CIRCUIT ANALYSIS: Find the equivalent resistance seen by the source

[VinnyCee]

1. The problem statement, all variables and given/known data

Using series/parallel resistance combination, find the equivalent resistance seen by the source in the circuit below. Find the overall dissipated power.

http://img300.imageshack.us/img300/6132/chapter2problem34au4.jpg

2. Relevant equations

For resistors in series: R_{eq}\,=\,R_1\,+\,R_2\,+\,\cdots\,+\,R_n

For resistors in parallel: R_{eq}\,=\,\frac{1}{R_1}\,+\,\frac{1}{R_2}\,+\,\cd ots\,+\,\frac{1}{R_n}

Also for resistors in parallel: R_{eq}\,=\,\frac{R_1\,R_2}{R_1\,+\,R_2}


3. The attempt at a solution

Adding the right three resistors that are in series to get an equivalent one that is 40 Ohms.

http://img205.imageshack.us/img205/1186/chapter2problem34part2dk8.jpg

Using the formula above for parallel resistors: R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega )\,+\,(40\Omega)}\,=\, 20\Omega

http://img391.imageshack.us/img391/7092/chapter2problem34part3ur9.jpg

Again, combining the resistors on the right that are in series.

http://img205.imageshack.us/img205/5549/chapter2problem34part4jn1.jpg

Using the parallel formula again: R_{eq}\,=\,\frac{(40\Omega)\,(40\Omega)}{(40\Omega )\,+\,(40\Omega)}\,=\, 20\Omega

http://img397.imageshack.us/img397/9507/chapter2problem34part5su9.jpg

Finally, adding the last two resistors in series, I get an R_{eq} of 40\Omega. Does this seem correct?

http://img300.imageshack.us/img300/2751/chapter2problem34part6ma8.jpg

Then, to find the Power, I get the current first.

i\,=\,\frac{v}{R}\,=\,\frac{12\,V}{40\Omega}\,=\,0 .3\,A

The I use the p = vi equation.

p\,=\,v\,i\,=\,\left(12\,V\right)\,\left(0.3\,A\ri ght)\,=\,3.6\,W

Right?

[doodle]

Or equivalently, P = V^2/R which gives the same answer.