Question

Use kirchhoff's rule to write the expression for current i1,i2,i3 in the circuit diagram

Answer 1

I1 = 2/13A

I2 = 1 - 3I1 = 7/13A

I3 = I1 + I2 = 9/13A

Explanation:

Applying Kirchhoff’s first rule, we get:

 I3 = I1 + I2  as algebraic sum of currents meeting at any junction is zero.

Applying Kirchhoff’s second rule, we get:

 -2 - 4I1 + 3I2 +1= 0

 4I1 - 3I2 = - 1 ------------(1)

Also, as per second rule, we get:

  -2 - 4I1 - 2I3 + 4 = 0

  2I1 + I3 = 1

Writing I3 as sum of I1 and I2, it becomes:

 2I1 + (I1 + I2 ) = 1

 3I1 + I2 = 1 ------------(2)

Solving 1 and 2, we get:

I1 = 2/13A

I2 = 1 - 3I1 = 7/13A

I3 = I1 + I2 = 9/13A

Answer 2

Answer:

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Explanation:

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