# Potential Division

Discussion in 'Homework Help' started by ghola, Oct 28, 2012.

1. ### ghola Thread Starter New Member

Sep 14, 2012
9
1
Hi, I am trying to calculate voltage across 3 resistors and I was wondering if someone could point me in the right direction for the formula I should be looking at? please. I was OK whilst calculating voltages across individual resistors in series then things took a turn for the worse

The circuit is a constant dc supply leading to 2 resistors in parallel which then feed into a single resistor and back to the supply.

I have been looking at Kerchoff and Thevenin but I seem to have just confused myself and too be honest I'm not convinced that I'm even on the right track.

Feb 17, 2009
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3. ### ghola Thread Starter New Member

Sep 14, 2012
9
1
Hi,

Attached is the circuit. ( not shown in circuit but Voltage supply and all Resistor values are known )

I have calculated total circuit current by combining R1 and R2 then treating it as series with R3.

I then calculated individual current.

However then using Ohms law to try and get Voltages across each doesn't add up.

I'm not sure which tree I'm barking up There is only one voltage supply so I don't think I should be trying to work in Thevenin

Thanks again.

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Last edited: Oct 28, 2012
4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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You doing everything as it should be, so I don't know why you get bad result.
Show as step by step your calculations for this values.
R1 = R2 = 200Ω ; R3 = 100Ω and V1 = 10V.

5. ### ghola Thread Starter New Member

Sep 14, 2012
9
1

I'll set out below how I approached the values that you set. It appeared to go OK but when I later tried to replicate my steps with different R values, it all went pear shaped. Here's the steps:

10v/200Ω = 0.05

Rv1 = (0.05 x 100 ) / (200 + 200 ) = 0.0125

0.0125 x 200 = 2.5v

Rv2 = Same method as above

Rv3 = (10v x 100 ) / (100 + 100) = 1000/200 = 5v

I think I am confusing adding combined resistances or actual resistor values when trying to calculate voltage in the parallel resistors although this probably isn't clear from the above given the inter changeability of the values. I'm pretty sure the method for V3 is OK.

Thanks again for looking.

Last edited: Oct 29, 2012
6. ### crutschow Expert

Mar 14, 2008
13,478
3,364
You can use Kirchoff's laws to determine the values.
First calculate the equivalent parallel value of R1 and R2.
Then calculate the current through this equivalent value in series with R3.
Then, using that value of current, calculate the voltage drop across R3.
From that you can calculate the voltage across R1 and R2.

7. ### ghola Thread Starter New Member

Sep 14, 2012
9
1
Hi,

I think from my previous post these are the steps I took. Although I did step 4. then step 3. Would this make a difference ?

Above I calculated voltage drop across R1 and R2 by:

Rv1 = (0.05 x 100 ) / (200 + 200 ) = 0.0125

0.0125 x 200 = 2.5v

Rv2 = Same method as above.

Can anyone confirm if the following is correct, for parallel resistors, as I can't find any of my class notes on this point:

RV1 = (I x Combined Resistance Value) / (Actual Value of R1 + Actual Value of R2) = Answer
Then, Answer x Actual Value R1 = RV1 The voltage across R1

Apologies if it's glaring me in the face. Thanks for your continued help.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
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First step is to find equivalent resistance for R1 and R2 connect in parallel.
To find equivalent resistance for the parallel resistors we use this formula:
1/Req = 1/R1 + 1/R2 ... + 1/Rn

But when we have only two resistor connect in parallel the equation will be look like this:

1/Req = 1/R1 + 1/R2 = R2/(R1 * R2) + R1/(R1 * R2) = (R2 + R1 )/ (R1 * R2)

So Req = (R1 * R2)/(R1+R2)

But since R1 = R2 = R = 200Ω we find Req = R/n = 200Ω/2 = 100Ω

Now we can find the total current that is flow in this circuit.
Itot = 10V/(Req + R3) = 10V/200Ω = 50mA

Voltage across R3 resistor is equal

V3 = Itot*R3 = 5V

Step 3
Resistor R1 and R2 are connect in parallel and in parallel circuit the voltage is the same across elements but the current is splitting between the branches. So we have a current divider.

The situation look like this

And since R1 = R2 Itot is split in haft.

I1 = Itot*(R2/(R1+R2)) = 50mA * 200Ω/400Ω = 50mA * 0.5 = 25mA

And

I2 = Itot - I1 = 25mA First Kirchhoff's law or

I2 = Itot * (R1/(R1 + R2))

http://en.wikipedia.org/wiki/Current_divider

Step 4.

V1 = I1 * R1 = 5V
V2 = I2 *R2 = 5V

But we could use II Kirchhoff's law.

