# Kirchoff's Circuit Law

1. Jan 23, 2014

### yaro99

1. The problem statement, all variables and given/known data
Determine the values of the other currents in Figure P1.37, given that ia = 2 A, ic =−3 A,
ig = 6A,and ih= 1A.

2. Relevant equations
∑currents entering a node = ∑currents exiting the node

3. The attempt at a solution

I used Kirchoff's Circuit Law on each of the outer nodes:

if = ig + ih = 6 + 1 = 7A

ia + id = if
2 + id = 7
id = 5A

ic + ih= ie
-3 + 1 = ie
ie = -2A

ib = ia + ic = 2 - 3 = -1A

I think I may have went wrong somewhere, because when I apply KCL to one of the inner nodes I get this:

ie + ig = id (?)
-2 + 6 ≠ 5

2. Jan 23, 2014

### Staff: Mentor

There's only one "inner node" where B, D, G, and E meet; despite there being two connection dots there, they are all one single conducting path with no components in between.

So, your node equation needs to account for $i_b$, too.

3. Jan 23, 2014

### yaro99

Aha, in that case it will be
-2 + 6 = -1 + 5
4 = 4

How do I know that the node includes $i_b$, is that current not affected by the current $i_d$?

Is the point connecting B and D not a node?

4. Jan 23, 2014

### Simon Bridge

The two inner nodes are actually one node.
You missed a current: two currents go into the node and two currents go out of the node.

gah: too slow!

5. Jan 23, 2014

### Staff: Mentor

So long as there is a continuous conducting path it is all one node. A node is not a dot on the diagram, it's the entire continuous conducting path.

To make it more clear, you can shift component E over a bit as follows:

The wiring in green is continuously connected, so it's all one node.

#### Attached Files:

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6. Jan 23, 2014

### yaro99

Ah, yes that confused me a bit. It makes sense now thanks!

7. Jan 23, 2014

### Simon Bridge

Great - the other way to look at it is to add an extra current arrow $\small{i_0}$ in the gap between the "two" center nodes. You can tell which direction it has to point in by looking at the other currents - but it actually doesn't matter:

Current in = current out:
... so on the left node you get $i_0 = i_b+i_d$
... and for the right node you get $i_0=i_g+i_e$

... eliminate the $\small{i_0}$ term gives you $i_b+i_d = i_g+i_e$

and it all comes out in the wash ;)

It's just easier to realize that the nodes for Kirkoffs laws actually include the entire wire between components. Thus two drawn node-symbols, on the diagram, that have no components between them, are one and the same.