125 - Basic Electrical Engineering

through the branch AB. Similarly, applying KCL at node C we see that the
sum of currents entering the node is equal to the current coming out of the
node. Thus current distribution in the various branches is perfectly done.

Now, we will apply KVL to the loop ABCA and loop ACDA.
From loop ABCA we can write the voltage equation as

−1(2 − I − I ) − 2(2 − I − I ) + 5 I = 0 (i)
1
2
1
2
2
or, 3I + 8I = 6
1
2
From loop ACDA we can write

−5I + 3I + 4I = 0
1
1
2
or, 7I + 5I = 0 (ii)
2
1
To solve eqs. (i) and (ii), multiply eq. (i) by 7 and (ii) by 3 and subtract as

21I + 56 I = 42 (i)
2
1

21I − 15 I = 0 (ii)
2
1

From which

and