# Current in a branch

by esmeco
Tags: branch, current
 P: 145 I was wondering,when we have a voltage source connected to a node in series with a resistance connected to the ground like this: ,when determining the current on the branches through node method is the current on that branch 5/1k or v1+5/1k? http://i75.photobucket.com/albums/i2...nchcurrent.jpg and a resistance co
 P: 838 The current accross the resistor is: $$\frac{V_1-V_g}{1k} | V_g = 0$$ Ground is a node voltage also, you just define it as $0V$ On a side note, if someone reads this thread...how do I make the "|" larger in LaTeX?
PF Gold
P: 100
 Quote by FrogPad The current accross the resistor is: $$\frac{V_1-V_g}{1k} | V_g = 0$$ Ground is a node voltage also, you just define it as $0V$ On a side note, if someone reads this thread...how do I make the "|" larger in LaTeX?
Test:

$$\left| \frac{V_1-V_g}{1k} \, \right| \, V_g = 0$$

hmm...was trying the \left and \right command which works for brackets \left( content \right). I only got it to work with a leading \left| command to go with the \right|.

Test2:

$$\frac{V_1-V_g}{1k} \, \Bigg| \, V_g = 0$$

Test3, finally got it

$$| \big| \Big| \bigg| \Bigg|$$

Mentor
P: 39,722

## Current in a branch

 Quote by nazzard Test: $$\frac{V_1-V_g}{1k} \, \right| \, V_g = 0$$ hmm...was trying the \left and \right command which works for brackets \left( content \right)
I must be brain dead (well, it is Friday afternoon, and it's been a tough week). What equation are you guys trying to write (a power?), and what is the "|" character supposed to be? Sorry if I'm being totally lame. Maybe I should go home....
 P: 145 So,the current that passes through the branch where is the voltage source in series with a resistance is (5v-0v)/1k?
Mentor
P: 39,722
 Quote by esmeco So,the current that passes through the branch where is the voltage source in series with a resistance is (5v-0v)/1k?
Nope, not necessarily. If you erase the little stub at the lower left that is labelled I1, and the little stub off the lower right that goes nowhere, then the loop current is 5V/(R1+R2). But if you inject some extra I1 current in that lower left stub (without showing it in a standard way, BTW) or something-something at the lower right stub, then you need to take that into account....
P: 838
 Quote by nazzard Test: $$\left| \frac{V_1-V_g}{1k} \, \right| \, V_g = 0$$ hmm...was trying the \left and \right command which works for brackets \left( content \right). I only got it to work with a leading \left| command to go with the \right|. Test2: $$\frac{V_1-V_g}{1k} \, \Bigg| \, V_g = 0$$ Test3, finally got it $$| \big| \Big| \bigg| \Bigg|$$

thanks man!

Yeah I was trying the \left and \right commands, but wasn't getting it either. Well cool. I didn't know about the big commands. Thanks again :)
P: 838
 Quote by berkeman I must be brain dead (well, it is Friday afternoon, and it's been a tough week). What equation are you guys trying to write (a power?), and what is the "|" character supposed to be? Sorry if I'm being totally lame. Maybe I should go home....
$$f(x,y,z,t) = t(x+y+z)$$

$$f(x,y,z,t) \, \Bigg| \, \begin{array}{c} x=y=z=1 & t =4 \end{array} \,\,\, = 2(1+1+1)$$

I believe it means evaluated at, or where...
P: 838
 Quote by FrogPad The current accross the resistor is: $$\frac{V_1-V_g}{1k} | V_g = 0$$ Ground is a node voltage also, you just define it as $0V$ On a side note, if someone reads this thread...how do I make the "|" larger in LaTeX?

