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## Homework Statement

15. (II) A close inspection of an electrical circuit reveals that a 480-Ω resistor was inadvertently soldered in the place where a 320-Ω resistor is needed. How can this be fixed without removing anything from the existing circuit?

16. (II) 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 1.6-kΩ, what is the resistance of the other?

## Homework Equations

P=IV

V=IR

Rt (series) = R1+R1...

Rt (paralell) = (1/R1 + 1/R2...)^-1

## The Attempt at a Solution

http://img362.imageshack.us/img362/3694/problems1mf2.th.jpg [Broken]

This isn't homework; just going through every problem in my Physics book for the upcoming exam =p.

For 15, the back of the book says it's 960-Ω, not sure why? I'm getting a different answer. Maybe someone can clarify as to why.

For 16, I'm pretty sure I did it completely wrong heh =p.

P=IV, I = V / Rt

P(series) = 4*P(parallel)

(V/Rts) * V = 4*[ (V/Rtp) * V ]

Rts = 1600-Ω + x-Ω

Rtp = [(1/1600-Ω) + (1/x-Ω) ]^-1

V^2 / Rts = 4*[V^2/Rtp]

Solving for X, Maple spits out --> -1400-200*I*sqrt(15), -1400+200*I*sqrt(15) Which both seems extremely highly unlikely =p.

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