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

In summary, The conversation revolves around a question in a math class that asks for the change in power when certain values of voltage and resistance are altered. The original approach was to simply plug in the given values and find the difference in power, but the correct answer was obtained by taking the partial derivatives of the equation. This shows that P is not a linear function of E or R. The questioner is surprised by this and questions why the more precise approach was not taken in the first place.
  • #1
Livethefire
51
0

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, I've 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.
 
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  • #2
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.
 

1. What is power dissipation in a resistor?

Power dissipation in a resistor is the amount of energy that is converted into heat as current flows through the resistor. In other words, it is the rate at which a resistor dissipates energy.

2. How is power dissipation calculated in a resistor?

Power dissipation in a resistor can be calculated using the formula P = I²R, where P is power in watts, I is current in amps, and R is resistance in ohms. This formula is known as Joule's Law.

3. What is the difference between power dissipation and power rating in a resistor?

Power dissipation is the actual amount of energy that is converted into heat in a resistor, while power rating is the maximum amount of power a resistor can safely dissipate without overheating. Power rating is typically indicated on the resistor's packaging or datasheet.

4. How does the value of resistance affect power dissipation in a resistor?

The higher the resistance value of a resistor, the lower the power dissipation will be. This is because as resistance increases, the current flowing through the resistor decreases, resulting in less energy being converted into heat.

5. Can a resistor dissipate more power than its power rating?

No, a resistor should never be allowed to dissipate more power than its power rating. Doing so can cause the resistor to overheat and potentially fail. It is important to always choose a resistor with a power rating that is equal to or greater than the expected power dissipation in a circuit.

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