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Electrical Engineers, your assistance is needed-R/2R ladders

Polandspring88

Polandspring88

Superman
Mar 31, 2004
3,066
7
Broomfield, CO
So we had a quiz with the following problem on it.



Please explain to me why the voltage drop at Va is 1/2 Vs, Vb is 1/2 Va and so on. I have sat here staring at this problem for 2 to 3 hours and nothing has come to me. Any help would be greatly appreciated.

Ok, so apparantly the image does not work. Now it does work but is super small. Unless you have microscopes for eyes, see the problem it is here...
http://ece.wpi.edu/~fjlooft/3601_c06/class/quiz1_soln.pdf
Problem #4
 
N8 v2.0

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
Polandspring88 said:
So we had a quiz with the following problem on it.



Please explain to me why the voltage drop at Va is 1/2 Vs, Vb is 1/2 Va and so on. I have sat here staring at this problem for 2 to 3 hours and nothing has come to me. Any help would be greatly appreciated.

Ok, so apparantly the image does not work. If you want to see the problem it is here...
http://ece.wpi.edu/~fjlooft/3601_c06/class/quiz1_soln.pdf
Problem #4
Click to expand...

M'kay... looking at it now...
 
D

druber

Chimp
May 23, 2005
90
0
Start from the end.

The equivalent resistance from Vb to ground is (1/(1/(R+R) + 1/2R)) = R.

Repeat for equivalent resistance from Va to ground and you get (1/(1/(R+R) + 1/2R)) = R...exactly the same as before.

Therefore, the voltage at Va = 1/2Vs = 5V.

Apply the same reasoning to Vb and Vc and you get the same thing.

Hard to explain things over the internet though....

Plus...it's been many, many years since that class.
 
urbaindk

urbaindk

The Real Dr. Science
Jul 12, 2004
4,819
0
Sleepy Hollar
Think equivalent circuits. Va is easy. If you take all the resistors to the left of Va and reduce all the resistors into 1 equivalent resistor you'll end up with a R resistor in edit: series with another R resistor. The voltage at the node is then 1/2V. I think you just keep going like that for Vb and Vc.



edit: Got a little confused there. That's what I get for not drawing a picture.
 
D

druber

Chimp
May 23, 2005
90
0
jdschall said:
Think equivalent circuits. Va is easy. If you take all the resistors to the left of Va and reduce all the resistors into 1 equivalent resistor you'll end up with a 2R resistor in parallel with another 2R resistor. The voltage at the node is then 1/2V. I think you just keep going like that for Vb and Vc.
Click to expand...
I think you mean one R resistor in series with one R resistor with Va in the middle. Right?
 
D

druber

Chimp
May 23, 2005
90
0
jdschall said:
Around 10 for me. I had to get out the book.
Click to expand...
10 for me too. I looked for the book, but couldn't find it. :(

Doing logic design now...I haven't had to think about a resistor in a long time. :)
 
N8 v2.0

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
druber said:
Start from the end.

The equivalent resistance from Vb to ground is (1/(1/(R+R) + 1/2R)) = R.

Repeat for equivalent resistance from Va to ground and you get (1/(1/(R+R) + 1/2R)) = R...exactly the same as before.

Therefore, the voltage at Va = 1/2Vs = 5V.

Apply the same reasoning to Vb and Vc and you get the same thing.

Hard to explain things over the internet though....

Plus...it's been many, many years since that class.
Click to expand...
Yep, that's what I was getting at... Req behind node Va is = to R.

So you are left with a simple series circuit with two R value resistors, so the current stays the same and the voltage drop thru the first R + the voltage drop thru the second R must equal the source voltage.
 
N8 v2.0

N8 v2.0

Not the sharpest tool in the shed
Oct 18, 2002
11,003
149
The Cleft of Venus
BSEE for me in Mar 1994.

Haven't work a problem like that since Circuits I in 1989... :p

Power factor problems using KVAR on the other hand... :thumb:
 
D

druber

Chimp
May 23, 2005
90
0
N8 said:
BSEE for me in Mar 1994.

Haven't work a problem like that since Circuits I in 1989... :p

Power factor problems using KVAR on the other hand... :thumb:
Click to expand...
BSEE in 1999.

I also haven't worked a problem like that since Circuits I (1995).
 
Polandspring88

Polandspring88

Superman
Mar 31, 2004
3,066
7
Broomfield, CO
Thanks a lot guys. Just figured out how to do all of the resistor equivalence and circuit reduction. With your help I shall boost my quiz grade into A range.

I am an ME and circuits have always been my nemesis, so this kind of stuff is the devil to me.
 
Jou

Jou

Monkey
May 16, 2004
235
0
Powell/Laramie, Wyoming
i took circuits last semested and i have no bleeding idea. My teached was friggen Vandersex which didn't help.

I'm ME also and feel that EE and computer Eng are out of thier minds.
 
WarEagle2K

WarEagle2K

Chimp
Feb 28, 2005
92
0
Tucson, AZ
I got my BSEE in 2000. I am sure there is a nifty trick involved. But if all else failed, and they let you use a calculator, I would set up a nodal equation using arbitrary values for the R's and solve it with the matrix solver. You will see the same voltage relationships. A note of encouragement...I am currently working on an MSEE, and I haven't had to do "much" circuit analysis. You can simulate all that stuff and worry about the really important things.
 
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