Question

Calculate each of the unknown currents I₁, I₂, and I₃ for thecircuit.

I1 =
I2 =
I3 =

Answer 1

For this solution, Kirchhoff's Rules will be applied. There are three unknowns so three equations must be set up and then do a lot ofsubstitution.

Equation 1 will be the junction equation.

I₁ = I₂ + I₃ (equation 1)

Equation 2 is chosen to loop clockwise in the upper loop (Remember that in any loop, the total potential difference must be zero). Refer toKirchhoff's rule if you do not understand why I used + and - where I did.

24 - I₁R₁ - 4I₁ - 3I₃ = 0 or 24 - 2.4I₁ - 4 I₁ - 3I₃ = 0

Simplify to get 24 - 6.4I₁ - 3I₃ = 0 (equation 2)

Equation 3 is chosen to loop clockwise in the lower loop. (Same rules apply)

V +3I₃ - R₂I₂ - 5I₂ = 0 or 10 + 3I₃ - 1.3I₂ - 5I₂ = 0

Simplify toget10 + 3I₃ - 6.3I₂ = 0 (equation 3)

Now to start substituting...

We know that I₁ = I₂ + I₃ from equation 1. I will substitute that into equation 2

24 - 6.4(I₂ + I₃) - 3I₃ = 0

Distribute and simplify to get

24- 6.4I₂ - 9.4I₃ = 0 (call this equation 4)

From equation 3, I will solve for I₂

6.3I₂ = 10 + 3I₃ or, after dividing both sides by 6.3, you getI₂ = 1.59 +.48I₃ (Call this equation 5)

I will take equation 5 and substitute what I know for I₂ into equation 4 and therefore be left with only one unknown (I₃)

24 - 6.4(1.59 + .48I₃) - 9.4I₃ = 0

Distribute and simplify to get

13.82 -12.48I₃ = 0

Solve for I_3, I₃ = 1.11 A

Substitute I₃ back into equation 4 to find I₂

24 - 6.4I₂ - (9.4)(1.11) = 0

Solve for I₂, I₂ = 2.12 A

Finally substitute I₂ and I₃ back into equation 1 to find I₁

I₁ = 3.23 A