- #1
lowcard2
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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
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