Question

Apply Kirchhoff's rules to find the currents I1, I2, I3 (passing through 2 ohm, 1 ohm, 4 ohm respectively) in the circuit shown in the attachment.
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Answer 1

Answer:

Current in the three branches are

$I_1 = 2.57 A$

$I_2 = -0.86 A$

$I_3 = 1.71 A$

Explanation:

As per Kirchoff law we know

$12 = 4 I_3 + 2 I_1$

$6 = 4 I_3 + 1 I_2$

also by junction law of current we have

$I_1 + I_2 = I_3$

now from above all equations we have

$12 = 6I_3 - 2I_2$

so we have

$24 = 14 I_3$

$I_3 = 1.71 A$

now we have

$I_2 = -0.86 A$

so we have

$I_1 = 2.57 A$

#Learn

Topic : Kirchoff's law