A simple diagram of two heat resistors.

AI Thread Summary
The discussion revolves around solving for two heat resistors, R1 and R2, using given power, voltage, and time data. The participants clarify the calculations for power and resistance, noting that the power for the parallel circuit is calculated as P = W / t, resulting in 1.76 kW. They emphasize the need for additional equations to find R1 and R2, suggesting the use of both parallel and series resistance formulas. A key insight is that the total resistance for the series diagram is 146.67 Ohm, which can help determine the individual resistances. Ultimately, the conversation highlights the importance of correctly applying circuit formulas to solve for unknown resistances.
KayVee
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Here is a problem I have. I need to find the resistors R1 and R2, and here is what I know:

http://triton.imageshack.us/Himg12/scaled.php?server=12&filename=varmalikam.jpg&xsize=640&ysize=480

How do I set up the equation? I am really lost.

Please ask, if you are having troubles reading the diagram. The first one is the original one.
 
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Welcome to PF!

Hi KayVee! Welcome to PF! :smile:

Ignore the first diagram.

In the second diagram, you're given the heat and the time and the voltage …

so what is the power?

and so what can you say about R1 and R2?
 
I see that I need to put up my own work/study on the circuit, but now I see that I was going no where :P

So, the power for the parallel is: P = W / t = 2112kJ / 1200s = 1,76 kW.

Then the (sum of the?) parallel resistors: R = V~2 / P = 220~2 / 1,75 kW = 27500 kOhm.

Can I then say that each resistor is 27500 / 2 ?

This pice is kinda hard, since we have had so little about power in school.
 
Hi KayVee! :smile:

(why did you use W for energy … that's really confusing :frown:)

(and the LaTeX for squared is ^2 not ~2 … though it's quicker to write, and easier to read, and easier on the PF server, if you just use the X2 tag above the Reply box :wink:)
KayVee said:
So, the power for the parallel is: P = W / t = 2112kJ / 1200s = 1,76 kW.

yes, except I think it's 396 kJ for that diagram
Then the (sum of the?) parallel resistors: R = V~2 / P = 220~2 / 1,75 kW = 27500 kOhm.

yes (but it would be better if you stated what formula you're using :wink:)
Can I then say that each resistor is 27500 / 2 ?

Noooo :frown:

you need another equation for R1 and R2 … try the other diagram! :smile:
 
Sorry tiny-tim, for the late reply. But I cracked the case :D I found the total ampere for
the second diagram, and then just setting the information I had in: U=I*R.

I knew that the total resistance in the serial diagram var 146,67 Ohm. There is even a simpler way to find the res. for each resistor. Find the total resistance for the parallel and and total for the serial, then set it up in the : (R1 * R2)/(R1 + R2)= Rtotal.
 
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