- #1

dillonmhudson

- 49

- 0

## Homework Statement

Two questions:

1.

2.

## Homework Equations

P(power)=V(voltage)*I(current)

P=v^2/R

1/((1/R_1)+(1/R_2)+(1/R_n))

## The Attempt at a Solution

1. I have simplified the bottom two resistors (60[tex]\Omega[/tex], 40[tex]\Omega[/tex]) into one resistor using the equation 1/((1/R_1)+(1/R_2)+(1/R_n)) and have a single resistor of 24[tex]\Omega[/tex]. Then I do the same with the top two (20[tex]\Omega[/tex], 30[tex]\Omega[/tex]) and get a result of 12[tex]\Omega[/tex]. I then combine the 12[tex]\Omega[/tex] and the 5[tex]\Omega[/tex] on the left by simple addition and get 17[tex]\Omega[/tex]. Then, I combine the 17[tex]\Omega[/tex] in parallel with the 24[tex]\Omega[/tex] and get a simplified result of approx 9.95[tex]\Omega[/tex]. I don't believe this to be right, however. My prof states that the answer is 9.6 - so I'm close, but no cigar.

2. I use the equation P(power)=V(voltage)*I(current)

I substitute I for V/R (ohm's law) and get P=v^2/R. Using this equation I plug and chug 120V and either 1280W or 240W and get 11.25[tex]\Omega[/tex], and 60[tex]\Omega[/tex], respectively. I don't know what to do next however.

I tried the following:

1/((1/60[tex]\Omega[/tex])+(1/x))=11.25[tex]\Omega[/tex]

Then using my TI-89 I solve for x and get a result of 13.85[tex]\Omega[/tex]. I can therefore put 60[tex]\Omega[/tex] and 13.85[tex]\Omega[/tex] into parallel with a voltage source and get a power of either 1280W or 240W, correct?

The answers for 2 are 15[tex]\Omega[/tex] and 45[tex]\Omega[/tex] - wrong again.

Thanks for any help.