Electrical Engineering - Branch, Loop and Node Analyses - Discussion

Discussion Forum : Branch, Loop and Node Analyses - General Questions (Q.No. 9)
9.

Find I1.

4I1 + 4I2 = 2
6I1 + 7I2 = 4
0.5 A
50 mA
–0.5 A
–50 mA
Answer: Option
Explanation:
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Discussion:
1 comments Page 1 of 1.

N.mikitharao. said:   1 decade ago
let us take 4I1+4I2=2.....eqn. no (1) & 6I1+7I2=4....eqn. no (2)
multiply the eqn no 1 by 7 and eqn no 2 by 4 then the eqns will be as follows 28I1+28I2=14....eqn no (3) & 24I1+28I2=16....eqn no (4)
solve eqns 3 & 4 den u get 4I1= -2
therfore I1= -0.5A

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