Two Resistance Problems (Circuits)

AngeloG

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
P=IV
V=IR
Rt (series) = R1+R1...
Rt (paralell) = (1/R1 + 1/R2...)^-1

3. The attempt at a solution


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.

hage567

Well for question 15, you used 420 ohms instead of 480 ohms in your calculation!

AngeloG

Doh! Haha, that could be the problem =p.

It helps to read carefully :s.

hage567

I read it as P(series) = 0.25*P(parallel), which is the other way around from what you have. I have a solution, here's how I set it up:

For series circuit:

Since the current will be the same in both resistors, the power can be expressed as

[tex]P = I^2R = I_s^2(R_1 + R_2)[/tex] where Is is the current in the series circuit.

For the parallel circuit:

The voltage across each resistor will be the same, so the total power will be (with a little simplifying)

[tex]P_p =\frac{V^2}{R_1} + \frac{V^2}{R_2} = \frac{V^2(R_1+R_2)}{R_1R_2}[/tex]

You will also need to use the relation [tex]V=I_sR_T = I_s (R_1+R_2)[/tex]

Since the power used by the series circuit is 0.25 that of the parallel, you can relate the two expressions and solve for R2. No Maple should be needed!