And to clarify this above, it would be:
$$I_{R1} = \frac{V_1-V_g}{1k}=\frac{V_1 -(0)}{1k} = \frac{5}{1k}$$
 P: 145 But the equation for the node on the left(since the other node is ground) is: 1 + (v1-0)/1k + 5/1k=0 Or is: 1 + (v1-0)/1k + (v1-5)/1k=0 Edit:Thanks for the reply!
Mentor
P: 39,722
 Quote by FrogPad $$f(x,y,z,t) = t(x+y+z)$$ $$f(x,y,z,t) \, \Bigg| \, \begin{array}{c} x=y=z=1 & t =4 \end{array} \,\,\, = 2(1+1+1)$$ I believe it means evaluated at, or where...
Ohhhhh! Evaluated at! Now I get what you were trying to do. But the evaluation condition would normally be in smaller font, down at the bottom of the long vertical line, I believe. At least that's how I've seen it before.

I feel much better now. Have a good weekend, all.
P: 838
 Quote by berkeman Ohhhhh! Evaluated at! Now I get what you were trying to do. But the evaluation condition would normally be in smaller font, down at the bottom of the long vertical line, I believe. At least that's how I've seen it before.
Yeah. I don't know how to do that in LaTeX though

 Quote by berkeman I feel much better now. Have a good weekend, all.
You too man, have a good one!
PF Gold
P: 100
 Quote by FrogPad $$f(x,y,z,t) = t(x+y+z)$$ $$f(x,y,z,t) \, \Bigg| \, \begin{array}{c} x=y=z=1 & t =4 \end{array} \,\,\, = 2(1+1+1)$$ I believe it means evaluated at, or where...
Another test-session:

$$f(x,y,z,t) \, \Big|_{\substack {x=y=z=1 \\ t =4}} \, = 2(1+1+1)$$

_{\substack{content}} does the trick. Not sure if it could be placed further down, so that the equals sign is placed more to the left.
P: 838
 Quote by esmeco But the equation for the node on the left(since the other node is ground) is: 1 + (v1-0)/1k + 5/1k=0 Or is: 1 + (v1-0)/1k + (v1-5)/1k=0 Edit:Thanks for the reply!
Where is the 1 from!?

 Quote by esmeco But the equation for the node on the left
I don't know what this means. Label the node you are talking about, and post that circuit.

For sake of clarity. Lets just assume that your circuit is a single loop. Thus those little branches at the bottom are not connected to anything, so they therefore have no current running through them.

so your circuit looks something like this:

____/\/\/______
|                     |
|                     |
( + )                  |
( - )                   |--|||
|                     |
|____/\/\/_____|
Now if you want to use nodal analysis, you have use a super node. So you label TWO nodes. Both are on the left hand side. One you have already labeled V1, the other lets call V0 (and this is in the bottom left hand corner).

The super node expression is:
$$\frac{V_0-V_g}{R_2}+\frac{V_1-V_g}{R_1}=0$$

You write the expression for the voltage:
$$V_1-V_0 = 5$$

And also note that Vg is defined as 0V so:

$$V_g = 0$$

THUS,
$$\frac{V_0}{R_2}+\frac{V_1}{R_1}=0$$
$$V_1-V_0 = 5$$

Does that make sense?
P: 838
 Quote by nazzard Another test-session: $$f(x,y,z,t) \, \Big|_{\substack {x=y=z=1 \\ t =4}} \, = 2(1+1+1)$$
dude you are the man at LaTeX!

To be really picky do you know how to left-justify the expression at the evaluated-at symbol?
PF Gold
P: 100
 Quote by FrogPad dude you are the man at LaTeX! To be really picky do you know how to left-justify the expression at the evaluated-at symbol?
I'm just clicking on various examples in the following thread

$$f(x,y,z,t) \, \Big|_{\begin{subarray}{l} x=y=z=1 \\ t =4 \end{subarray}} \, = 2(1+1+1)$$
 Quote by nazzard I'm just clicking on various examples in the following thread http://physicsforums.com/showthread.php?t=8997 $$f(x,y,z,t) \, \Big|_{\begin{subarray}{l} x=y=z=1 \\ t =4 \end{subarray}} \, = 2(1+1+1)$$