Posted by **k** on Wednesday, March 20, 2013 at 4:35pm.

The figure below shows a collection of wires. The currents at the numbered points are: I1 = 2.4 mA right; I2 = 3.6 mA down; I3 = 0.60 mA right; I4 = 4.0 mA up; and I5 = 1.6 mA up.

i. imgur. com/SCyljHG. png

a. What is the current at point A (size and direction)?

b. What is the current at point B (size and direction)?

- physics -
**Elena**, Wednesday, March 20, 2013 at 6:18pm
We’ll use Kirchhoff’s Current Law: At any node (junction) in an electrical circuit, the sum of currents flowing into that node is equal to the sum of currents flowing out of that node.

Node: I₁, I₂ and I₁₂ currents:

I₁+I₂-I₁₂= 0

I₁₂=I ₁+I₂ = 2.6+3.4=6 mA (to the right)

Node: I₁₂, I₃ and I(A) currents:

I₁₂ - I₃ + I(A)=0

I(A) =I₃-I₁₂= 0.6-6=- 5.4 mA =>upward

Node: I₃, I₄, I₅, I₆:

I₃+ I₄- I₅ + I₆=0

I₆=-I₃- I₄+ I₅=-0.6-4+1.6=-3 A=> to the right

Node: I(A), I₅ and I₇:

I(A)+I₅-I₇=0

I₇=I(A)+I₅=5.4+1.6=7 mA => to the right

Node: I(B), I₆ and I₇:

I₆ + I₇-I(B)0

I(B)= I₆ + I₇=3+7=10 mA (upward)

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