Six resistors R1=9.00 Ω, R2=3.00 Ω, R3=12.0 Ω, R4=10.0 Ω, R5=10.0 Ω, and R6=8.00 Ω are mounted as shown in the figure.?

Find the current leaving the battery by combining resistances:

[(3||12)+9] || [10+ (8||10)] = [36/15+135/15] || [180/18+ 80/18] =

57/5 || 130/9 = (57*130)/45]/(513+650)/45 = 7410/1163 = 6.37Ω

The current leaving the battery = 11630/7410 = 1.5695A

Now use a current divider to determine the current into the top 10Ω

1.5695 *57/5/(57/5+130/9) = 0.6923A

Use another current divider to find how much enters the top 8Ω

0.6923*(10/18) = 0.3846

Power = 0.3846²*8 = 1.183W or about 1.2W with just 2 s.f.

R1, R2 and R3 don't figure in so you can ignore them.

Voltage x current = power. All you have to do is find the voltage across R6 and the current through it and multiply them together. To do that you need to find the equivalent resistance of R5 and R6 so use this formula for parallel resistors :

R5 x R6

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R5 + R6

So that would be 80 / 18 = 4.444 Ω

So 4.44 Ω in series with R4 (10Ω) = 14.4Ω So then the total resistance of R4 + R5R6 = 14.4Ω

Now we need to find the current through that total resistance which is 10v / 14.4Ω = .69 amps.

That means .69 amps is flowing through R4 so the voltage dropped across that is 10Ω x .69 amps = 6.9 volts.

That leaves 10v - 6.9v = 3.1 volts across R5R6. Now that we know there is 3.1 volts across R6 we can figure the power dissipated as follows:

3.1 v / 8 Ω = .39 amps. So .39 amps is flowing through R6. Now Power = 3.1 * .39 amps = 1.2 watts dissipated in R6. The combined way to do that would have been to say Power = V² / R so 3.1 x 3.1 / 8Ω = 1.2 watts.

R1, R2, and R3 don't matter. R5 & R6 have a resistance of 1/R = 1/8 + 1/10, so R = 4.444 ohms. The total resistance in the upper branch is 10 + 4.444 = 14.444. With 10 volts, the current is 0.6923 amps, because 10 = 0.6923 * 14.4444.

The voltage drop on R4 is 0.6923 * 10 = 6.923 volts. That leaves 3.077 V across R6 and R5. Now you have the voltage and the resistance of R6. The rest is up to you.

just go thru it. 

8| parallel with 10 = 80/18 = 4.44

in series with 10 that is 14.44

total I in upper branch = 10/14.44 = 0.692 amps

that produces a voltage across the parallel pair of

E = IR = 0.692 x 4.44 = 3.08 volts

and power = E²/R = 3.08²/8 = 1.18 watts