You can solve using mesh analysis if you wanted, but I'd condense that middle branch into an equivalent resistance first.
Find the voltage V and the power absorbed by the 12-ohm resistor.
p = vi
v = ir
The Attempt at a Solution
Tried using KVL and KCL in various ways. This book gives horrid examples.
The current is [tex].4[/tex] and the voltage is [tex]4.8[/tex]. However, [tex].4 \times 4.8[/tex] is not [tex]9.6[/tex]. It's [tex]1.92[/tex](as the book says). Also, [tex]I^2 r = (.4)^2 12 = 1.92[/tex] as well if you wanted to avoid that extra calculation of voltage.I found the current in the last loop to be .4 A. Using V = IR to solve to get the voltage, I get 4.8 V at the 12-ohm resistor. I then used P = VI to get a power of 9.6 W, but the book has 1.92 W. Did I do anything wrong or is the book wrong again?