- #1

- 1,039

- 2

## Homework Statement

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

## Homework Equations

[tex]\frac{1}{R_{eqp}} = \Sigma \frac{1}{R_{i}}[/tex]

[tex]R_{eqs} = \Sigma R_{i}[/tex]

[tex]P = \frac{V^{2}}{R} = VI = I^{2}R[/tex]

## 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:

[tex]\frac{1}{R_{1}} = \frac{1}{8} + \frac{1}{R}[/tex]

[tex]\frac{1}{R_{2}} = \frac{1}{8} + \frac{1}{R} + \frac{1}{3} = \frac{11R+24}{24R}[/tex]

[tex]R_{total} = \frac{24R}{11R+24}[/tex]

Now, since [itex]P = \frac{V^{2}}{R}[/itex] :

[tex]295 = 48^{2}(\frac{11R+24}{24R})[/tex]

[tex]7080R = 25344R + 55296[/tex]

[tex]-18264R = 55296[/tex]

[tex]R = -3.03 \Omega[/tex]

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?