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Need help on series circuit

btsmith

Jul 4, 2018
3
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Need help solving the following for "R". It seems like not enough information to solve?
upload_2018-7-4_15-30-6.png
 

Kabelsalat

Jul 5, 2011
182
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Jul 5, 2011
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Yes, you got everything you need.

One method is to makke out the formula of voltage divider - what voltage over R with respect of the value of R.
Then insert this formula (replace by it where "R" would be) in the formula where you would solve the Wattage with respect of R/inserted formula.

Then you reverse the formula so you get R by respect of W.
 

Ylli

Jun 19, 2018
402
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402
P = E * I

P = 24 [given dissipation of R]
E = 24R/(R + 6) [voltage divider to determine voltage across R]
I = 24/(R + 6) [Current through the loop per Ohms law)

Substitute into the first equation and solve for R. It does end up giving you an easily solvable quadratic equation.
 

Kiwi

Jan 28, 2013
471
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Jan 28, 2013
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471
"quadratic equation" crikey that takes me back over 40 years to my high school days.

Might have to try solving this to see if I still remember how to do them.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
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Here's a way that doesn't require any magic.

Assume R=0. What is the current and then the total dissipation of R1 and R2.

Power dissipation is proportional to current squared, and this pair of resistors just happens to dissipate 4 times the power that the target R needs to dissipate.

Given that this means the current would halve to make this pair of resistors dissipate 24W, what does that tell you about the value of R and the dissipation it would have?

Simple approaches like this (whilst not generally applicable) can often work well for problems that are designed to have "nice" solutions. At worst, they provide a method to estimate a value or double-check your answer.
 

btsmith

Jul 4, 2018
3
Joined
Jul 4, 2018
Messages
3
P = E * I

P = 24 [given dissipation of R]
E = 24R/(R + 6) [voltage divider to determine voltage across R]
I = 24/(R + 6) [Current through the loop per Ohms law)

Substitute into the first equation and solve for R. It does end up giving you an easily solvable quadratic equation.
Got it, thanks for the help.
 

btsmith

Jul 4, 2018
3
Joined
Jul 4, 2018
Messages
3
Yes, you got everything you need.

One method is to makke out the formula of voltage divider - what voltage over R with respect of the value of R.
Then insert this formula (replace by it where "R" would be) in the formula where you would solve the Wattage with respect of R/inserted formula.

Then you reverse the formula so you get R by respect of W.
Thanks for the help
 
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