How do you solve for power i parallel circuit?

Question

here is the problem http://img151.imageshack.us/img151/459/untitlednv3.jpg

Answer 1

The 1 Ω and 2 Ω resistors are going to split the 2 amps of current is a ⅔ to ⅓ ratio.

You can compute this with the current divider formula.
First combine the two parallel resisters to find the equivalent resistance.

Equivalent R = 1/(1/R1 + 1/R2 + 1/R...)
Equivalent R = 1/(1/2 + 1/1)
Equilvalent R = 1/(1.5) = ⅔ Ω. (labeled R total below)

I of a resistor = (I total * R total) / R of the resistor

I of 1Ω resistor = (2 amps * ⅔ Ω) / 1 Ω = 1⅓ amps

The current flowing through the 1 Ω resistor is 1⅓ amps. The power through the resistor is I²R.
(1⅓ amps)² x 1 Ω = 1.77 watts.

Answer 2

The current source draws 2A through the parallel combination of 1 ohm and 2 ohm resistors.

This current will divide in inverse proportion to the resistances, so the current through the 1 ohm resistor is 2/3 × 2A = 4/3 A.

The power dissipated when current I flows through resistance R is I^2R. So the power in the 1 ohm resistor (4/3)^2 × 1 = 1.78 W.

Answer 3

Voltage on 2 ohm resistor and voltage over 1 ohm resistor are the same, say V. Then the current through 2 ohm is V/2, and though 1 ohm is V/1. The sum of those currents is 2 A. So:
V/2 + V/1 = 2 => V = 4/3 V, and current through 1 ohm I = 4/3 A.
The power over 1 ohm resistor, then, is (4/3)^2 W.