# Can't figure it out

Discussion in 'Homework Help' started by emagdnim47, Mar 6, 2013.

1. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
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2) A parallel circuit contains the following resistor values:
R1= 270k Ω, R2= 360k Ω, R3=430k Ω, R4= 100k Ω, IT= 0.006 A

Find the following missing values:

RT= Ω

I1= A

I2= A

I3= A

I4= A

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2. ### MrChips Moderator

Oct 2, 2009
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This is homework help. We don't do your homework for you. Show us what you have done so far.

3. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
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I've attached my current work on this problem (pdf format), it's likely my sig fig is totally off balance here.

Do you keep the prefixes as is when doing the math, or do break it down to their original values and then proceed w/ computation?

Feb 17, 2009
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Jul 16, 2011
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Feb 17, 2009
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7. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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Yikes! Changed it to 10μΩ, and the R3 is actually 430k not 300k. Apologies w/ the mix up.

I'm still having a field day w/ finding the RT, not sure if it's 102k or 10.2k.

Getting the applied voltage is tuff when I can't seem to find the right RT....I check my work, once I get the branch current values (IT= I1+ I2 + I3 +I4) I add them all up but it never adds up to 0.006 A.

Last edited: Mar 6, 2013
8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,501
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Hmm

1/430K = 2.32μΩ
1/100K = 10μΩ

So

RT = 1/ (3.7μΩ + 2.77μΩ + 2.32μΩ + 10μΩ) = ??

9. ### TM The Fool New Member

Mar 6, 2013
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Hey, Do you read in A-levels?

10. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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Exactly, I've just tried that and came up w/ 18.79 or 18.8kΩ

RT = 0.06A x 18.8kΩ = 112.8kΩ or 113kΩ ??

11. ### Jony130 AAC Fanatic!

Feb 17, 2009
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OK i made a error in the units

1/1Ohm is equal to ONE Siemens .

1/100KΩ = 10μS

so

RT = 1/ (3.7μS + 2.77μS + 2.32μS + 10μS) = ??

12. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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I'm still coming up w/ RT= 18.79, is this what you get too?

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Yes
RT in Ohms is equal to = 1/18.79μS = ??

14. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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Okay, I rounded the RT to 18.8μΩ, after dividing 1 over 18.8μΩ I'm getting this :

RT = 0.053 or 53kΩ ?

E = IT x RT

0.006A x 53k

0.006A x 53000Ω

E = 318 V ? Then tried using this applied voltage value and on those four individual branches,
branch values are way off. To check my work I added all the branch values, but but sum doesn't add up to 0.006 A. In fact it's not even close, the applied voltage
has to be lower than 318 V right??? I'm missing something here......

Last edited: Mar 6, 2013
15. ### Jony130 AAC Fanatic!

Feb 17, 2009
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ET = RT * Itot = 6mA * (232200/4367)KOhm = 319V

16. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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Where'd did you get these values (232200/4367)KOhm ?

17. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Simply I solve this
1/(1/270 + 1/360 + 1/430 + 1/100) = 232200/4367 = 53.1715

18. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
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Apologies on my part. I see. That's what I get too, RT = 53kΩ right???

So we take the RT and multiply it with the IT to get applied voltage.

E = IT x RT

0.006A x 53k

0.006A x 53000Ω

E = 318 V This can't be right, can it??????

19. ### WBahn Moderator

Mar 31, 2012
23,149
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Again, given an answer (612V from your pdf) you can check if it is correct or not. Let's do a quick sanity check. Your smallest resistance is 100kΩ and so that branch will have the most current. That means that your total current for all four branches has to be less than four times the current in this branch.

I4 = 612V/100kΩ = 6.120mA.

Let's ignore that for the moment and push on.

That means that your current has to be between 6.12mA (if all the other three branches were near-infinite resistance) and 24.48mA (if all the other three were near-100kΩ resistance).

On the other end, your highest resistance branch is 430kΩ and so we know that all of the other branches have at least whatever current is flowing in this branch, which is

I3 = 612V/430kΩ = 1.423mA so the minimum total current is 5.69mA, which is less than we have in the I4 branch so that takes precedence.

Hence, with a 612V applied, we know we have between 6.12mA and 24.5mA of current.

Let's go ahead and solve for the other two currents and get the actual total for this voltage (carrying an extra couple sig figs to avoid roundoff error):

I1 = 612V/270kΩ = 2.2667mA
I2 = 612V/360kΩ = 1.7000mA
I3 = 612V/430kΩ = 1.4233mA
I4 = 612V/100kΩ = 6.1200mA.

This gives a total current of

It = 11.5099mA

Since you actual current is 6mA, and since this is a linear circuit, you know that your 612V is too high by a factor of 11.5099mA/6mA = 1.918, meaning that the actual voltage is 319.0V.

Do you see how, using this approach, you didn't need to even calculate your original answer of 612V. You could have just said, "Let's assume the applied voltage is 100V (or 1000V or 1V or whatever). What would the corresponding current be?" Then you take that answer and divide it by the known total current to get a correction factor and you divide your assumed voltage by the correction factor (or, equivalently, divide the known answer by the assumed answer and multiply the assumed voltage by the result).

Now, in your pdf, you have some units issues.

1/resistance is NOT resistance, it is conductance. 1/1Ω is equal to 1S (S is for Siemen).

So your work should have been:

$
R_T\;=\;\frac{1}{\frac{1}{270k \Omega}\;+\;\frac{1}{360k \Omega}\;+\;\frac{1}{300k \Omega}\;+\;\frac{1}{100k \Omega}}
$

Notice that you had an expression, not an equation. An equation involves two things being set equal to each other by way of an equals sign. Try to avoid dangling expressions.

Notice that you are using a value of R3=300kΩ instead of the 430kΩ given in the post. Which is it? I'll go with the number you have in the equation since the main point is to track your work.

$
\frac{1}{3.704\mu S\; +\; 2.778\mu S\; +\; 3.333\mu S\; +\; 10\mu S}\;=\;\frac{1}{19.815\u S}\;=\;50.46k\Omega
$

The problem is that you had 0.01mS (you had mΩ, but the 'm' is what tripped you up). This is correct, but 0.01mS is 10μS and everything else was in μS. You can't add things together unless the units are identical, including scaling prefixes. Just think about how you would add 1cm to 1mm, you would first convert one quantity to the other quantity's units. So you would either have 10mm + 1mm = 11mm or you would have 1cm + 0.1cm = 1.1cm. But you don't have 1+1=2 of anything.

Finishing out, your voltage would then be

$
E\;=\;I_T \dot R_T\;=\;6mA \dot 50.46k\Omega \;=\; 302.8V
$

This answer, of course, is wrong if R3 is supposed to be 430kΩ. But you can use the method described earlier to detect that it is wrong and to find the correct value. Though you should also go back through your work and find where the error was made, as well.

20. ### emagdnim47 Thread Starter New Member

Jul 16, 2011
21
0
Finally figured it out!!! Attached is my work, I hope I get validation from you as you've been most helpful.

Thanks for your guidance, this forum is amazing!

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