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)
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)?
1 answer