# parallel circuits with unknown values question

Discussion in 'Homework Help' started by Sam2002tii, Jul 19, 2013.

1. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
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0
3 resistors in parallel have an Resistance eq of 4 ohms. The total conductance is 0.25 Siemens using the formula Resistance eq = 1/Gt 4=1/Gt What is the resistance value in ohms of r1 r2 and r3?

the conductance, g1 is one forth that of g2 and one fifth that of g3.

I figured out Gt is = 0.25 but I can't figure out how to find 1/4 of G2 and 1/5 of G3. Simple Algebra, but I can't figure out how to properly break up 0.25

thanks

2. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
What is the expression of Gt in terms of g1, g2, and g3?

If I tell you what g1 is, can you find g2 and g3?

If so, then that must mean that you can write the value of g2 and g3 in terms of g1.

If so, then substitute these expressions into the expression for Gt. You now have one equation in one unknown.

3. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
All I know is the total value of g1 + g2 +g3 = 0.25

g1 = 1/4 of g2

g1 = 1/5 of g3

what is g1, g2 and g3?

so yes if I had g1 then g2 and g3 would be easy to figure out

thanks

Last edited: Jul 19, 2013
4. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
Track your units. What does it mean to have a conductance of 0.25? Who's taller, someone that is 32 or someone that is 72? Who's older, someone that is 20 or someone that is 15?

The units are a fundamental part of the value and can't just be ignored.

So you have

g1 + g2 + g3 = 0.25S

But I would leave everything in symbolic terms for as long as possible.

g1 + g2 + g3 = Gt

This is a verbal description of g1 in terms of g2 and g3. What you want are mathematical expressions that give g2 and g3 in terms of g1.

5. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
I figured out the answer by plugging in numbers until they worked with everything else. I still don't understand how to set up the algebra equation.

g1=.025 g2 =.10 g3 =.125

I understand "of " in math means multiply

So,

g1 = 1/4 x g2 g1 = .25g2 ???

g1 = 1/5 x g3 g1 = .20g3 ????

If you plug the numbers in then you get the right answers, but that's using the answers I already figured out just by guessing. If you have two unknow variables in the equation then how do you find the answer. The only known number is the total (Gt).

thanks

6. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
What grade level are in and what math background have you had up to this point. That will help me describe what you are doing in a more useful context.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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$\text{g_{tot}=g_1+g_2+g_3}$

$\text{g_{tot}=\frac{g_2}{4}+g_2+g_3}$

$\text{g_{tot}=\frac{g_3}{5}+g_2+g_3}$

Hence ...

$\text{\frac{g_2}{4}+g_2+g_3=\frac{g_3}{5}+g_2+g_3}$

or simply ....

$\text{\frac{g_2}{4}=\frac{g_3}{5}}$

Knowing this last relationship in addition to knowing g1 in terms of either g2 or g3, allows you to rewrite the original equation solely in terms of g2 or g3 - thereby giving your first value from which the other values are then deduced.

8. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
ELECT 101 Direct current is the community collage class. Intermediate Algebra is my next math class. The question is out of Grob's Basic Electronics

9. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
I think it is simpler to put g2 and g3 each in terms of g1 and then solve for g1. But I think the OP's algebra background is perhaps to shallow to see how to do this at this point. But before I walk through a step-by-step solution, I want to get a feel for how detailed the steps should be and how heavily documented so that the OP can get the most value from them.

The fact that the problem is dealing with resistances and conductances has probably led me to assume a stronger math background that I should have.

10. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
@ t_n_k

thank you, but I still don't understand how to solve the problem when you have

g2/4 = g3/5

What is the next step? You still don't know what g2 or g3 is so how do you continue the math?

11. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
I did test into intermediate algebra, but it's been about 10 yrs since I was in a high school algebra class...

12. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
So it sounds like you've had SOME kind of algebra. I don't know where the prereq to Intermediate Algebra would stop and Intermediate would start, but I think that the introductory course should be sufficient to understand the solution to this problem will little detailed explanation. But if you don't understand any of the following steps, just ask.

One of the first things you should get in the habit of doing is trying to estimate the answer or at least put some kind of min/max bounds on it.

