How to Simplify and Find Power in a Circuit?

[Paymemoney]

Homework Statement

[PLAIN]http://img802.imageshack.us/img802/9172/qns33.png

Simplifying the circuit?

Also find the power dissipated in the circuit?

Homework Equations

[tex]P=I^2R[/tex]

[tex]V=IR[/tex]

The Attempt at a Solution

I tired to simply the diagonal by making it parallel with (15+18ohms) and the (10+16ohms) and i get

[tex]R = (\frac{1}{33}+\frac{1}{48}+\frac{1}{16})^{-1} = 8.8ohms
[/tex]
Power of 15ohms is: [tex]1.3^2 * 15 = 25.35W[/tex]

Now how would i find the power dissipated for the rest of the resistors?

P.S

 

[SammyS]

Homework Statement

[PLAIN]http://img802.imageshack.us/img802/9172/qns33.png
...

I tried to simply the diagonal by making it parallel with (15+18ohms) and the (10+16ohms) and i get

[tex]R = (\frac{1}{33}+\frac{1}{48}+\frac{1}{16})^{-1} = 8.8ohms
[/tex]

The 48 Ω resistor is in parallel with the (10 Ω + 6 Ω ) combination.

[tex]\displaystyle R_{48,10,6}=\left({{1}\over{48}} + {{1}\over{16}} \right)^{-1}\Omega = 12\ \Omega[/tex]

This combination, [tex]\displaystyle R_{48,10,6}\,,[/tex] is in series with the 18 Ω resistor.

The 15 Ω resistor is in parallel with the combination of the other 4 resistors.

 

[Paymemoney]

So it will end up being 10ohms when you have simplified everything.

 

[gneill]

I tired to simply the diagonal by making it parallel with (15+18ohms) and the (10+16ohms) and i get

Whoa! The 15 and 18 ohm resistors have another branch coming off of their mutual junction; They won't simplify to being simply in parallel.

Now, you *could* get fancy (and complicated) and use something called a Delta-Y transformation on resistors 15, 18, and 48, but it is not necessary if you spot the fact that the 10 and 6 ohm resistors are already alone in series, so that their sum parallels the 48 ohm resistor directly.

 

[SammyS]

So it will end up being 10ohms when you have simplified everything.

Yes. The equivalent resistance for the circuit is 10 Ω.