Vbat = V1 + V3 sine V3 = 5V

V1 = V2 = Vbat - V3

Look at those imagine

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9. ### WBahn Moderator

Mar 31, 2012
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Here's your primary problem (and why we always ask people to post their attempts). Ohm's law relates the value of a resistance with the voltage across THAT resistor and the current through THAT resistor. Here you are grabbing just any old voltage because it happens to be the only voltage you happen to see and blindly throwing an equation at it.

I had a suspicion that something like this was going to be the case after reading just the first paragraph of your original post. You weren't asking for an explanation or to gain any understanding, but just to be pointed in the right direction of a formula. That's a poor mindset. Instead, you should be saying to yourself, "Hmmm, I know KVL and KCL and Ohm's Law. What does that tell me about the relationships between the various voltages and currents in this circuit?" and then trying to write down equations that express those relationships and then trying to solve for what you want to know and then presenting that work and asking questions when that process goes awry.

So take a step back and consider the basics.

Q1) What does KVL tell you about the voltages along any closed path through a circuit?

Q2) What does KCL tell you about the currents at each and every node of a circuit?

Q3) What do you know about the voltage and/or currents through elements that are in series?

Q4) What do you know about the voltage and/or currents through elements that are in parallel?

Now assign an unknown voltage and current to each of the resistors in your diagram, including polarity. Make sure that the current in a resistor enters the end of the resistor that you have assigned as the positive end of the voltage across that resistor. For the voltage source, assign an unknown current that is leaving the positive end of the source.

Now apply the relationships you identified above to these unknowns. Use Ohm's Law to convert between a resistor voltage and the resistor current as needed.

If you still need help, then post your work along these lines and we will help guide you along.

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10. ### ghola Thread Starter New Member

Sep 14, 2012
9
1
Thanks to all who replied for your help. And thanks for the added diagram Jony.

WBahn - I think you mean well but your reply shows a lack of understanding of differentiated learning. Actually from that I might deduce that you are one of my lecturers .

In asking for the formula I thought that I could then work round the circuit and come to an understanding. That approach has worked best for me when learning other detailed technical subjects. I don't think I would presume to tell someone else how they should learn.

I don't understand the quote :

" Here you are grabbing just any old voltage because it happens to be the only voltage you happen to see and blindly throwing an equation at it."

This seems to have been in respect of my writing :

10v/200Ω = 0.05

Yet for the same step Jony130 wrote:

10V/200Ω = 50mA

So I'm genuinely not sure what you are saying here, it seemed to be a point I was fairly sure about and Jony130's answer seemed to confirm that, but I'll definitely look at it again.

11. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,393
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If you considered the basics that wbahn asked you about, your Rv1 formula that you wrote you. Would not have been considered.

My theorum is diffrrential learning isn't.

Of course only you know if you were guessing .... some of us highly suspect it

Last edited: Oct 31, 2012
12. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
By all means, then feel free to ignore everything I say since it doesn't mesh with your learning style.

This is because I thought that, when you said "Here's the steps" you were starting at Step 1, not Step 6ish. Starting with Step 1, you have a circuit with three resistors, two of which are 200Ω and neither of which have 10V across them. But you were actually starting from further down the line after a bunch of steps not shown had given you an equivalent total resistance that, coincidentally, happened to be numerically equal to one of the individual resistances. Had you distinguished the two, such as indicating that you were starting with Req=200Ω, I would have caught that that was what you were working from. Had I not been rushed to get out the door I might have still noticed it, but I don't think people should have to look halfway through your work to figure out what you meant in line one. Jony130 either did that or, very possibly, spotted it right away since the values you were using were values he supplied and therefore he already knew that this particular numerical equation would appear at Step 6ish and just assumed that that's where your steps started.

13. ### ghola Thread Starter New Member

Sep 14, 2012
9
1
Ahhh! I see what you are saying - Yes perhaps that is correct or another possibility is that he read my earlier post where I wrote:

I have calculated total circuit current by combining R1 and R2 then treating it as series with R3.
I then calculated individual current.
However then using Ohms law to try and get Voltages across each doesn't add up.

But seriously Thanks for your input ( no I am not being sarcastic ) you obviously take a lot of time out to help people and I'm sure it is appreciated. I can only apologise for not being precise/clear in my questions or answers and causing confusion.

14. ### WBahn Moderator

Mar 31, 2012
18,085
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Being clear and precise in describing problems is a skill that improves with practice and which very few people every really get down completely (and I'm certainly not one of the few).

I also have to occassionally remind myself to unjade myself and not assume that everyone that look's like their heading down the same wrong path is actually heading down that same wrong path -- they may just be on a path that looks the same, but is perhaps an alternate path to the right place (though I am usually pretty good at spotting that) or they may be on a different wrong path.

It usually takes a misstep like this one to unjade me for a while.

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