Newbie needs help with basics

"hdjim69"

DO NOT POST AMBIGUOUS QUESTIONS !!!

....... Phil

Hey Phil, chill out !! The link should have taken right to the problem. It's Activity 1 exercise 2.

If the current is too high the bulb will burn out and the full supply voltage will appear across the failed (opened) bulb.

When the bulb is burnt out, no current will flow so there will be no voltage drop across the resistor.

If the lamp has a rating of 10W at 12V, you can calculate a resistance for the bulb from P = V^2/R. You then have an equivalent circuit consisting of two resistors in series. Note that real light bulbs have different resistance for different operating points (resistance depends upon the temperature of the filament). In this simplified example, they're probably ignoring this little detail.

You forgot to include the resistance of the bulb, or alternativly, if you assume that the bulb is operating at its correct operating point and dropping 12V at 0.833A (for a power dissipation of 10W), that leaves (15 - 12)V = 3V to drop across the resistor.

Hey Phil, chill out !! The link should have taken right to the problem. It's Activity 1 exercise 2.

Mr Googlegroups, anonymous "hotmail" S*****AD !!!!

SCUM of THE EARTH TROLL !!!

YOU f****ng ASSHOLE !!

....... Phil

Mr Googlegroups, anonymous "hotmail" S*****AD !!!!

SCUM of THE EARTH TROLL !!!

YOU f****ng ASSHOLE !!

....... Phil

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

(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

Good luck with your studies

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

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

^^ Oops... 0.974A

bulb from P = V^2/R.

--- 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

---

--- 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.

---

--- 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.

---

--- No, those will come out later.

---

--- See above.

-- John Fields Professional Circuit Designer

Thanks so much guys ! I think I can take it from here.

"Chris"

The cold resistance of a tungstsn filament bulb is close to 10% of its hot value.

For the bulb in question, this means about 1.4 ohms.

Bulbs that run exceptionally bright ( ie halogen projector lamps) can reach a ratio of 15.

......... Phil

You're right, of course. And it's already evident I needed another cup of coffee this morning when I posted.

Thanks again for the spot, sir.

Cheers Chris

Consider:

The applied voltage is 15 V so 3 V must be dropped by the current limiting resistor to light the lamp properly.

The closet resistor to this value is the 5 Ohm resistor

OR

3 V/ 100 Ohms = 30 MA - too low

"James Douglas"

The OP was an arrogant moron.

YOU are even WORSE !

......... Phil

Jeesh, go to your doctor and get the pine cone out of your ass! BTW, I read various newsgroups that relate to my hobbies and I think your dad, or maybe your bother, is in alt.design.grapics where his every reponse is angry and seems to be written by a high schooler that has spend 5-6 years there.

I would rather see 100+ simple questions than one of your posts, like you mom probably said, or maybe not, if you have nothing good to say, don't say anything at all.

To the original poster, sorry about this and simply ignore, there are plenty of good folks both here and other news groups.

JimD

Bye Bye potty mouth: