# Resistor Square Problem

Discussion in 'Homework Help' started by yan500, Jul 22, 2011.

1. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Hey all, hope you're having a good Friday. I've got a question on Question 18 of this worksheet: http://www.allaboutcircuits.com/worksheets/dc_sp.html

I can't understand why the answers are what they are. When they ask what is the resistance between A and B shouldn't it be 3k3 ohms?

2. ### blah2222 Well-Known Member

May 3, 2010
565
33
No because between points A and B, there is a 3k3 resistor in parallel with 4k7 + 1k5 + 2k7, which has to be accounted for.

Rab = 3k3 || (4k7 + 1k5 + 2k7) = 2k41

3. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Ohhhh, so you can essentially draw an imaginary line between the two leads then representing a wire?

4. ### blah2222 Well-Known Member

May 3, 2010
565
33
What do you mean by "imaginary line"?

5. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Sorry, I should have been clearer. What I mean is that point A essentially turns into a node that would split a current (if there was a current source) between AB and AC. And that would make it a parallel connection.

6. ### blah2222 Well-Known Member

May 3, 2010
565
33
Yes. If a current source were to be connected between nodes A and B. A current divider with two paths, A->B and A->C->D->B, would result.

yan500 likes this.
7. ### yan500 Thread Starter Member

Jul 12, 2011
48
0
Thank you so much for your help.