calculating unknown unequal resistor values in a parallel circuit/series circuit

Thread Starter

ngrobertson

Joined Mar 18, 2011
11
I have been assigned bonus questions by my ELF 190 Prof @ Seneca College (Fire Protection Engineering Technology Program). The solutions are not in our notes and we are to seek outside help in finding solutions to these problems and return them to class for bonus marks. I need all the help I can get, with the questions and the overall mark. Would this be the rite place to look for assistance with these questions?
 

t_n_k

Joined Mar 6, 2009
5,448
As long as you make an honest attempt at solving any particular problem you should get all the help you need from the forum members.
 

Thread Starter

ngrobertson

Joined Mar 18, 2011
11
I have been assigned bonus questions by my ELF 190 Prof @ Seneca College (Fire Protection Engineering Technology Program). The solutions are not in our notes and we are to seek outside help in finding solutions to these problems and return them to class for bonus marks. I need all the help I can get, with the questions and the overall mark. Would this be the rite place to look for assistance with these questions?
I do not know how to begin these equations due to lack of infromation provided deliberately by the prof. My first attempt In the first equation (series circuit) I was given 3 resistors each of different values, one being an unknown value in ohms, yet a value provided in watts. I was provided v (24V) so I divided R1 by Vs to give me I (12 A). Then I took the v divided by I to get 2 ohms. The second question is a parrellel circuit with 3 resistors, 2 values givens in ohms, the third in watts. No V given just I (19 A). I cannot come up with a formula to calculate R3. RT is not given nor is V. Any direction would be appreciated.
 

Thread Starter

ngrobertson

Joined Mar 18, 2011
11
the given information is R1 = 2 ohms, R2 = 4 ohms, R3 = ? ohms,-P3 = 24W. RT = ? I(S) = 19 A, V (E) = ?. Find the value of R3 ?
 

Jony130

Joined Feb 17, 2009
4,968
This value that you provided is for series circuit or for parallel ??

If for series circuit
Is = 19A must flow through all resistors.

P = V*I ---> V = 24W/19A = 1.26315789V
Or
P = V*I = I*R * I = I^2*R ---> R3 = P/I^2 = 24W/361A = 0.0664819945Ω
 

Jony130

Joined Feb 17, 2009
4,968
And IS=19A is the total current?
Is this the case

 

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Thread Starter

ngrobertson

Joined Mar 18, 2011
11
This value that you provided is for series circuit or for parallel ??

If for series circuit
Is = 19A must flow through all resistors.

P = V*I ---> V = 24W/19A = 1.26315789V
Or
P = V*I = I*R * I = I^2*R ---> R3 = P/I^2 = 24W/361A = 0.0664819945Ω
Thank you, but the unknown R3 value is in a parallel circuit
 

Jony130

Joined Feb 17, 2009
4,968
For parallel circuit we can replace R1 and R2 with equivalent resistor RT = 4/3Ω



So for this circuit we can write
19A = I1 + I3 (1)

I1/I3 = R3/RT (2)

24W = I3^2 * R3 (3)

So we have three unknowns and three equations.
So we are able to solve this system of linear equations

R3 = 2/27Ω or R3 = 24Ω
 

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Thread Starter

ngrobertson

Joined Mar 18, 2011
11
For parallel circuit we can replace R1 and R2 with equivalent resistor RT = 4/3Ω



So for this circuit we can write
19A = I1 + I3 (1)

I1/I3 = R3/RT (2)

24W = I3^2 * R3 (3)

So we have three unknowns and three equations.
So we are able to solve this system of linear equations

R3 = 2/27Ω or R3 = 24Ω
How did you get to this part of the equation? If i reaarange the 3rd equation do I not end up with 24W divided by I3 squared (19) equals R3?
 

Jony130

Joined Feb 17, 2009
4,968
How did you get to this part of the equation?
Which part?

