# Last questions on voltage dividers

Discussion in 'Homework Help' started by pilotnmech, Apr 23, 2005.

1. ### pilotnmech Thread Starter Member

Feb 26, 2005
27
0
I have a voltage divider that I tried to post, but it won't take it. It starts at the left where you tap off of the circuit 30mA at 300v, go through a resistor, tap off 10mA at 150v, go through another resistor, then all of the voltage is used up, and that end is grounded, go through another resistor (bleeder), tap off -3v at 0 A, go through another resistor, then tap off -5v at 0 A. The bias bleeder current is 60mA. The left side resistor is positive, and the right side (last) resistor is negative. How do I find the values of the resistors, and how can I have a negative voltage?

2. ### David Bridgen Senior Member

Feb 10, 2005
278
0
Try again. A picture is worth a thousand words, and your words don't make sense.

3. ### Firestorm Senior Member

Jan 24, 2005
353
0
try saving your picture as a gif.

4. ### n9xv Senior Member

Jan 18, 2005
329
1
Also, try right clicking your file and then "send to" a zipped folder and call up the file and attach as a .zip file. I think that method ultimately keept me from going insane trying to upload a file!

When looking at your voltage divider, the bleeder current flows through the entire series string. Each resistor drops X amount of voltage @ X amount of load current. So, mathematically the value of individual R would be;

(Voltage drop across individual R) / (Bleeder current + Load current through individual R)

Here's a spread sheet method of your voltage divider problem. When loaded, use "Ctrl + F" and select the "Go to" tab and type in A470 for the cell reference to voltage dividers. Try pluging in your numbers and see if helps at all - just a thought anyway.

I for one cant really comment on the negative voltages without seeing the circuit.

5. ### pilotnmech Thread Starter Member

Feb 26, 2005
27
0
Does this make sense?

6. ### dragan733 Senior Member

Dec 12, 2004
152
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A simple bmp picture occupies 80Mb!!
You scanned the drawing with a very big resolution 5000x5456 pixels and with 24 bits color. You had to scan the drawing with a resolution for example: 300x300 pixels and with scan mode: lineart and then the bmp picture will occupy only some kb.

R4=Ur4/Ib=-2/60mA=33,33 Ω
R3=Ur3/Ib=-3/60mA=50 Ω
Ir1=150/R1
Ir2=150/R2
Ir1=Ir2-10
150/R1=150/R2-10
R1*R2/(R1-R2)=15
IA=Ir1+30=150/R1+30
If we know IA, then one can calculate R1 and R2

7. ### n9xv Senior Member

Jan 18, 2005
329
1
Yea, try again pilotnmech. I thought it was just my crappy old compaq but it must have been as dragan733 stated above.

8. ### dragan733 Senior Member

Dec 12, 2004
152
0
pilotnmech needs not scan again the drawing

Here is the schematic drawn with ASCII

|...............|................|...............|...............|
|+300V.....|+150V......|0V..........|-3V...........|-5V
|30mA.......| 10mA.......|..............|0mA.........|0mA
|...............|.................|..............|................|
|...R1=?....|...R2=?......|...R3=?....|...R4=?...|
|--/\/\/\---+--/\/\/\-----+---/\/\/\--+---/\/\/\--+
|A............B.............C.|............... D.............E
| |......................GND ---
| |
| +

9. ### pilotnmech Thread Starter Member

Feb 26, 2005
27
0
I see how R3 and R4 are figured. Since there is no current tap at R3 and R4 the same .06A goes through both of them. Besides that there is no way to figure what R1 and R2 should be? Is there really such a thing as negative voltage, or does it only read negative because of the placement of my leads? Would there have to be a battery at the bottom of R4 that isn't shown?

10. ### David Bridgen Senior Member

Feb 10, 2005
278
0
Your diagram is either not complete or is erroneous. (or both)
It has 3 connections. One each at points A, C and E.
The following is based upon that.

The 60mA through R3 has to come from somewhere. The only place it can come from is point E, yet you have shown zero current from the -5V. This is impossible.
The same 60mA has to go somewhere too. There is a 40mA discrepancy between it and the sum of the currents at A and B.
If your figures for current at A and B are correct, then R1 is 7k5.
Yes.
You have two voltage sources in series. It doesn't matter whether either or both of them is a battery or a "power supply."

11. ### pilotnmech Thread Starter Member

Feb 26, 2005
27
0
David Bridgen, please don't answer any of my posts again. I'm sorry that you think you are a supreme being and don't want to be bothered by mere mortals. I would doubt you have many friends.

12. ### pebe AAC Fanatic!

Oct 11, 2004
628
3
I think you are being unfair to David. Far from thinking he is a 'supreme being', he has given a great deal of help to people on this, and several other, forums. Experts and newbies alike have all benifited from his knowledge.

I tried to download your image in ADC without success, and I think David probably had no luck either and was giving you direct answers to your questions - as he saw them - based on dragan733's ASCII diagram.

Would you have preferred that he waffled on like a politician with meaningless drivel that sounded nice?