# Finding an unknown resistor and current value.

Discussion in 'Homework Help' started by pAntz, Nov 7, 2015.

1. ### pAntz Thread Starter New Member

Nov 7, 2015
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Evening all, first post here and have searched similar threads but can't seem to find anything that resembles my question. And it appears most OPs don't provide too much information.

Please find attached a copy of the question I am unable to solve.

From the diagram I can see input voltage is 100v and RT is 25.59ohms. From this I can work IT to be 3.91amps.

If I'm correct, I make the voltage across the 10ohm resistor to be 39.1volts, therefore, the output voltage from that resistor being 60.9volts. (100-39.1)

From here, I have no idea why, but I'm going round in circles. I know this must be so simple but my figures are just absurd. I know I'm going to kick myself when I figure it out. Every questions up until this one has been find power or current of a resistor and all values have been given. I am more than happy to work out RT (when it isn't given) and I'm fine with the voltage divider rule. But now this question has been presented I'm stumped.

Any help/guidance would be greatly appreciated.

Thank you.

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2. ### shteii01 AAC Fanatic!

Feb 19, 2010
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You have a voltage source and total resistance of the circuit at 25.59 Ohm. Now you work backwards to when you had 10 Ohm resistor and "Resistor 1". What do you know about Resistor 1? You know that it is in series with 10 Ohm resistor. What ability do series resistors have? They add together. So? So, 25.59 Ohm resistor is 10 Ohm + Resistor 1 = 25.59 Ohm. This way you find that Resistor 1 is 25.59-10=15.59 Ohm.

Keep working this way until you have R and 80 Ohm resistor in parallel.

3. ### WBahn Moderator

Mar 31, 2012
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Welcome, pAntz.

Yes, most OPs don't provide anywhere near enough information. That is a never-ending problem.

Before we get started, try not to post such large files when they aren't needed. For instance, I used Paint to crop your image and reduce it to 500 pixels wide and, even after marking it up, it is now only 39 kB, less than 3% of your file. Many folks have slow connections and won't download needlessly large files. Making the image a reasonable size also lets you insert it directly into the post where it is easier for readers to work with.

The first thing you should always do it define all of the variables you have used (or plan to use). This includes identifying the common node (often called 'ground') which is usually defined to be 0V. Then labeling all of the nodes in the equation. Finally, indicate all of the currents, including a label and a polarity (direction) for each. Now you are in a position to communicate unambiguously with us, with the grader, and with yourself.

Now start with what you want to know and ask yourself what else you need to know to find it.

You want to know IL and R. If you knew Vc and I3, could you find IL and R?

If so, then you have a new problem -- finding Vc and I3.

If you knew Vb and I3, could you find Vc?

If so, you have a new problem -- finding Vb and I3.

If you know I1 and Vb, could you find I3?

If so, you have a new problem -- finding Vb and I1.

If you know Va and I1, can you find Vb?

If so, you have a new problem -- finding I1. Can you do that?

4. ### pAntz Thread Starter New Member

Nov 7, 2015
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Thanks for the swift reply, and the tips on future posts.

I have taken aboard what you said and started to work through it, and have come to a standstill yet again. For some reason I just can't get my head around something which should be so simple.

So I have a total voltage and resistance. I can work out from this that the total current for the circuit is 3.9 amps.

I can work out the voltage drop across the 10ohm resistor as I know the current going in is the same as the total current. So 3.9x10 gives me 39volts. Thus at point B I have a voltage total of 100-39 = 61volts.

I can also back this up with the voltage divider, 100x15.59/25.59 = 60.92 or 61v.

For some reason now, I can't get my head around what to do next, am I being a complete idiot and missing the basics.... More than likely.

5. ### WBahn Moderator

Mar 31, 2012
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So you know I1 and you know Vb.

Is this enough information to determine either what I2 is or what I3 is?

If you can find one of them, will you then have enough information to find the other?

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,387
497
You will make good fisherman.

7. ### pAntz Thread Starter New Member

Nov 7, 2015
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Sorry, I'm missing something here. I understand how frustrating this may be to read for those that understand, but I am meant to be going from zero to hero in a week and with all topics I'm covering I just can't get my head around it all.

My biggest problem at the minute is I'm simply learning a calculation process, not the what, why or how.

Is a breakdown or explanation possible so I can attempt to tackle other questions? I fully appreciate answers aren't meant to be provided, but at the minute I need some serious guidance!

8. ### WBahn Moderator

Mar 31, 2012
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You know that Vb = 61V, right?

What is the voltage across the 20 Ω resistor?

Can you find the current, I2, through the 20 Ω resistor?

9. ### pAntz Thread Starter New Member

Nov 7, 2015
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Nope, I can't find anything. I'm using basic ohms law and my figures are stupid. (Mr Ohm will be turning in his grave) Im clearly not understanding the basic circuitry here, and it's only when I have an unknown resistor in a question that I get completely thrown.

10. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,387
497
So, you can not reduce resistor network to a single equivalent resistor and you can not expand it back into its original resistor network.

This problem is a bit beyond Ohm's Law. Here you are dealing with serial and parallel resistors, and application of voltage and current dividers.

11. ### JoeJester AAC Fanatic!

