Hobby Electronics Basics Newbie needs help with basics

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Newbie needs help with basics hdjim69 03-16-06
Posted by Phil Allison on March 16, 2006, 5:53 pm
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"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



Posted by Chris on March 16, 2006, 6:15 pm
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Phil Allison wrote:
> ** 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

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


Posted by John Fields on March 16, 2006, 8:40 am
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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?

---
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.


--
John Fields
Professional Circuit Designer

Posted by John Fields on March 16, 2006, 9:31 am
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On Thu, 16 Mar 2006 07:40:31 -0600, 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

>resistance is 14.4 ohms, the voltage the meter _should_ be reading
>is:
>
> E = IR = 0.974A * 14.4R ~ 14.0V
>

--
John Fields
Professional Circuit Designer

Posted by ehsjr on March 17, 2006, 11:59 pm
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John Fields wrote:
> 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?
>
>
> ---
> 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

You have to look closely at the bulb. :-(
They show it burned out. Ed

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