Question
Question: In the following network of 5 branches, the respective currents are i<sub>1</sub>, i<sub>2</sub>, i<...
In the following network of 5 branches, the respective currents are i1, i2, i3 etc. Given that i1 = – 0.5A, i4 = 1A and i5 = 0.5A, the remaining currents are (figure shown below) –
A
i2 = – 1.5A, i3 = 0.5A, i6 = 0.5A
B
i2 = 1.5A, i3 = – 0.5A, i6 = 0.5A
C
i2 = 1.5A, i3 = 0.5A, i6 = – 0.5A
D
i2 = 1.5A, i3 = 0.5A, i6 = + 0.5A
Answer
i2 = 1.5A, i3 = – 0.5A, i6 = 0.5A
Explanation
Solution
i1 = – 0.5A, i2 = 1A, i5 = 0.5A
i2 = 1.5A, i3 = – 0.5A, i6 = 0.5A