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    ELECTRICAL AND ELECTRONICS Instrumentation

    1. The current in a 20 ohm resistor is given by I=2+4sin314t. The current is measured by a hot wire ammeter. What is the measured value?

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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Should be about 3.414 amps. Unless this is a trick question?



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by kripacharya View Post
    Should be about 3.414 amps. Unless this is a trick question?
    That's it.

    2 + 2*0.707 = 2+1.414 = 3.414 A
    Last edited by Akanimo; 15th July 2019 at 18:07.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Yay. I pass. What's the prize ??



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Hi.

    The voltages should be added by squaring them beforehand then take the square root.

    I see 2V DC and
    A sine wave with 4Vp which is 2.828V RMS.

    [Edit: the following calculation is wrong, please see correct calculation in post#12]

    2^2 + 2.828^2 = 4 + 10 = 14
    Sqrt(14) = 3.742V RMS.

    3.742V RMS / 20 Ohms = 0.1871A.

    Am I wrong?

    Klaus
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by Akanimo View Post
    That's it.

    2 + 2*0.707 = 2+1.414 = 3.414 A
    OMG!

    This was weird of me. It should be 2 + 4*0.707 = 2 + 2.828 = 4.828A.

    I got affected by the fact that kripacharya changed his answer from 2.828A to 3.414A at the instance I quoted it.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    KlaussT !!!!!

    First and foremost - the equation given is for current I, not voltage. So resistance is irrelevant.

    Second, the sinewave component requires rms calculation, but not the DC component.

    You have been tricked !



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Klaus,

    The sine wave current has a +2A DC offset. The DC is already providing it's heat. The AC component provides additional heat in form of RMS that the equivalent (Vp*0.707)Vdc would provide.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by Akanimo View Post
    OMG!

    This was weird of me. It should be 2 + 4*0.707 = 2 + 2.828 = 4.828A.

    I got affected by the fact that kripacharya changed his answer from 2.828A to 3.414A at the instance I quoted it.
    Actually you're right. I used 4 instead of +/- 4. So 4.828 is correct.

    - - - Updated - - -



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by KlausST View Post
    ...
    Am I wrong?

    Klaus
    The squaring you're doing is computed with instantaneous peak values every cycle of the AC waveform.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by Akanimo View Post
    OMG!

    This was weird of me. It should be 2 + 4*0.707 = 2 + 2.828 = 4.828A.

    I got affected by the fact that kripacharya changed his answer from 2.828A to 3.414A at the instance I quoted it.
    LoL.. !
    I did that especially to confuse you.

    - - - Updated - - -

    Rajasekharn... Please post more puzzles like this. It will keep us awake at nights.



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Hi,

    KlaussT !!!!!

    First and foremost - the equation given is for current I, not voltage. So resistance is irrelevant.

    Second, the sinewave component requires rms calculation, but not the DC component.

    You have been tricked !
    I agree, the equation gives current ... no voltage....no restor needed to calculate the current.

    But "2A" is the DC component, it's the RMS current at zero frequency and has to be treated the same as any other frequency.

    4A is the amplitude of the sine and thus the RMS of the sine is 2.828A

    The "hot wire" measures total RMS current.
    Thus - and here I'm pretty sure - you need to add both RMS currents with the square and square root calculation.

    But I had another error in my (mind) calculation. 2.828^2 = 8
    Thus the total RMS current = sqrt (2^2 + 2.828^2) = sqrt (4 + 8) = sqrt (12) = 3.464A

    I just confirmed this with an Excel simulation that calculates P(t)

    Klaus
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    I just simulated with LTspice.

    Click image for larger version. 

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    Klaus, you are right.

    This is so revealing. Really something to think about.

    Thanks for the question.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    How did you get 3.414 ,explain clearly

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    how did you get 3.414 , can u please explain clearly



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    KlaussT has done it! I concur, there was an error in calculation originally. Now it's fixed. 3.464 amps



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Quote Originally Posted by rajasekharnbkrist View Post
    How did you get 3.414 ,explain clearly

    - - - Updated - - -

    how did you get 3.414 , can u please explain clearly
    See post #3. I thought I saw a 2Ap instead of a 4Ap. That was influenced by kripacharya.

    I clicked on the "reply with quote" button bearing in mind to discuss the 2.828A that I saw. As at then I had 4.828A just as I showed in my latter post. When the field loaded, I saw 3.414A. Oops! I posted 4.828A initially. If you were on the thread at the time, you would have seen it. Then I had a quick thought over what just happened, and I was like "2+2sin (314t) A can't give above 4A. So I edited what I wrote initially. You can see the 2+2 *0707 in that post.

    At that time I didn't even picture 2+4sin (314t) anymore in my thoughts. I wrote nothing on paper though.

    Funny though.
    Last edited by Akanimo; 16th July 2019 at 07:07.
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    I am also getting 3.414, but answer is 3.46,how did you get 3.46 amp

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    Here what is the purpose of performing square and square root



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    See post #12 as written by Klaus. Klaus had already shown the calculation steps.
    -------------
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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    I’m not sure I follow this. By superposition the current should be:

    2 + .707*4=2+2.8=4.8, right?

    RMS+RMS=RMS. Why all the extra squaring/square-rooting?

    Calculating power with excel is irrelevant.



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    Re: ELECTRICAL AND ELECTRONICS Instrumentation

    Hi,

    RMS+RMS=RMS. Why all the extra squaring/square-rooting?
    RMS+RMS = RMS is only true for signals of same waveform, same frequency and sane phase.

    Now you ask: "Why all the extra squaring/square-rooting?"
    (I'm really not sure if you mean this seriously)
    RMS stands for Root-Mean-Square ... here you have root= square root and square.

    The RMS value represents the value that generates the same heat on a pure resistive load as DC.
    Examples:
    Any 3V RMS value generats the same heat on a resistor as 3V DC.
    Any 6.789A RMS generate the same heat as 6.789V DC

    Now heat is power.
    And power is proportional to the square of voltage.
    Voltage --> squaring --> power --> square root --> voltage

    Klaus

    Added:
    Calculating power with excel is irrelevant.
    Could you please explain why you say so.

    It shows I(t), it shows P(t), it shows the average of P(t) over a given time, it calculates the square root of the average value. The results should not differ from the result of any electronics simulator tool.
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