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)?
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)
a. Ah, point A, the source of all currents! Let me consult my calculations... *puts on thinking hat* So, if I1 is going right, I3 is also going right, but I4 and I5 are going up. Since A is where all the currents gather, let's sum them up...2.4 mA + 0.60 mA - 4.0 mA - 1.6 mA = -2.6 mA. So, the current at point A is 2.6 mA to the left, or should I say, "going against the flow." Such a rebel, that point A!
b. Ah, the mysterious point B! Well, let's do some more math wizardry. Since I2 is going down, that means we have a downward current. But what about the other currents? Well, I4 and I5 are both going up, while I1 and I3 have nothing to do with B. So we simply add them up...3.6 mA - 4.0 mA - 1.6 mA = -2.0 mA. Voila! The current at point B is 2.0 mA downwards, breaking all the rules, just like the rebellious teens.
Unfortunately, I am unable to view and process images. However, I can still explain how to solve this problem step-by-step.
To determine the current at point A and point B, we need to use Kirchhoff's current law (KCL). According to KCL, the sum of currents entering a node (such as point A or point B) must be equal to the sum of currents leaving that node.
Let's start with point A:
a. To determine the current at point A, we need to find the sum of currents entering and leaving point A. The currents entering point A are I1 and I3, both flowing to the right. The currents leaving point A are I2 and I4, both flowing down. So, the equation for point A using KCL is:
I1 + I3 = I2 + I4
Substituting the given values: I1 = 2.4 mA, I2 = 3.6 mA, I3 = 0.60 mA, and I4 = 4.0 mA:
2.4 mA + 0.60 mA = 3.6 mA + 4.0 mA
Simplifying the equation:
3.0 mA = 7.6 mA
Therefore, the current at point A is 3.0 mA to the right.
b. Now let's move on to point B.
To determine the current at point B, we can again use KCL. The currents entering point B are I2 and I4, both flowing down. The currents leaving point B are I5 flowing up. So, the equation for point B using KCL is:
I2 + I4 = I5
Substituting the given values: I2 = 3.6 mA, I4 = 4.0 mA, and I5 = 1.6 mA:
3.6 mA + 4.0 mA = 1.6 mA
Simplifying the equation:
7.6 mA = 1.6 mA
Therefore, the current at point B is 7.6 mA down.
Please note that without the visual reference, it might be challenging to accurately interpret and solve the problem.
To determine the current at point A and point B in the given circuit, we need to apply Kirchhoff's current law (KCL), which states that the sum of currents entering a node is equal to the sum of currents leaving that node.
Let's analyze the currents at each individual point:
i. At point A, there are two currents that contribute to the overall current. One wire carries a current of 2.4 mA right (I1), and the other wire carries a current of 0.60 mA right (I3). Since both currents are in the same direction, we can simply add them up to find the total current at point A:
Total current at A = I1 + I3
= 2.4 mA + 0.60 mA
= 3.0 mA (right)
Therefore, the current at point A is 3.0 mA, and it is flowing towards the right.
ii. At point B, there are three currents to consider. One wire carries a current of 3.6 mA down (I2), another wire carries a current of 4.0 mA up (I4), and the third wire carries a current of 1.6 mA up (I5). Since the two currents are flowing in opposite directions, we need to subtract the upward current from the downward current to find the net current at point B:
Total current at B = I2 - I4 - I5
= 3.6 mA - 4.0 mA - 1.6 mA
= -2.0 mA (down)
Therefore, the current at point B is 2.0 mA, and it is flowing downwards.
Note: The negative sign in front of the result indicates the opposite direction of the net current compared to the direction of the individual currents.