Current up to a node?

[chopnhack]

I understood all of them except question g. - Shouldn't the current at node I and H be the same since the current hasn't passed through the node yet? I.E. the voltage drop across the resistor is what determines how much current is distributed across this parallel network. Shouldn't the current be the same up to that point? Is the question poorly worded or am I misunderstanding this?

 

[Ratch]

View attachment 26582

I understood all of them except question g. - Shouldn't the current at node I and H be the same since the current hasn't passed through the node yet? I.E. the voltage drop across the resistor is what determines how much current is distributed across this parallel network. Shouldn't the current be the same up to that point? Is the question poorly worded or am I misunderstanding this?

You are misunderstand the question. How can H and I be the same when I takes the load of both H and F?

Ratch

 

[chopnhack]

Thank you Ratch, I follow your logic to a point - i.e. if the question asked I vs G - obviously, I would be greater since the current is distributed through and after the resistor. I just got caught up in the line of demarcation. Which I still wonder, just where exactly does that occur? If I were to measure current between I and H, would I find what would be apportioned through the 5Ω resistor?

 

[Ratch]

Thank you Ratch, I follow your logic to a point - i.e. if the question asked I vs G - obviously, I would be greater since the current is distributed through and after the resistor. I just got caught up in the line of demarcation. Which I still wonder, just where exactly does that occur?

At the node points.

If I were to measure current between I and H, would I find what would be apportioned through the 5Ω resistor?

Voltage is measured between nodes. Current is measured within branches of the nodes. In other words, you cannot put a ammeter between H and I for meaningful results.

Ratch

 

[chopnhack]

In other words, you cannot put a ammeter between H and I for meaningful results.

I think I understand - there is no potential difference between I and H - I and H should have the same potential difference, the difference will drop across the node.
Thanks again!!

 

[hevans1944]

Thank you Ratch, I follow your logic to a point - i.e. if the question asked I vs G - obviously, I would be greater since the current is distributed through and after the resistor. I just got caught up in the line of demarcation. Which I still wonder, just where exactly does that occur? If I were to measure current between I and H, would I find what would be apportioned through the 5Ω resistor?

Current is measured at a point in the circuit, meaning that you would cut the circuit at the point and insert your ammeter there. Please do not confuse the points labeled on your diagram with nodes. KCL says all the current entering a node is equal to all the current leaving the node. This is a trivial statement for a node with only two connections! In your diagram the current at points A, B, C, D, I, J, K, and L are all equal. The current at point E is equal to the current at point F. The current at point G is equal to the current at point H.

Some confusion might occur if you consider the node where the wires carrying currents F, H, and I are connected together. This node is not identified as a point on your diagram. KCL says the algebraic sum of the currents at F, H, and I must be zero at this node, but there is no way you can measure that! Actually, KCL says the algebraic sum of the currents at any node will always sum to zero, no matter how many currents are entering or leaving the node. All this statement means is charge does not accumulate at a circuit node. Since current is charge in motion, whatever charge (or current) enters a node must be equal to the charge (or current) that leaves the node. All you can do is measure each current individually and observe that their algebraic sum is zero. There is no possible way to "cut the wire" at the node where three wires (or more) join together to insert an ammeter to measure the current at the node.

If I were to measure current between I and H, would I find what would be apportioned through the 5Ω resistor?

If you cut the connection between points I and H and insert an ammeter, you would measure the current at H, which is also the current through the 5 Ω resistor. However, the connection between F and I would remain intact, so you would not be measuring the current in the 3 Ω resistor.

 

[chopnhack]

Thanks for the clarification Hop. I shouldn't have used the word node, point would have been a better choice.

 

[hevans1944]

A node is a point where two or more wires come together. If you place a point on a wire, that point becomes a two-wire node... a trivial construct. If you place a point where three (or more) wires are connected in common, then that becomes a three (or more) wire node. And so on, for as many wires as you care to add to that node. The whole point of KCL is that charge does not accumulate at nodes, so the currents in the wires connected to a node always algebraically sum to zero. Any node, any number or wires.

How can you determine if a wire is connected to a node? There will always be zero potential between that wire and the other wires connected to the same node. This all goes hand-in-hand with KVL, which states that the voltage drops around any circuit loop must algebraically add to zero. Together, they form a powerful analysis tool for series, parallel, and combinations of series and parallel circuits.

Throw in Ohm's Law and Thevenin Equivalent Circuits and you pretty much have DC circuit analysis in your tool bag.

 

[chopnhack]

Throw in Ohm's Law and Thevenin Equivalent Circuits and you pretty much have DC circuit analysis in your tool bag.

Easier said than done! LOL

I have been practicing Ohm's Law and its derivations on combination circuits and have been getting the hang of it, conceptualizing to find the ESR of a parallel grouping, etc. My current text of choice is Tab Electronics Guide to Understanding Electricity and Electronics. Was able to pick it up for $4.24 shipped from Abebooks.com worth every penny