Newbie needs help with basics

[hdjim69]

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

 

[Phil Allison]

"hdjim69"

http://www.electronicsworkbench.com/understandelectricity/ewb.html

** Which EW **activity** in the above URL are you FUCKING on about
!!!!!!!!


DO NOT POST AMBIGUOUS QUESTIONS !!!






........ Phil

 

[hdjim69]

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

 

[Greg Neill]

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 ?

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

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.

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

And if you connect the 100
ohm resistor, the voltmeter reads 1.88v. How did they come up with
that number ?

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.

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 ?

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.

 

[hdjim69]

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

 

[Phil Allison]

Hey Phil, chill out !!

** WELL FUUUCK YOU !!!

Mr Googlegroups, anonymous "hotmail" SHITHEAD !!!!

SCUM of THE EARTH TROLL !!!

The link should have taken right to the
problem.

** Would anyone complain if it FUCKING DID ??

YOU fucking ASSHOLE !!





........ Phil

 

[Phil Allison]

Hey Phil, chill out !!

** WELL FUUUCK YOU !!!

Mr Googlegroups, anonymous "hotmail" SHITHEAD !!!!

SCUM of THE EARTH TROLL !!!

The link should have taken right to the
problem.

** Would anyone complain if it FUCKING DID ??

YOU fucking ASSHOLE !!



........ Phil

 

[Chris]

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

 

[John Fields]

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?

---
See below.
---

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,

---
If the lamp weren't in the circuit you'd be right, but since it is
and you neglected to add its resistace into the string, that's where
you got 3 amps, which is wrong. See below.
---

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

---
Electronics workbench is a POS simulator, and this "tutorial" seems
to follow suit.

Looking at the first example you cite, the circuit looks like this:


+15V
|
+<-------+
| |
[LAMP] [METER]
| |
+<-------+
|
[1R]
|
GND

Calculating the resistance of the lamp from:

E²
P = ---
R

we rearrange and solve for R:

E² 12V²
R = --- = ----- = 14.4 ohms
P 10W

Now, since the total resistance in the circuit is that of the lamp
in series with the 1 ohm resistor, and series resistances add, it's:


Rt = 14.4R + 1R = 15.4R


therefore, the current through the string must be:


E 15V
I = ---- = ------- = 0.974 amperes
Rt 15.4R

Finally, since there's 1A passing through the lamp, and its
resistance is 14.4 ohms, the voltage the meter _should_ be reading
is:

E = IR = 0.974A * 14.4R ~ 14.0V

So, since it's a 12V lamp it'll light up. But too brightly, so it'll
have a reduced life.


In your second example, since we already know the resistance of the
lamp, the circuit will look like this:


+15V
|
+<-------+
| |
[14.4R] [METER]
| |
+<-------+
|
[100R]
|
GND


So, the current in the circuit will be:


E 15V
I = ---- = ------- ~ 0.131 amperes
Rt 114.4R


and the drop across the lamp:


E = IR = 0.131A * 14.4R ~ 1.887V

Therefore, the lamp won't light.


In your third example, since we already know the resistance of the
lamp, the circuit will look like this:


+15V
|
+<-------+
| |
[14.4R] [METER]
| |
+<-------+
|
[5R]
|
GND


So, the current in the circuit will be:


E 15V
I = ---- = ------- ~ 0.773 amperes
Rt 19.4R


and the drop across the lamp will be:


E = IR = 0.773A * 14.4R ~ 11.13V


So the lamp will light, albeit not as brightly as it would with 12V
across it, but it'll last longer.


In this exercise the resistance of the lamp filament has been
assumed to be constant regardless of the voltage across it, but such
isn't the case. What really happens is that as the filament heats
up its resistance increases, and its value changes about an order of
magnitude from its cold value to its hot value. But... that's
probably best left alone 'til later.

 

[Chris]

Chris wrote:

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.

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

 

[hdjim69]

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

 

[John Fields]

therefore, the current through the string must be:


E 15V
I = ---- = ------- = 0.974 amperes
Rt 15.4R

Finally, since there's 1A passing through the lamp, and its

^^
Oops... 0.974A

 

[John Fields]

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.
---

 

[hdjim69]

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

 

[Phil Allison]

"Chris"

(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.

** Nonsense.

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.

The filament resistance increases
dramatically as it heats up.

** By a factor of 9 or 10.

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




.......... Phil

 

[Chris]

** Nonsense.

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.




** By a factor of 9 or 10.

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

 

[Dorian McIntire]

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 ?

Consider:



10 W (Watts) / 12 V (Volts) = 0.833 A (Amps) required by the bulb to light
without burning out or burning to dimly.



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



3 V / 0.833 A = 3.6 Ohms required to drop 3 V at the required 0.833 A.



The closet resistor to this value is the 5 Ohm resistor



OR



3 V / 1 Ohm = 3 A - too high



3 V/ 100 Ohms = 30 MA - too low



3 V / 5 Ohms = .6 A - about right

 

[Phil Allison]

"James Douglas"

Jeesh, go to your doctor and get the pine cone out of your ass!

** **** OFF - IDIOT !!

The OP was an arrogant moron.

YOU are even WORSE !




.......... Phil

 

[James Douglas]

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

 

[Greg Neill]

"James Douglas"



** **** OFF - IDIOT !!

The OP was an arrogant moron.

YOU are even WORSE !

Bye Bye potty mouth:

*PLONK* *FLUSH*