small question about Ohm's law - Electronic Circuits Projects Diagrams Free

TronicBrain · 26th September 2007, 08:05 PM

Hi
I have a small question about Ohm law

Suppose I have 4 identical resistors
50 Ohms 1 Watt
And I want to connect them in order to achieve 30 Ohme resistor with 4 Watts

Is this scheme correct?

**Nigel Goodwin** · 26th September 2007, 08:55 PM

I can't be bothered to work out the resistance, but the total wattage won't be 4W, it will be less than two watts.

ecerfoglio · 26th September 2007, 09:04 PM

Yes

and no

The resulting resistance will be 30

hm: as you calculated:

R1 in paralel with R3 = 25 hm:,
25hm: in series with R2 = 75 hm:,
75 hm: in paralel with R4 = 30 hm:.

BUT those resistors don't share the power evenly.

You must calculate the power dissipated by each one:

R4 will dissipate V^2/50

hm: = 1 W,
so V should be less or equal than the square root of (1 W * 50

hm

= 7.07 V.

With a 7.07 V supply, the 75

hm: branch will have a current of 94 mA, which gives:
0.444 W in R2 ( = 94 mA^2 * 50

hm: ), and
0.111 W each in R1 and R3 (with half the current, 94/2 mA^2 * 50

hm: )

So, as you must limit the power dissipated in every resistor to 1W,
the "new" 30hm: resistor will be able to dissipate less than 2 W (in fact, just 1.666 W):

1 W + 0.444 W + 2 x 0.111 W = 1.666 W, or

7.07 V ^2 / 30

hm: = 50/30 W = 1.666 W

**Nigel Goodwin** · 26th September 2007, 09:13 PM

Thanks for the 1.66W calculation, like I said - less than 2W - but I couldn't be bothered to do the sums, other than the 'instant' approximation of under 2W

audioguru · 26th September 2007, 10:12 PM

Quote:

Originally Posted by ecerfoglio

R4 will dissipate V^2/50

hm: = 1 W

No.
R4 will dissipate 2W and since it is rated for a max of only 1W then it will smoke and burn.

rmn_tech · 26th September 2007, 10:31 PM

Quote:

Originally Posted by Nigel Goodwin

I can't be bothered to work out the resistance, but the total wattage won't be 4W, it will be less than two watts.

I'm sorry But I thought the idea of this forum was to help people. But if you can't be bothered sorry to have disturbed you.

Why answer if you're not bothered ?

audioguru · 26th September 2007, 10:45 PM

It is not a technical question.
It is arithmatic for 10 year old kids.

Most of us can't be bothered to do simple arithmatic.

Leftyretro · 26th September 2007, 11:13 PM

Quote:

Originally Posted by audioguru

It is not a technical question.
It is arithmatic for 10 year old kids.

Most of us can't be bothered to do simple arithmatic.

Well I think RMN TECH response was that it takes more effort to post about not being bothered to answer then to not respond with a post at all. One chooses to respond to a post or not.

Lets be honest, it was a grouchy response. We all have bad days at times but doesn't hurt to try and be civilized, especially to newer posters.

Lefty

audioguru · 27th September 2007, 01:36 AM

I think the little kids couldn't be bothered to ask their teacher about the simple arithmatic. So they asked us.

Leftyretro · 27th September 2007, 01:42 AM

Quote:

Originally Posted by audioguru

I think the little kids couldn't be bothered to ask their teacher about the simple arithmatic. So they asked us.

Probably correct, but some here don't mind responding to their requests for help.

Lefty

Tomble · 27th September 2007, 04:54 AM

Quote:

Originally Posted by rmn_tech

I'm sorry But I thought the idea of this forum was to help people. But if you can't be bothered sorry to have disturbed you.

Why answer if you're not bothered ?

If you look at his post again, he said that he couldn't be bothered to work out the resistance, but he answered very very clearly that it would not be able to take 4W. So he did in fact answer, he simply didn't answer ALL of it. And why should he work out half of the answer if he already knew and said that the other half was "it won't work"?

Now can we stop all the damn stupid poo-flinging, everybody?

