Calculating Dissipation of Electric Power in a Circuit

[indojo24]

Homework Statement

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.

Homework Equations

R=V²/P
R=(Ωhm)
P=(Watt)
V=(Volt)

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

 

[lewando]

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

 

[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?

 

[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 .

 

[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.

R_series=R₂+R₃=2R

1/R_parallel=(1/R_series)+(1/R₄)
1/R_parallel=(1/2R)+(1/R)
1/R_parallel=(1/2R)+(2/2R)
1/R_parallel=3/2R
R_parallel=2R/3

P₁=36 W
P_parallel=2P₁/3=24W
P_series : P₄ = 1/2R : 1/R
P_series : P₄ = 1/2 : 1
P_series : P₄ = 1 : 2
P_parallel=P_series+P₄=x + 2x=3x

x=(24/3)W=8W
P₄=2x=2(8W)=16 W(answer is b)

Is this correct?

 

[haruspex]

Quite correct.