Circuit Problem?

Resistors in series simply add

Resistors in parallel can be solved with 1/Rt = 1/R1 + 1/R2 + 1/R3....

or if only two resistors, then Rt = (R1 * R2)/(R1 + R2)

Once you combine all the resistors into one equivalent resistor across the battery you can use Ohm's law to determine the amount of current in the resistor. 

Power = Voltage * Current.

Req=2+2=4//4=2+10=14//12=168/26=6.46+4=  10.46Ω

P=E^2/R=25/10.46= 

2.39W

R6 & R7 are in series:

=>Resultant(Rr) = R6 + R7 = 2 + 2 = 4Ω

Now Rr & R5 are in parallel:

=>1/Rr = 1/4 + 1/4 => Rr = 2Ω

Now Rr and R4 are in series:

=>Rr1 = 2 + 10 = 12Ω

& R2 & R3 are also in series:

=>Rr2 = 4 + 8 = 12Ω

Now Rr1 & Rr2 are in parallel:

=>1/Rr = 1/12 + 1/12 =>Rr = 6Ω

Now Rr & R1 are in series:

=>R(net) = 4 + 6 = 10Ω

Thus by P = V^2/R

=>P(total) = (5)^2 /10 = 2.5 W

R6 and R7 and in series and add up to 4

that is in parallel with R5, and combine to 4/2 = 2

that is in series with R4, total 12

R2,3 in series add up to 12

that in parallel with the other 12 equals 6

and that in series with 4 equals 10 Ω

total current = 5/10 = 0.5 amps

total power = 5 x 0.5 = 2.5 watts

for R1, voltage drop is 2 volts, so power is 2x0.5 = 1 watt

0.5 amps into the network to the right of R1, which is 6 Ω, produces a voltage of 3 volts, which gives you a current into R2 and R3 of 3/12 = 1/4 amp

etc

etc

5,6,7 are in parallel

1/4 + 1/4 = 2/4 = 1/R so R = 2 ohm

that parallel combo is in series with 4

R = 2 + 10 = 12 ohm

all the above is in parallel with 2 and 3

1/12 + 1/12 = 2/12 = 1/R so R = 6 ohm

all the above is in series with 1

4 + 6 = 10 ohm total R

then

P = V I = 5v * 10 ohm = 50 watts

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