**Nigel Goodwin** · 27th September 2007, 08:44 AM

Quote:

Originally Posted by rmn_tech

I'm sorry But I thought the idea of this forum was to help people. But if you can't be bothered sorry to have disturbed you.

Why answer if you're not bothered ?

I answered one major part of the question, with didn't need any sums doing - as suggested we're not here to do primary school arithmetic for people.

arunb · 27th September 2007, 11:51 AM

'TronicBrain' you should study some basic electronics books, as the question is quite a simple one.

Internet is a great idea.. but I think a book will go a long way in solving some of the fundamentals of electronics.

HiTech · 27th September 2007, 01:17 PM

I can understand both Nigel's initial response as well as TronicBrain's question. To a knowledgable technician, it's quickly obvious that the wattage is under 2 watts. However, TronicBrain apparently is trying to understand Ohm's Law as it applies to his diagram despite the elementary math needed to solve it. I can certainly add 2 + 2 x 2, but if I don't know how to apply it to a specific situation, it's meaningless.

audioguru · 27th September 2007, 01:28 PM

Quote:

Originally Posted by HiTech

it's quickly obvious that the wattage is under 2 watts.

No.
R4 is burning up and is trying to dissipate 2.0W. The remaining resistors dissipate 1.33W so the total dissipation is 3.33W, not less than 2W.

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26th September 2007, 08:55 PM	(permalink)
Nigel Goodwin	I can't be bothered to work out the resistance, but the total wattage won't be 4W, it will be less than two watts. __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

26th September 2007, 09:04 PM	(permalink)
ecerfoglio	Yes and no The resulting resistance will be 30 hm: as you calculated: R1 in paralel with R3 = 25 hm:, 25hm: in series with R2 = 75 hm:, 75 hm: in paralel with R4 = 30 hm:. BUT those resistors don't share the power evenly. You must calculate the power dissipated by each one: R4 will dissipate V^2/50hm: = 1 W, so V should be less or equal than the square root of (1 W * 50hm = 7.07 V. With a 7.07 V supply, the 75 hm: branch will have a current of 94 mA, which gives: 0.444 W in R2 ( = 94 mA^2 * 50hm: ), and 0.111 W each in R1 and R3 (with half the current, 94/2 mA^2 * 50hm: ) So, as you must limit the power dissipated in every resistor to 1W, the "new" 30hm: resistor will be able to dissipate less than 2 W (in fact, just 1.666 W): 1 W + 0.444 W + 2 x 0.111 W = 1.666 W, or 7.07 V ^2 / 30 hm: = 50/30 W = 1.666 W __________________ E Cerfoglio Buenos Aires Argentina

26th September 2007, 09:13 PM	(permalink)
Nigel Goodwin	Thanks for the 1.66W calculation, like I said - less than 2W - but I couldn't be bothered to do the sums, other than the 'instant' approximation of under 2W __________________ PIC programmer software, and PIC Tutorials at: http://www.winpicprog.co.uk

26th September 2007, 10:45 PM	(permalink)
audioguru	It is not a technical question. It is arithmatic for 10 year old kids. Most of us can't be bothered to do simple arithmatic. __________________ Uncle $crooge

27th September 2007, 01:36 AM	(permalink)
audioguru	I think the little kids couldn't be bothered to ask their teacher about the simple arithmatic. So they asked us. __________________ Uncle $crooge

27th September 2007, 11:51 AM	(permalink)
arunb	'TronicBrain' you should study some basic electronics books, as the question is quite a simple one. Internet is a great idea.. but I think a book will go a long way in solving some of the fundamentals of electronics.

27th September 2007, 01:17 PM	(permalink)
HiTech	I can understand both Nigel's initial response as well as TronicBrain's question. To a knowledgable technician, it's quickly obvious that the wattage is under 2 watts. However, TronicBrain apparently is trying to understand Ohm's Law as it applies to his diagram despite the elementary math needed to solve it. I can certainly add 2 + 2 x 2, but if I don't know how to apply it to a specific situation, it's meaningless. __________________ Don't make me reach through this monitor to slap you a good one!