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Current in a branch 
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#1
Jul1406, 05:56 PM

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 


#2
Jul1406, 06:22 PM

P: 835

The current accross the resistor is:
[tex] \frac{V_1V_g}{1k}  V_g = 0 [/tex] Ground is a node voltage also, you just define it as [itex] 0V[/itex] On a side note, if someone reads this thread...how do I make the "" larger in LaTeX? 


#3
Jul1406, 06:27 PM

PF Gold
P: 100

[tex] \left \frac{V_1V_g}{1k} \, \right \, V_g = 0 [/tex] 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: [tex] \frac{V_1V_g}{1k} \, \Bigg \, V_g = 0 [/tex] Test3, finally got it [tex] \big \Big \bigg \Bigg[/tex] 


#4
Jul1406, 06:37 PM

Mentor
P: 40,717

Current in a branch



#5
Jul1406, 06:50 PM

P: 145

So,the current that passes through the branch where is the voltage source in series with a resistance is (5v0v)/1k?



#6
Jul1406, 06:58 PM

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P: 40,717




#7
Jul1406, 07:09 PM

P: 835

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 :) 


#8
Jul1406, 07:13 PM

P: 835

[tex] f(x,y,z,t) \, \Bigg \, \begin{array}{c} x=y=z=1 & t =4 \end{array} \,\,\, = 2(1+1+1) [/tex] I believe it means evaluated at, or where... 


#9
Jul1406, 07:16 PM

P: 835

And to clarify this above, it would be: [tex] I_{R1} = \frac{V_1V_g}{1k}=\frac{V_1 (0)}{1k} = \frac{5}{1k} [/tex] 


#10
Jul1406, 07:21 PM

P: 145

But the equation for the node on the left(since the other node is ground) is:
1 + (v10)/1k + 5/1k=0 Or is: 1 + (v10)/1k + (v15)/1k=0 Edit:Thanks for the reply! 


#11
Jul1406, 07:24 PM

Mentor
P: 40,717

I feel much better now. Have a good weekend, all. 


#12
Jul1406, 07:27 PM

P: 835




#13
Jul1406, 07:39 PM

PF Gold
P: 100

[tex] f(x,y,z,t) \, \Big_{\substack {x=y=z=1 \\ t =4}} \, = 2(1+1+1) [/tex] _{\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. 


#14
Jul1406, 07:42 PM

P: 835

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:
The super node expression is: [tex] \frac{V_0V_g}{R_2}+\frac{V_1V_g}{R_1}=0[/tex] You write the expression for the voltage: [tex] V_1V_0 = 5 [/tex] And also note that Vg is defined as 0V so: [tex] V_g = 0 [/tex] THUS, [tex] \frac{V_0}{R_2}+\frac{V_1}{R_1}=0[/tex] [tex] V_1V_0 = 5 [/tex] Does that make sense? 


#15
Jul1406, 07:44 PM

P: 835

To be really picky do you know how to leftjustify the expression at the evaluatedat symbol? 


#16
Jul1406, 07:54 PM

PF Gold
P: 100

http://physicsforums.com/showthread.php?t=8997 [tex] f(x,y,z,t) \, \Big_{\begin{subarray}{l} x=y=z=1 \\ t =4 \end{subarray}} \, = 2(1+1+1) [/tex] 


#17
Jul1406, 08:15 PM

P: 835




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