# Resistors in combination circuit help

Discussion in 'Electronics Homework Help' started by Je55ica, Feb 19, 2018.

1. ### Je55ica

4
0
Feb 19, 2018
I have the attached circuit and am required to calculate the resistance of R1 and R2.

I approached this by calculating the current through R2 as 0.5A as R2 and R3 have to equal 2.0 A

For R1 I used ohms law with the supply 12V and 2.0 amp current giving me a value of 6 ohms

I then used 12v and 0.5 amp to obtain R2 at 24 ohms

Can someone please point out my mistake?

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2. ### Tha fios agaibh

2,215
738
Aug 11, 2014
First part good.
When you divide 12v by 2a you get the total resistance of the entire circuit. R1 is only part of that resistance.
Try to calculate the voltage of R4, then the parallel resistance of R2 and R3, then when you know the parallel resistance (you already know .5a flows through R2) you should then know the voltage drop across it, and the voltage dropped across R4, and then R1 voltage drop should be known because the total voltage of the circuit has to equal 12v.

Hope that helps

3. ### Je55ica

4
0
Feb 19, 2018

So calculating V4 I get 4 V

would I use the 4 volts that I have calculated to be flowing through R4 for the calculation in finding R2?

4 v / 0.5A ? giving me resistance value 8 ohms for R2?

4. ### Tha fios agaibh

2,215
738
Aug 11, 2014
R3 is 4Ω and has 1.5a flowing through it. Whats the voltage? This same voltage will be across R2. Use this voltage and the .5a to figure that resistance. Now, whats the parallel resistance?

5. ### WHONOES

1,190
322
May 20, 2017
OK. You don't need to calculate any voltages.
Total circuit resistance is defined as 12V / 2A which = 6Ω
R3 is passing 1.5A which is 75%, or 3/4 if you like, of 2A.
Therefore 25% of the current is flowing in R2.
To calculate R2, divide 75% by 25% the result of which is 3.
To get the value of R2, multiply R3 x 3 = 4 x 3 = 12. Therefore R2 = 12Ω.
To resolve R1, you now need to know the combined value of R2 and R3. This is done using the standard formula:
R2 x R3 / R2 + R3 = 4 x 12 / 4 + 12 = 48 / 16 =3.
Therefore combined value of R2 and R3 is 3Ω.
We now know that the combined values of R2, R3 and R4 is 5Ω. Therefore R1 = 6 - 5 = 1Ω.
Thus:
R1 = 1Ω
R2 = 12Ω
R3 = 4Ω
R4 = 2Ω

OK
Hope you followed that.

Last edited: Feb 19, 2018
6. ### Je55ica

4
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Feb 19, 2018
Ok awesome thank you so now I have R2 value of 12 ohms

1/4 + 1/ 12 = 0.17 ohms

would I then add this to R4 and deduct the total from Rtotal leaving 3.83 ohms?

7. ### WHONOES

1,190
322
May 20, 2017
No. Re read my reply. The only sums you need to do are those described. My description is the entire solution.

8. ### Tha fios agaibh

2,215
738
Aug 11, 2014
Giving the entire solution isn't the best way for someone to learn.

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9. ### Je55ica

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Feb 19, 2018
Thank you both for all your help.

10. ### WHONOES

1,190
322
May 20, 2017
No but the other proposed solutions were getting a bit misleading. At least she will know what to do next time in a similar situation.

11. ### Tha fios agaibh

2,215
738
Aug 11, 2014
You understand?
1/4=(.25)+ 1/12=(.0833)=.3333 or 3 Ω
Reciprocal 1/.3333=3
So if the parallel resistors are 3 and R4 is 2 that leaves 1 ohm for R1. Remember the total resistance is 6 ohms. It can also be figured out by adding up the voltage drops. R4 drops 4v the parallel resistors drop 6v which leaves 2v for R1. Get it? 12volts

12. ### WHONOES

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May 20, 2017
Yes but the whole point of my reply was that you don't need to calculate any voltages at all. The original question had no requirement to do so. Your bit of maths is correct, I just couched it such that is was easier to understand.

13. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
I would certainly calculate the voltages. It seems the most simple and direct way.

14. ### WHONOES

1,190
322
May 20, 2017
Sorry but you simply don't need to calculate the voltages to get the correct answer. I think the method used is the quickest and least confusing. If you wish to know the voltages it may be done afterwards with no confusion. That is my last word on the subject.

15. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,501
2,841
Jan 21, 2010
In my post above I should have said the method I would have used seems simpler and easier to me.

I don't think it's the easiest, but that doesn't necessarily apply to others, and it clearly doesn't apply to you

People find different methods easier or harder, and finding a different way to explain something is often what it takes.

I will take a closer look at your method. I certainly do something similar (I think) when intuitively approximating the required parallel resistance, but it's not the method I've used for numerically exact values.

Tha fios agaibh likes this.