What is the power dissipated in the 5 ohm resistor in this complex circuit?

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The circuit consists of a 12-volt battery connected to a 4-ohm resistor in series with two parallel branches: one containing a 2-ohm and a 1-ohm resistor in series, and the other containing a 5-ohm and a 1-ohm resistor in series. To determine the power dissipated in the 5-ohm resistor, the total effective resistance of the circuit must first be calculated by simplifying the series and parallel resistances. After finding the total effective resistance, the total current can be determined, which will then split between the parallel branches according to Kirchhoff's Current Law. The voltage drop across each branch is necessary to calculate the current in the 5-ohm resistor. Finally, the power dissipated in the 5-ohm resistor can be calculated using the current found in that branch.
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ok there is a picture that goes along with it, i will try and draw it in as best i can.
We've got a 12 volt battery. a 4 ohm resistor is in series with this big thing I am going to describe next. A two ohm and 1 ohm resistor that are in series are in parallel with a 5 ohm and 1 ohm resistor that are also in series. What is the power dissipated in the 5 ohm resistor?

____2ohm__1ohm___
l--4ohm--l____5ohm__1ohm__l
l l
l l
—————12v---------------------

i doubt that will come out okay but w/e
 
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dang it..
 
You need to find the currents and voltages of the circuit. Just keep simplifying the circuit by combining series and parallel resistances into effective resistances. Once you find the total effective resistance seen by the load, you can find the total current. This current, by Kirchoff's Current Law, must be split up between the parallel branches that you have. So you then need to know the voltage drop across the branches to find the currents in the individual branches. Once you have the current in the 5 Ohm branch, you can find the power.
 
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