Power Dissipation in Resistive Circuit Conceptual Question

  • Thread starter sushii9
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Hey guys, so I have a homework question that I have already solved one way, but when I try it another way I can't seem to get the correct answer.
First of all, the question is:
"A single resistor is wired to a battery as shown in the diagram http://session.masteringphysics.com/problemAsset/1011457/10/1011457A.jpg" [Broken] Define the total power dissipated by this circuit as P_0.
Now, a second identical resistor is wired in series with the first resistor as shown in the second diagram http://session.masteringphysics.com/problemAsset/1011457/10/1011457B.jpg" [Broken].
What is the power, in terms of P_0, dissipated by this circuit?"

Relevant equations:
P = V^2/R = I^2*R

So I have solved the equation by using V^2/R (P = V^2/(2R) = P_0/2), however the first way I tried to solve it was with I^2*R. I know this equation is correct, so why doesn't it give me the right solution? If anyone could give me an explanation I would be grateful. (My work is shown below)

P = I^2 * (R + R) = I^2 * 2R = 2 * P_0 ?

Homework Statement





Homework Equations





The Attempt at a Solution

 
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Answers and Replies

  • #2
Disconnected
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So I guess you are treating the two resistors as one resistor with double the resistance?
Would the current really be the same? I think you may find the V=IR, so I=V/R, if the V is constant and the resistance doubles then the current will half.
so I(f) the final current (after the addition of the second resistor) is half the initial current, I(s).

so

P(0)=I(s)^2*R

P(f)=I(f)^2*2R

=(1/2 I(s))^2*2R

=(1/4)I(s)^2*2R

=(1/2)P(0)


Edit to ask: Does that make sense?
 

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