which inductor current waveform is correct?

Discussion in 'General Electronics Chat' started by PG1995, Oct 31, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please tell me which current waveform is correct. Thank you for the help.

    Regards
    PG
     
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    Neither is correct. You may be getting confused by notation.

    The correct way to write the integral is as follows.

    i(t)={{1}\over{L}}\int_{t_o}^t v(\tau)d\tau +i(t_o)

    The difference is that the tau variable is a dummy integration variable, but the t variable represents real time. So, when you think about calculating i(t) it means finding current as a funtion of time. The time variable shows up as the upper limit in the integral, and hence you should not integrate over t, but over a dummy variable, which I called tau.

    In problems like these, you should try to visualize the physics to double check the answer. If you integrate a triangle or a ramp shape, what shape do you expect in the answer? If you integrate a constant value, what shape to you expect in the answer?
     
    PG1995 likes this.
  3. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, Steve.

    Actually, I'm also learning calculus.

    Isn't my answer, 2mA, correct for 0< t <0.1? That someone else has also got this answer.

    I don't know how else I can integrate the v(t)=-2/1000 over the interval 0.1< t <0.2. Could you please help me with it?

    Many, many thanks for helping me with so many queries. I'm much obliged.

    Best wishes
    PG
     
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I think 2 mA is correct if the input voltage is in milliVolts and not Volts. I wasn't sure what the units are. So basically, you seemed to have done a calculation for t=0.1 s correctly. Notice how you did the integral with 0.1 in the upper limit of the integral. That's why it came out correctly. So, now to find other values with t<0.1 s, you plug in different values of t. Rather than doing this over and over, you can simply solve using t in the upper limit and then your final formula is valid for all t in the range of 0 to 0.1 s.

    That one is even easier. Remember integral just means area under the curve. So, you can calculate the area of a rectangle with no trouble. However, to do it more formally, you would apply the same integral formula.

    Here, to=0.1 s, i(to)=2 mA, and v(t)=-2 mV

    Now, for future reference, you should remember that the integral of a constant value is a ramp, and the integral of a ramp is a parabola. I was able to instantly see that both you and your friend had the wrong answer just by noting this fact. You should be able to do the same from now on. Always check your answer for reasonableness in terms of general shapes and in terms of approximate values. In this case you should notice that the area under the first part of the curve is equal in magnitude and opposite in sign from the area under the second part of the voltage curve. This implies that the final voltage should equal the starting voltage.
     
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