# Circuit Question

1. May 6, 2007

### ianb

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

2. Relevant equations

Parallell: $$\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$$

Series: $$R_{eq} = R_{1} + R_{2}$$

3. The attempt at a solution

Part 1 - When the switch is not closed:

$$R_{eq_2} = R + ((90+10)^{-1} + (90+10)^{-1})^{-1}$$
$$R_{eq_1} = R + 50$$

Part 2 - When the switch is closed:

$$R_{eq_2} = R + (90^{-1} + 10^{-1})^{-1} + (90^{-1} + 10^{-1})^{-1}$$
$$R_{eq_2} = R + 18$$

Since $$R_{eq_1} = 2R_{eq_2}$$, then

$$R + 50 = 2(R + 18)$$
$$R + 50 = 2R + 36$$
$$R = 14$$

The answer in the back, though, is 24. What have I done wrong?

Last edited: May 6, 2007
2. May 6, 2007

### Mindscrape

Good question, it looks to me like you did it right. Are you sure you didn't make a numerical error (this part looks okay too, though I didn't actually calculate)? How trustworthy are the answers in your book?

3. May 6, 2007

### ianb

Thanks, mindscrape. I suppose the book could have misprinted the answer. One more thing, though, if you don't mind:

When the switch is closed, I calculated two parallel circuits. Are these two circuits, in turn, parallel to each other, or in series to each other (like I did above)?

Last edited: May 6, 2007
4. May 7, 2007

### mlowery

When the switch closes, it creates a node between the four resistors. The resulting parallel connections will be in series with each other - the way you calculated it.