Question

ce Fand the currents I, I and I, in the curcuit shown below 1091 2.0 Ω

Answer 1

I₁ + I₂ = I₃

loop abcda: 10 – 6I₁ – 2I₃ = 0

loop befcb: -4I₂ – 14 + 6I₁ – 10 = 0

-4I₂ + 6I₁ = 24

-2I₂ + 3I₁ = 12 ..........(1)

5 – 3I₁ – I₃ = 0

=> 5 – 3I₁ = I₁ + I₂

=> 5 = 4I₁ + I₂

=> I₂ = 5 – 4I₁ ............(2)

Using the above two equations we get:

-2(5 – 4I₁) + 3I₁ = 12

=> -10 + 8I₁ + 3I₁ = 12

=> 11I₁ = 22

=> I₁ = 2 Amps

I₂ = -3 Amps

I₃ = -1 Amps

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