In the following network of 5 branches, the respective currents are i 1 , i 2 , i 3 etc. Given that i 1 = – 0.5A, i 4 = 1A and i 5 = 0.5A, the remaining currents are (figure shown below) –
Text Solution
Verified by ExpertsThe correct answer is:
B
i 1 = – 0.5A, i 2 = 1A, i 5 = 0.5A
i 2 = 1.5A, i 3 = – 0.5A, i 6 = 0.5A
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