2 Resistors in Parallel versus in Series

AI Thread Summary
Two resistors connected in series to a 110 V line consume one-fourth the power compared to when they are connected in parallel. Given one resistor is 2.0 kΩ, the task is to find the resistance of the other resistor. Initial calculations suggest the second resistor's value is 1.56 kΩ, but the method used is incorrect. The correct approach involves calculating power using the formulas P=I^2R and P=V^2/R for both configurations. Ultimately, solving for the equivalent resistance in parallel requires using the quadratic equation for accurate results.
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Homework Statement


Two resistors when connected in series to a 110 V line use one-fourth the power
that is used when they are connected in parallel. If one resistor is 2.0 kΩ, what is the resistance of the other?


Homework Equations



V=IR

The Attempt at a Solution


v=ir

110=i(2000+x)

110=1/4(2000+x)

440=2000+x

x=440-2000

x=-1560

snce x= 1560

the value of another resistance is 1.56 kilo ohm

Did I do that right?
 
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Not quite. Okay, your series formula is fine. But now you have to get the power for the resistors in series. P=I^2 R and P=V^2/R.

Now that you have the power for the series resistors, find the power for the resistors if they were in parallel. 110=i*Req again, but different Req this time.
 
Can I just cancel the V's out and solve like this:

R2 + 2000 = 2*(2000*R2)/(2000 + R2)

Which looks like it would require the quadratic eqn
 
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