Apr 26, 2005
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Can you show your work so far? This problem is working with KVL, KCL, and Ohms Law.

Attached are two aids ... one is the diagram WBahn made with the resistors annotated with a designation. The other is a table of knowns with blanks to be filled in.

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Last edited: Nov 7, 2015
12. ### pAntz Thread Starter New Member

Nov 7, 2015
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Yes, given a combination circuit where all resistors have a value, I am more than capable of finding the total resistance. But working backwards for some reason throws me.

JoeJester, I shall fill the table in in the morning. It's 2:15 am here, I have text books thrown around in anger and my neighbours have heard every profanity going.

Guys, your help so far has been greatly appreciated.

13. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
You will need to expand that table to include the equivalent resistances of the group as you move along the circuit .... than apply Kirchoff's laws and Ohms law to fill in the equivalent blanks and the main blanks. Maybe that will get you back on track.

14. ### Theory New Member

Nov 3, 2015
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You can actually get it all now. You can just do kirchoffs laws as you expand your circuit.

15. ### Theory New Member

Nov 3, 2015
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Vb = Vc + V40Ω; u can have Vc with this equation. Use kcl to find I2, then ohms law for V40Ω

16. ### WBahn Moderator

Mar 31, 2012
17,743
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Look at the diagram I made.

You know that the voltage at Node B is 61 V, right?

You know that the voltage at Node G is 0 V, right?

What is the voltage across the 20 Ω resistor?

Remember, by definition, the voltage between two nodes is the voltage at one node minus the voltage at the other.

Use the standard subscript convention.

A single subscript, such as Va, means the voltage at Node A relative to the common reference (Node G in our case).

So Va = Va - Vg (but since Vg = 0V in our case, it can be dropped off).

A double subscript, such as Vab, means the voltage at the first node relative to the voltage at the second node.

This means that Vab = Va - Vb

So the voltage across the 20 Ω resistor is Vag = Va - Vg = Va. Right?

Given the voltage across a resistor and the value of the resistor, what is the value of the current flowing through that resistor?

17. ### pAntz Thread Starter New Member

Nov 7, 2015
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Ok, so I had a look at my question this morning, all fresh faced n'all!

I know the voltage at Vb is 61v therefore the voltage across the 20ohm resistor is also 61v. (20 plus 10 ohm resistor equal to the input voltage.)

From this, using ohms law I know the current across the 20ohm resistor to be 3.05 amps.

I now know that the current in the 40ohm resistor is the current at point B minus the current in the 20ohm resistor, so, 3.91-3.05 = 0.86amps.

If the current in the 40ohm resistor is 0.86amps, the voltage across it must be 34.4v

The voltage at Vb is the sum of Vc and the voltage across the 40ohm resistor, therefore, Vc = 61-34.4 = 26.6v

From here I know that the current in the 80ohm resistor must be 26.6/80 = 0.3325amps

Current in the 40ohm is equal to the sum of the currents in the 80ohm and the unknown resistor, therefore, the unknow must be 0.86-0.3325 = 0.5275amps.

The value of the unknown resistor is now simply ohms law again, which gives me a total of 49.289 ohms.

To confirm, I quickly used this value along with all other resistors to work out the total resistance.

18. ### JoeJester AAC Fanatic!

Apr 26, 2005
3,373
1,159
Ok ... Your instructions were written rounded to 2 decimal places.

Here is what I did in excel, the formulae was rounded to two decimal places and I used a 50 ohm resistor as the unknown. The other formulae pic of the totals were extended to more than 2 decimal places.

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19. ### pAntz Thread Starter New Member

Nov 7, 2015
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Roger that regarding the decimal places. Had a crack at more of these throughout the day and not had a single drama. Just needed a push in the right direction!

I would like to thank everybody for their help and taking the time to reply to my original post! Many thanks chaps!!

20. ### WBahn Moderator

Mar 31, 2012
17,743
4,795
Good. Your reasoning is just fine.

I also particularly like that you checked your answer. That is SO rare these days. Most engineering problems, even in the real world, have a multitude of ways to check the correctness of an answer, yet so few students are taught/expected to do so.

Now that you know one way to do it, you might consider other ways. Let's look at just the resistances.

You are given that the total resistance is 25.59 Ω. That must equal the series equivalent of 10 Ω plus everything to the right of it, so everything to the right of it must be 15.59 Ω.

The 15.59 Ω is the parallel resistance of 20 Ω and everything to the right of it, so everything to the right of it must be 70.70 Ω.

The 70.70 Ω is the series resistance of 40 Ω and everything to the right of it, so everything to the right of it must be 30.70 Ω.

The 30.70 Ω is the parallel resistance of 80 Ω and the unknown resistor, so the unknown resistor must be 49.83 Ω.

Note that I kept the intermediate results in the calculator, so this answer (reported as 49.8 Ω) has very little round off error in it. If I use the intermediate results as reported above the final answer comes out to 49.82 Ω.

If you aren't going to keep intermediate results in your calculator (or other computing device), then always include at least two sig figs more than you will report the answer to in order to prevent runaway roundoff error accumulation.

As a rule, engineering results should be reported to three sig figs (not that this is different that the number of places to the right of the decimal point, which is irrelevant), four if the leading digit is a one.

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