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Calculus I homework question about rate of change of resistance in an RC circuit

Discussion in 'Electronics Homework Help' started by dietermoreno, Mar 2, 2013.

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  1. dietermoreno


    Dec 30, 2012
    Calculus I homework question about rate of change of resistance in an RC circuit.

    I think I understand the calculus better than I understand what an RC circuit is.

    So I won't post all of the information that the homework problem gave yet.

    The homework problem gave values in ohms of two resistors R1 and R2. The homework problem gives how fast the rate of change is of R1 in ohms/second and it gives how fast the rate of change of R2 is in ohms/second. The homework problem wants me to find the rate of change of resistance R that is R1 and R2 in parallel.

    How does a resistor have a rate of change? That doesn't make any sense that a resistor is changing resistance?

    I know that 1/R = 1/R1 + 1/R2 ,
    but it is in a calculus text book, so I think that it is not so simple as using the equation for resistance of a parallel circuit. Either that, or I don't understand the given information in the problem and even worse I don't understand that I don't understand the given information in the problem.

    I'm lost.

    Any hints?

    Please and thank you.

    P.S.: There is a typo in the thread title, it should be "change of resistance" instead of "change or resistance". When I edit the post it won't let me edit the thread title. Name now edited :)

    Attached Files:

    Last edited by a moderator: Mar 2, 2013
  2. duke37


    Jan 9, 2011
    You are right, the resistance will not change.

    Surely they want either the rate of change of voltage or current.
  3. john monks

    john monks

    Mar 9, 2012
    Maybe the book is referring to thermistors or light dependent resistors.
    This seems like a strange problem for calculus I.
    Perhaps you should just post the problem.
  4. dietermoreno


    Dec 30, 2012
    It says in the sticky that you can't give answers to homework, but I guess I'll just give you the numbers and you can help me reach an answer with out giving me the answer.

    The author of my textbook, Stewart, appears to like to use "the discovery method of teaching". He doesn't give any examples about projectile motion, thermodynamics, and circuits, and then he has homework problems about those. My professor isn't any better. My professor only gave an example of solving a problem with a water trough. Stewart gave examples with a water trough too and gave several homework problems with the dumb water trough. Maybe Stewart is a farmer?

    Well enough with my rant, here is the homework problem. It is #35 in Section 3.9 of Stewart Single Variable Calculus Early Transcendentals 7E:
    Yep, Stewart wants me to find the rate of change of resistors, which is non-sensical, like many of Stewart's teaching methods are non-sensical.

    Attached Files:

  5. duke37


    Jan 9, 2011
    I do not see what the diagram has to do with the problem.

    Suppose you have a resistor wire dipped into a conducting liquid, the the resistance will go up as the liquid level goes down so you have a way of measuring what is in the tank. Not so daft as it seems at first.

    You first need to calculate R.
    Tip1 With resistors in parallel, the resistance will be less than the smallest resistance so will be less than ?
    Tip2 Two identical resistors in parallel will be half the resistance of one resistor. Here one resistor is 80 ohm and the other is 100 ohm so R will be around half of ? = ?

    Get your Babbage engine fired up and calculate the exact value of R.
    Find the value of R1 and R2 one second later.
    Recalculate R
    Calculate the change of R in one second.
  6. dietermoreno


    Dec 30, 2012
    Oh I get it now thanks!

    Probably the circuit diagram in the text book was what was confusing me more than anything else, because I thought it was some sort of filter or oscillator since it had a capacitor in it.

    Apparently the problem means it could be a resistor network measuring how much gas is in your car.

    I don't know what the capacitor has to do with anything.

    Maybe Stewart just put the bad diagram in the text book to mess with students because he wants students to think about what they are doing before jumping to doing the calculus.
  7. Ratch


    Mar 10, 2013

    That looks like a straight forward problem.

    OK, the total resistance Rt is the product of R1 and R2 divided by the sum of R1 and R2. So find the derivative of Rt with respect to time in terms of R1 and R2. It's easy, you can do it. Then just plug and chug in the numbers to get an answer.

  8. Laplace


    Apr 4, 2010
    As was stated previously the problem is to find the derivative dR/dT{(R1xR2)/(R1+R2)} and substitute the given values to get the answer. There is a numerical method to check for the correct answer. One millisecond prior to the point of interest R1=79.9997 & R2=99.9998 while one millisecond after the point of interest R1=80.0003 & R2=100.0002. So calculate the parallel resistance before and after, take the difference and divide by two milliseconds to get a close approximation of dR/dT at the point of interest.
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