# Power in network of resistors?

## Homework Statement

Find the resistance of R, such that the network dissipates 295 Watts.

## Homework Equations

$$\frac{1}{R_{eqp}} = \Sigma \frac{1}{R_{i}}$$
$$R_{eqs} = \Sigma R_{i}$$
$$P = \frac{V^{2}}{R} = VI = I^{2}R$$

## The Attempt at a Solution

What I did was first find the equivalent resistance of the whole network. The 4 ohm resistors are in series so I added them into one single 8 ohm resistor. Then that 8 ohm can be combined with R using the rule for resistors in parallel, then once again, that can be used to find the total resistance:

$$\frac{1}{R_{1}} = \frac{1}{8} + \frac{1}{R}$$

$$\frac{1}{R_{2}} = \frac{1}{8} + \frac{1}{R} + \frac{1}{3} = \frac{11R+24}{24R}$$

$$R_{total} = \frac{24R}{11R+24}$$

Now, since $P = \frac{V^{2}}{R}$ :

$$295 = 48^{2}(\frac{11R+24}{24R})$$

$$7080R = 25344R + 55296$$
$$-18264R = 55296$$
$$R = -3.03 \Omega$$

The book claims this to be 12.1 ohms, but no matter how I work it out I get some number less than one, or a negative? I tried making 295, since it says the power is dissipated, even though that didn't really make sense, still no luck. What am I missing here?

Related Introductory Physics Homework Help News on Phys.org
R1 is in series with the 3 ohm resistor, not in parallel.

Oh wow,

Okay so:

$$R_{tot} = \frac{11R+12}{R+8}$$

Then into P = V^2/R

$$295 = \frac{48^{2}(R+8)}{11R+12}$$
$$(11R+12)295 = 48^{2}(R+8)$$
$$3245R +3540 = 2304R +18432$$
$$-14892 = -941R$$
$$R = 15.8 \Omega$$

Which is still a bit off from the 12.1 ohms it's supposed to be?

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

Find the resistance of R, such that the network dissipates 295 Watts.

## Homework Equations

$$\frac{1}{R_{eqp}} = \Sigma \frac{1}{R_{i}}$$
$$R_{eqs} = \Sigma R_{i}$$
$$P = \frac{V^{2}}{R} = VI = I^{2}R$$

...
Why not use: $\displaystyle P = \frac{V^{2}}{R_{eq}}$ to find the equivalent resistance for the circuit, then use series & parallel analysis to find R ?

Why not use: $\displaystyle P = \frac{V^{2}}{R_{eq}}$ to find the equivalent resistance for the circuit, then use series & parallel analysis to find R ?
So like:
$$R = \frac{V^{2}}{P} = \frac{48^{2}}{295} = 7.81$$

$$7.81 = \frac{8R}{R+8} + 3$$

$$4.81R + 38.48 = 8R$$

$$38.48 = 3.19R$$

$$R = 12.1$$

What the heck, how is it that the other method I was using did not work?

Also, Thanks!