Power in network of resistors?

  • #1
1,039
2

Homework Statement


40xzd.jpg

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?
 

Answers and Replies

  • #2
56
0
R1 is in series with the 3 ohm resistor, not in parallel.
 
  • #3
1,039
2
Oh wow, :blushing:

Okay so:

[tex]R_{tot} = \frac{11R+12}{R+8}[/tex]

Then into P = V^2/R

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

Which is still a bit off from the 12.1 ohms it's supposed to be?
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012

Homework Statement


40xzd.jpg

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]

...
Why not use: [itex]\displaystyle P = \frac{V^{2}}{R_{eq}}[/itex] to find the equivalent resistance for the circuit, then use series & parallel analysis to find R ?
 
  • #5
1,039
2
Why not use: [itex]\displaystyle P = \frac{V^{2}}{R_{eq}}[/itex] to find the equivalent resistance for the circuit, then use series & parallel analysis to find R ?
So like:
[tex]R = \frac{V^{2}}{P} = \frac{48^{2}}{295} = 7.81[/tex]

[tex]7.81 = \frac{8R}{R+8} + 3[/tex]

[tex]4.81R + 38.48 = 8R[/tex]

[tex]38.48 = 3.19R[/tex]

[tex]R = 12.1[/tex]

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

Also, Thanks!
 

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