Whoops, maybe R_{eq}\,=\,7.5\Omega?
http://img90.imageshack.us/img90/1933/chapter2problem38part3jr5.jpg
Combining the two resistors in parallel that are circled in green.
\frac{(12\Omega)\,(6\Omega)}{(12\Omega)\,+\,(6\Omega)}\,=\,\frac{72}{18}\,\Omega\,=\,4\Omega
http://img111.imageshack.us/img111/9434/chapter2problem38part4um9.jpg
Now combining the two resistors circled in green that are in parallel.
\frac{(80\Omega)\,(20\Omega)}{(80\Omega)\,+\,(20\Omega)}\,=\,\frac{1600}{100}\,\Omega\,=\,16\Omega
http://img441.imageshack.us/img441/9001/chapter2problem38part5rb8.jpg
Combine the two resistors circled in green that are in series.
http://img20.imageshack.us/img20/8697/chapter2problem38part6bu7.jpg
Combine the two resistors circled in green that are in parallel.
\frac{(60\Omega)\,(20\Omega)}{(60\Omega)\,+\,(20\Omega)}\,=\,\frac{1200}{80}\,\Omega\,=\,15\Omega
http://img294.imageshack.us/img294/7270/chapter2problem38part7ac4.jpg
Combine the last two resistors circled in green that are in parallel.
\frac{(15\Omega)\,(15\Omega)}{(15\Omega)\,+\,(15\Omega)}\,=\,\frac{225}{30}\,\Omega\,=\,7.5\Omega
http://img363.imageshack.us/img363/1525/chapter2problem38part8yh6.jpg
So how do I solve for i_0 if that is the right R_{eq}?