Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

ELECTRICAL AND ELECTRONICS Instrumentation

Status
Not open for further replies.

rajasekharnbkrist

Junior Member level 2
Joined
Nov 10, 2013
Messages
21
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
182
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
Joined
Dec 28, 2012
Messages
1,209
Helped
182
Reputation
360
Reaction score
175
Trophy points
1,343
Location
New Delhi
Activity points
0
Should be about 3.414 amps. Unless this is a trick question?
 

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
20,790
Helped
4,492
Reputation
8,996
Reaction score
4,557
Trophy points
1,393
Activity points
137,268
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

Advanced Member level 3
Joined
Aug 30, 2016
Messages
717
Helped
127
Reputation
254
Reaction score
127
Trophy points
43
Activity points
4,920
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
Joined
Dec 28, 2012
Messages
1,209
Helped
182
Reputation
360
Reaction score
175
Trophy points
1,343
Location
New Delhi
Activity points
0
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

Advanced Member level 3
Joined
Aug 30, 2016
Messages
717
Helped
127
Reputation
254
Reaction score
127
Trophy points
43
Activity points
4,920
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
Joined
Dec 28, 2012
Messages
1,209
Helped
182
Reputation
360
Reaction score
175
Trophy points
1,343
Location
New Delhi
Activity points
0
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 - - -
 

kripacharya

Banned
Joined
Dec 28, 2012
Messages
1,209
Helped
182
Reputation
360
Reaction score
175
Trophy points
1,343
Location
New Delhi
Activity points
0
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:

- - - Updated - - -

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

KlausST

Super Moderator
Staff member
Joined
Apr 17, 2014
Messages
20,790
Helped
4,492
Reputation
8,996
Reaction score
4,557
Trophy points
1,393
Activity points
137,268
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
 

Akanimo

Advanced Member level 3
Joined
Aug 30, 2016
Messages
717
Helped
127
Reputation
254
Reaction score
127
Trophy points
43
Activity points
4,920
I just simulated with LTspice.

Capture.PNG

Klaus, you are right.

This is so revealing. Really something to think about.

Thanks for the question.
 

rajasekharnbkrist

Junior Member level 2
Joined
Nov 10, 2013
Messages
21
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
182
How did you get 3.414 ,explain clearly

- - - Updated - - -

how did you get 3.414 , can u please explain clearly
 

kripacharya

Banned
Joined
Dec 28, 2012
Messages
1,209
Helped
182
Reputation
360
Reaction score
175
Trophy points
1,343
Location
New Delhi
Activity points
0
KlaussT has done it! I concur, there was an error in calculation originally. Now it's fixed. 3.464 amps
 

Akanimo

Advanced Member level 3
Joined
Aug 30, 2016
Messages
717
Helped
127
Reputation
254
Reaction score
127
Trophy points
43
Activity points
4,920
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:

rajasekharnbkrist

Junior Member level 2
Joined
Nov 10, 2013
Messages
21
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Activity points
182
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
 

Akanimo

Advanced Member level 3
Joined
Aug 30, 2016
Messages
717
Helped
127
Reputation
254
Reaction score
127
Trophy points
43
Activity points
4,920
See post #12 as written by Klaus. Klaus had already shown the calculation steps.
 

barry

Advanced Member level 5
Joined
Mar 31, 2005
Messages
5,344
Helped
1,134
Reputation
2,280
Reaction score
1,152
Trophy points
1,393
Location
California, USA
Activity points
29,215
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
Joined
Apr 17, 2014
Messages
20,790
Helped
4,492
Reputation
8,996
Reaction score
4,557
Trophy points
1,393
Activity points
137,268
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.
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Top