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Question
Using Kirchhoff’s laws of electrical networks, calculate the current I3.

Numerical
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Solution
I3 = I3 + I2
Applying KVL to the loop ABCDEFA,
2 I1 + 8 (I1 + I2) + 1 I1 = 33
2 I1 + 8 I1 + 8 I2 + 1 I1 = 33
11 I1 + 8 I2 = 33
Applying KVL to the loop BCDEB,
6 I2 + 8 (I1 + I2) + 2 I2 = 26
6 I2 + 8 I1 + 8 I1 + 2 I2 = 24
8 I1 + 16 I2 = 24
On solving, we get
I1 = 3A, I2 = 0
∴ I3 = I1 + I2 = 3A
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