What is the Resistance R for 295 Watts Dissipation in a Network of Resistors?

[QuarkCharmer]

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?

 

[Norfonz]

R1 is in series with the 3 ohm resistor, not in parallel.

 

[QuarkCharmer]

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]

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 ?

 

[QuarkCharmer]

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!