1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Current in RL circuit problem

  1. Nov 26, 2011 #1
    I have the following calculations to do.
    An inductor of 10H and 25 ohm resistance is connected in series with a 75 ohm resistor and Vs=240 V DC.
    1. the time for current to develop maximum current.
    so what I did is that:
    Max current is I= V/R= 240/(25+75)= 2.4 A
    And time constant T= 0.1 s

    i=I(1-e^(-t/τ) )
    2.4=2.4(1-e^(-t/0.1) )
    0= e^(-t/0.1)
    and now I cannot take natural log to find t. What is wrong? In simulation in Multisim it goes up to 2.4 eventually.

    2. calculate the initial rate of change of current

    Now idea what it is about so far.

    3. Time when current reaches 1.5 A
    1.5=2.4(1-e^(-t/0.1) )
    0.625=(1-e^(-t/0.1) )
    0.375=e^(-t/0.1), and taking natural log of both ides we have
    t= 98.08ms
    But in Multisim it shows it should be 98.6 so, is my calculation right? All components have 0 % tolerance etc. so I would assume it should be more accuret to what I calculated.

    Any help is welcome
  2. jcsd
  3. Nov 26, 2011 #2


    User Avatar

    Staff: Mentor

    If you consider the function [itex] f(t) = 1 - e^{-\frac{t}{\tau}} [/itex], it never actually reaches 1 in finite time. It gets arbitrarily close to 1, but never actually reaches 1 as long as t is a finite value. This is why you can't solve for a precise value of t in a meaningful way -- the ln(0) is undefined (and heads off to negative infinity in the limit).

    So what to do?

    In practical terms the main action for the exponential is over and done with after 5 time constants (5 times [itex] \tau [/itex]). The function reaches within 99% of its final value then. This is used as a practical rule of thumb for circuit design.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook