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15th July 2019, 17:46 #1
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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?

15th July 2019, 17:51 #2
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
Should be about 3.414 amps. Unless this is a trick question?

15th July 2019, 18:01 #3
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15th July 2019, 18:12 #4
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
Yay. I pass. What's the prize ??

15th July 2019, 18:36 #5
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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?
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.

15th July 2019, 18:41 #6
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15th July 2019, 18:45 #7
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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 !

15th July 2019, 18:51 #8
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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.
Akanimo.

15th July 2019, 18:54 #9
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15th July 2019, 18:55 #10
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15th July 2019, 19:01 #11
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15th July 2019, 22:59 #12
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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 !
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)
KlausPlease don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
1 members found this post helpful.

15th July 2019, 23:48 #13
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16th July 2019, 05:38 #14
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
How did you get 3.414 ,explain clearly
   Updated   
how did you get 3.414 , can u please explain clearly

16th July 2019, 06:28 #15
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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

16th July 2019, 07:00 #16
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
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.

Akanimo.

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17th July 2019, 18:26 #17
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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
   Updated   
Here what is the purpose of performing square and square root

17th July 2019, 18:51 #18
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
See post #12 as written by Klaus. Klaus had already shown the calculation steps.

Akanimo.

20th July 2019, 23:13 #19
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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/squarerooting?
Calculating power with excel is irrelevant.

21st July 2019, 00:16 #20
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Re: ELECTRICAL AND ELECTRONICS Instrumentation
Hi,
RMS+RMS=RMS. Why all the extra squaring/squarerooting?
Now you ask: "Why all the extra squaring/squarerooting?"
(I'm really not sure if you mean this seriously)
RMS stands for RootMeanSquare ... 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.
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.Please don´t contact me via PM, because there is no time to respond to them. No friend requests. Thank you.
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