Dissipation of Electric Power

1. Dec 14, 2012

indojo24

1. The problem statement, all variables and given/known data

I want to ask about a question which is related to dissipation of electric power.
Here's a quick picture of my problem:

R1=R2=R3=R4
R1 dissipates an electrical energy of 36 W
Question: What is the electrical energy dissipated by R4?

Please look at the circuit I've attached below.

This is a Junior High physics questions which I don't understand. Thanks in advance.

2. Relevant equations
R=V2/P
R=(Ωhm)
P=(Watt)
V=(Volt)

3. The attempt at a solution
I haven't made any attempt because it's confusing.
Possible choices:
a. 18 W
b. 16 W
c. 9 W
d. 4 W

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2. Dec 14, 2012

lewando

Another equation that might help is P=I2R. Is there anything in particular that you find confusing?

3. Dec 14, 2012

CWatters

I suspect it appears confusing because not enough info is provided to calculate currents and voltages. Try thinking of it in terms of ratios.

For example how does the current flowing in R4 compare to that in R2&3?

4. Dec 14, 2012

lewando

To CW's point: if the notion of a "current divider" is in your toolbox, now would be a good time to bring it out :tongue2:.

5. Dec 14, 2012

indojo24

Well, not enough information is confusing, but I did it this way.

First, I consider R1=R2=R3=R4=R

Then, I calculate the resistance of the parallel circuit, that is, R2, R3, and R4.

Rseries=R2+R3=2R

1/Rparallel=(1/Rseries)+(1/R4)
1/Rparallel=(1/2R)+(1/R)
1/Rparallel=(1/2R)+(2/2R)
1/Rparallel=3/2R
Rparallel=2R/3

P1=36 W
Pparallel=2P1/3=24W
Pseries : P4 = 1/2R : 1/R
Pseries : P4 = 1/2 : 1
Pseries : P4 = 1 : 2
Pparallel=Pseries+P4=x + 2x=3x

x=(24/3)W=8W