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 ?
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.
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.
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.
--- 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.
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.
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