Power Dissipated for 2 different circuits

AI Thread Summary
To determine the ratio of R1 to R2 for equal power dissipation in circuits A and B, the currents through R2 must be set equal, leading to the equation (2EMF)/(2R1+R2) = (EMF)/[(R1^2/2R1) + R2]. The discussion highlights uncertainty about simplifying the equation by removing EMF, which is indeed constant. For part B, the condition under which power dissipated in circuit A exceeds that in circuit B is debated, with R2 > R1 being a proposed solution. Participants express confusion regarding calculations and seek confirmation on their findings, particularly when substituting values for R2.
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Part A:
For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?

current for R2 of circuit A is (2EMF)/(2R1+R2)
current for R2 of circuit B is (EMF)/[(R1^2/2R1) + R2)

Part B:
Under which of the following conditions would power dissipated by the resistance in circuit A be bigger than that of circuit B? Choose the most restrictive answer.
R2<R1
R2>R1
R2<2R1
R2<.5R1

Equations used:
P = IV = I^2 x R = V^2 /R

Attempt:
I understand since R and P is same for both i only have to make the 2 currents equal.
(2EMF)/(2R1+R2) = (EMF)/[(R1^2/2R1) + R2)
I haven't taken physics or math class in 2 years so I'm kinda unsure on how to continue.
Since EMF is constant can I just remove it?

For part B just by plugging in numbers R2>R1 works but I am not certain
 
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Diagram please.
 
totally forgot but here it is

4km5h.jpg
 
deleting my post
 
if i can take out EMF and make R2 = 1 i come out with 4 as my answer. can anyone confirm?
 

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