If the resistors are all the same value, then the total value of the three in parallel would be R/3, which would have to equal 4Ω. Thus, the maximum value that the SMALLEST resistor can have is 12Ω, since if ALL of the resistors are greater than 12Ω, the parallel combination would be over 4Ω.

Similarly, the minimum value that the SMALLEST resistor can have is 4Ω, since the total resistance will always be less than the smallest resistance.

So the smallest resistor has to lie somewhere between 4Ω and 12Ω. This will be our first sanity check to see if the answer makes sense.

----------------------
GIVEN
----------------------

(1) Rt = 4Ω

(2) g1 = 0.25*g2

(3) g1 = 0.20*g3

----------------------
BY DEFINITION
----------------------

(4) g1 = 1/r1

(5) g2 = 1/r2

(6) g3 = 1/g3

(7) Gt = 1/Rt

----------------------
Because the resistors are in parallel:
----------------------

(8) Gt = g1 + g2 + g3

Multiply both sides of (2) by 4 (or divide both sides by 0.25)

(9) g2 = 4*g1

(10) g3 = 5*g1

----------------------
FIND R1
----------------------

Substitute (9) and (10) into (8)

(11) g1 + 4*g1 + 5*g1 = Gt

Factor out g1

(12) g1(1 + 4 + 5) = Gt

(13) 10*g1 = Gt

Divide both sides of (13) by 10

(14) g1 = Gt/10

Substituting (4) and (7) into (14)

(15) 1/r1 = (1/Rt)/10

Multiplying both sides of (15) by (r1*Rt)

(16) Rt = r1/10

(17) r1 = 10*Rt

----------------------
FIND R2
----------------------

Substituting (4) and (5) into (9)

(18) 1/r2 = 4*(1/r1)

Multiplying both sides of (18) by(r1*r2)

(19) r1 = 4*r2

Dividing both sides of (19) by 4

(20) r2 = r1/4

Substituting (17) into (20)

(21) r2 = (10*Rt)/4

(22) r2 = 2.5*Rt

----------------------
FIND R3
----------------------

Substituting (4) and (6) into (10)

(23) 1/r3 = 5*(1/r1)

Multiplying both sides of (23) by(r1*r3)

(24) r1 = 5*r3

Dividing both sides of (24) by 5

(25) r3 = r1/5

Substituting (17) into (25)

(26) r3 = (10*Rt)/5

(27) r3 = 2*Rt

----------------------
FIND R1, R2, and R3
----------------------

Apply (1) to (17), (22), and (27)

(28) r1 = 10*4Ω = 40Ω

(29) r2 = 2.5*4Ω = 10Ω

(30) r3 = 2*4Ω = 8Ω

----------------------
CHECK THE RESULTS
----------------------

First, does it pass our sanity check based on our estimate? Yes, because the smallest resistor is 8Ω, which is between 4Ω and 12Ω.

Substituting (28), (29) and (30) into (4), (5), and (6)

(31) g1 = 1/40Ω = 0.025S

(32) g2 = 1/10Ω = 0.1S

(33) g3 = 1/8Ω = 0.125S

Substituting (31), (32), and (33) into (8)

(34) Gt = 0.025S + 0.1S + 0.125S = 0.25S

Using (7) and (34) to check against (1)

(35) 4Ω =? 1/0.25S = 4Ω √

Using (31) and (32) to check against (2)

(36) 0.025S =? 0.25*0.1S = 0.025S √

Using (31) and (33) to check against (3)

(36) 0.025S =? 0.20*0.125S = 0.025S √

Now, this was done in excruciatingly more detail than you should be able to do it. With a little experience, you should be able to get the answers in just a few lines.

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13. ### Sam2002tii Thread Starter New Member

Jul 19, 2013
13
0
Thank you so much. This makes sense now. I guess I should take my required math classes next quarter before I continue with more Electronics classes...

14. ### WBahn Moderator

Mar 31, 2012
18,277
4,952
That might be a real good idea. If you don't have a strong handle on the necessary math, you will struggle so much with the math that you will have a hard time learning the electronics concepts.