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Power Dissipated in a Resistor (really basic, but confused)

  • #1

Homework Statement



I understand the maths... I'm here to ask WHY we have to do it this way.

The question states:
"The power dissipated in a resistor is given by [itex]P= E^2/R[/itex]. If [itex] E=200[/itex] and [itex] R=8 [/itex], find the change in [itex] P [/itex] resulting in a drop of [itex] 5 Volts [/itex] in [itex] E [/itex] and an increase of [itex] 0.2 Ohms [/itex] in [itex] R [/itex]."

Homework Equations


Above.


The Attempt at a Solution



Physically I was thinking, okay plug in [itex] 200 [/itex] and [itex] 8 [/itex] then subtract from that answer the power calculated when [itex] 195 [/itex] and [itex] 8.2 [/itex] are input into the equation.

This gives Change in power[itex]\approx362.8W [/itex]

My line of thought was, well if I have a resistor of 8 Ohms and a voltage of 200 across it the power will be a certain value. Then if I had a similar resistor of resistance 8.2 Ohms and a Voltage across if of 195 V then the difference when these values are put into the equation will be the change in power.

Why is this NOT the case? Namely the true answer is apparently: 375W,

You get this by doing the partial derivative of the equation with respect to E and R, ive done the math and it checks out to that answer alright, but as stated- What is wrong with what I have done?

What is my fatal assumption?
Is it because the changes are small and thus calculus needs to be involved?

Thanks for any responce.
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,585
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Your answer is correct. The precise change in power is 362.8 W. If you take the rate of change of power as a function of voltage x change in voltage + the rate of change of power as a function of resistance x change in resistance, you will only get an approximate answer since P is not a linear function of E or R.

AM
 
  • #3
Thanks for the quick reply.

This is kind of ironic though- that question was in a math class. Ussually they try to be the precise ones, and physicists make the approximations :P.

Im not confused at the question anymore, rather why they would do it that way if they have all the information to get a better answer.

Regardless, thanks for clearing that up.
 

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