ELECTRICAL AND ELECTRONICS Instrumentation

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rajasekharnbkrist

Junior Member level 2 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?

kripacharya

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

Akanimo 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

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kripacharya

Banned Yay. I pass. What's the prize ?? :grin:

KlausST

Super Moderator
Staff member 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

Akanimo 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.

kripacharya

Banned KlaussT !!!!! :lol:

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 ! :wink:

Akanimo 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.

kripacharya

Banned 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.

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Akanimo ...
Am I wrong?

Klaus

The squaring you're doing is computed with instantaneous peak values every cycle of the AC waveform.

kripacharya

Banned 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. :wink:

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

KlausST

Super Moderator
Staff member 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

• Easy peasy

Easy peasy

Points: 2

Akanimo I just simulated with LTspice. Klaus, you are right.

This is so revealing. Really something to think about.

Thanks for the question.

rajasekharnbkrist

Junior Member level 2 How did you get 3.414 ,explain clearly

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how did you get 3.414 , can u please explain clearly

kripacharya

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

Akanimo How did you get 3.414 ,explain clearly

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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.

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rajasekharnbkrist

Junior Member level 2 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

Akanimo See post #12 as written by Klaus. Klaus had already shown the calculation steps.

barry 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.

KlausST

Super Moderator
Staff member 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

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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