Hobby Electronics Basics Newbie needs help with basics

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Subject Author Date
Newbie needs help with basics hdjim69 03-16-06
Posted by hdjim69 on March 16, 2006, 8:49 am
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>If the lamp has a rating of 10W at 12V, you can calculate a resistance for the
bulb from P = V^2/R.

Not sure I understand. this formula (P = V^2/R) finds P. I guess the
issue is, I'm not sure how to find the R value of the bulb when only V
& P are known for it. So, if I use the formula above, P = V^2/R, to
find R it would read R = V^2/P ? And are we using the true value of
the source voltage ? or do we have to include all the drops before it
gets to the bulb ?

If I could see the calculations & results using the values in the
example I can follow it.

TIA


Posted by John Fields on March 16, 2006, 10:01 am
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wrote:

>>If the lamp has a rating of 10W at 12V, you can calculate a resistance for the
bulb from P = V^2/R.
>
>Not sure I understand. this formula (P = V^2/R) finds P. I guess the
>issue is, I'm not sure how to find the R value of the bulb when only V
>& P are known for it.

---
From the tutorial, the lamp is rated to dissipate 10 watts when
there's 12 volts across it.

Since we know that:


P = IE

we can rearrange to sove for I by dividing both sides of the
equation by E, like this:


P IE
--- = ----
E E

Simplifying, the E's on the right hand side cancel and we're left
with:


P
--- = I,
E

which is the same as:

P
I = ---
E

Solving for the current through the lamp, then, we have

P 10W
I = --- = ----- ~ 0.833 ampere
E 12V

Now, since we know what the voltage across the lamp and the current
through it will be when it's dissipating 10 watts, we can solve for
the resistance of the filament like this:


E 12V
R = --- = -------- ~ 14.4 ohms
I 0.833A

---

>So, if I use the formula above, P = V^2/R, to
>find R it would read R = V^2/P ?

---
Yes.

Try it.

E² 12V * 12V 144
R = --- = ----------- = ----- = 14.4 ohms
P 10W 10w

which is the same as we what we got the long way.
---

>And are we using the true value of
>the source voltage ?

---
If you're solving for the resistance of the filament, the true value
of the source voltage is immaterial since all you need to know
about is the rated power dissipation at the specified _filament_
voltage.
---

>or do we have to include all the drops before it
>gets to the bulb ?

---
No, those will come out later.
---

>If I could see the calculations & results using the values in the
>example I can follow it.

---
See above.

--
John Fields
Professional Circuit Designer

Posted by hdjim69 on March 16, 2006, 10:12 am
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Thanks so much guys ! I think I can take it from here.


Posted by Chris on March 16, 2006, 8:38 am
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hdjim69 wrote:
> Hi, I'm trying to learn electronics, I'm right at the beginning,
> just learned Ohms law and Watts law. I understand both. I understand
> how to calculate both. However, I went to this tutorial on
> Electronicsworkbench.com and I'm confused about how they come up with
> these numbers.
>
> In this example there is a series circuit with a 15v battery, a 10w/12v
> light bulb with a voltmeter reading the voltage across the bulb and you
> have to choose between 3 different resistors that will enable the bulb
> to light - 1ohm, 100ohm and 5ohm. If you connect the 1ohm, the
> voltmeter reads 15volts. How can that be ? In a series circuit each
> load has some resistance and the sum of all the loads must equal the
> source. So, what about the resistor ? It has to drop some voltage but
> according to this, it's dropping zero. And if you connect the 100
> ohm resistor, the voltmeter reads 1.88v. How did they come up with
> that number ? And the 5ohm reads 12v across the voltmeter. But
> according to Ohms law, E = IxR, if the resistor is 5ohms and the source
> is 15v, then 15/5 = 3amps, that part I get but they are saying it's
> dropping 3 volts ?? huh? Again, how did they get that value ?
>
>
> http://www.electronicsworkbench.com/understandelectricity/ewb.html
>
> TIA,
>
> J

Hi, J. Activity 1, Exercise 2 has an error which you noticed.

Here's how it should have worked:

The bulb is 12V 10 watts. That means .833 amps should be going through
the bulb when it's got 12V across it:

Power = Volts * Amps 10 watts = 12 volts * I amps I = 0.833

Using Ohm's law, you can then infer that the resistance of the hot bulb
is 14.4 ohms

R = V / I R = 12V / 0.833 amps R = 14.4 ohms

Now, if you connect a 15V battery with the 14,4 ohm resistor and the 1
ohm resistor in series:

Total resistance = 14.4 ohms + 1 ohm = 15.4 ohms

By Ohms Law:

I = 15V / 15.4 ohms = .974 amps

Which means the voltage across the 1 ohm resistor should be .974 volts

V = I * R V = .974 amps * 1 ohm = .974 volts

That leaves 14.026V for the light bulb. The point of the exercise is
that, unless you choose the correct series resistance, you'll put too
much voltage across the bulb, which might burn it out.

The funny thing is, the answer for the 5 ohm resistor is wrong, too. A
5 ohm resistor in series with the 14.4 ohm bulb would produce only 8.9V
across the bulb, by the same method shown above.

(The actual circuit is a little more complicated than this, because the
resistance of a light bulb is very dependent on temperature. If you
measure the cold resistance of a 10W automotive bulb, you'll find it's
more like 0.1 to 0.2 ohms. The filament resistance increases
dramatically as it heats up. Something to keep in mind as you learn
more about electronics.)

A couple of lessons here:

* If this website is trying to sell Electronics Workbench as a way to
get the answers, they could have done a better job.

* When it comes to any math on the web, trust but verify. There are
no editors here to catch stuff like this (and I provide no guarantees
about the math shown above, either. Check it yourself.)

* When you're learning something new, take advantage of the fact that
publishers of technical books hire editors to carefully check the copy
for obvious problems and errors. There are exceptions, and some real
howlers do occasionally get through, but that's generally true. If you
want a good really basic introduction to electronics, you coould do a
lot worse than "Getting Started in Electronics" by Forrest M. Mims III.
It's available at Amazon and some libraries. If you're serious about
learning, get a book, too.

* Welcome to newsgroups, J.

Good luck with your studies
Chris


Posted by Chris on March 16, 2006, 8:45 am
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Chris wrote:
<snip>
> The funny thing is, the answer for the 5 ohm resistor is wrong, too. A
> 5 ohm resistor in series with the 14.4 ohm bulb would produce only 8.9V
> across the bulb, by the same method shown above.
<snip>
Good luck with your studies
> Chris

And there it is. Mr. Fields was kind enough to work through the 5 ohm
series resistor, and lo and behold, I made an arithmetic error.

Trust, but verify ;-). Even the best-intentioned can go astray (and
thanks again, Mr. Fields)

Cheers
Chris


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