Kindly note:
The information given in the question is not sufficient to solve the question.
Adjusting the conditions of the question to:
Find the current through the branch AB in the circuit diagram below.
Solution:
In loop XYBA:
3I2−1I1=0I1=3I2
In loop WZBA
4(I2+Iab)−2(I1−Iab)=0
6Iab=2I1−4I26Iab=2(3I2)−4I2I2=3Iab
In loop ZQPY
50−[3I2+4(I2+Iab)]=050=7I2+4Iab
=7(3Iab)+4Iab=25Iab
Iab=2A
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