From first equation find I1
19A = I1 + I3 (1)

From second equation find R3
I1/I3 = R3/RT (2)

then substitute I1 from (1) to (2) and R3 from (2) to (3) and solve for I3

24W = I3^2 * R3 (3)
 
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Thread Starter

ngrobertson

Joined Mar 18, 2011
11
Which part?

From first equation find I1
19A = I1 + I3 (1)

From second equation find R3
I1/I3 = R3/RT (2)

then substitute I1 from (1) to (2) and R3 from (2) to (3) and solve for I3

24W = I3^2 * R3 (3)
Thank you for your assistance. You are more experienced than I, I am struggling to absorb these types of equations and I cannot rework ohms law to work thru steps 1 & 2 and end up with the final answer you posted. I have scheduled extra help with my Prof later this week. Thank you again.
 

Jony130

Joined Feb 17, 2009
4,968
Your, circuit look like this



And it is clear that R1 and R2 are connect in parallel.
So we can replace R1 and R2 with a single resistor RT
RT = R1||R2 = (R1*R2)/(R1+R2) = (4Ω*2Ω)/(4Ω+2Ω) = 8Ω/6Ω = 4/3 = 1.33333333Ω

We also know that input current is equal Itot = 19A.

So from the first Kirchhoff's law we can write:
19A = I1 + I3 (1)
I1 = 19 - I3
I3 = 19 - I1
Now let's try find I1 and I3 currents
I1 = V/RT
I3 = V/R3
V = I1*RT = I3*R3 (In a parallel circuit, the voltage is the same across each resistors)

First let's find I1 current

I1*RT = I3*R3 (divided both side by RT)
I1 = (I3*R3)/RT

Know from KCL

Itot = I1 + I3
I3 = Itot - I1
I1 = (I3*R3)/RT = (Itot - I1)* R3/RT

I1 = (Itot - I1)* R3/RT (multiply both sides by RT/R3)
I1*RT/R3 = Itot - I1 (add I1 to both side)
I1*RT/R3 + I1 = Itot
(I1*Rt + R3*I1)/R3 = Itot
I1*(RT+R3)/R3 = Itot (divided both sides by R3/(RT+R3))

And finally we end-up with the answer
I1 = Itot * R3/(RT+R3) (current divider rule)

If we do the same thing for I3 current we get

I3 = Itot * RT/(RT+R3)

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

So
I1/I3 = (Itot*R3)/(RT+R3) * (RT+R3)/ (Itot*RT) = R3/RT

I1/I3 = R3/RT (2)

Next, we also know that power dissipated by R3 is equal 24W
P = V*I = I*R * I = I^2*R
24W = I3^2*R3 (3)

So we end up with three equations

19A = I1 + I3 (1)
I1/I3 = R3/RT (2)
24W = I3^2 * R3 (3)

And this is the end of a job for electronics.
The rest is just a Math

From first equation we get
I1 = 19A - I3
From second equation we get R3
I1/I3 = R3/RT (multiply both sides by RT)
I1/I3 * RT = R3
R3 = (I1 * RT)/I3 (substitute I1 from (1) I1 = 19 - I3)
R3 = ((19A - I3)/I3 * 4/3
R3 = 4*(19 - I3)/3I3 = (76 - 4*I3)/ (3*I3) (know we substitute this to (3) )

24W = I3^2 * R3

24 = I3^2*(76 - 4*I3)/(3*I3)

So we end-up with quadratic equation

\[\frac{4}{3} I3^2 - \frac{76}{3} I3 + 24 = 0\]

And for example if we use this aplet withe this code
4/3 *x^2 -76/3*x +24 = 0
http://www.wolframalpha.com/

We get
I3 = 1A or I3 = 18A
So if I3 = 1A
R3 = 4*(19 - I3)/3I3 = ( 76 - 4)/ 3 = 24Ω
And If I3 = 18A
R3 = (76 - 72)/54 = 4/54 = 2/27 = 0,0740740741Ω

And this is the end of your question
 
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