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Newbie needs help with basics

R

redbelly

Dorian said:
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

The real question is, why don't they just provide the correct 3.6 ohms
as one of the choices?

Mark
 
H

hdjim69

So to summarize what I have learned:

1) Find total resistance in curcuit
1a) use formula R=E^2/P if R is unknown but P & E are known
2) Find (total) current in circuit by using RT and VT
3) Find voltage drop of each load and ensure it's adequate for each
load

Excellent ! I learned a lot from this post !!

One small question...why do we square the voltage in the formula
R=E^2/P ? Why can't we just use the actual voltage ?

TIA
 
R

redbelly

hdjim69 said:
One small question...why do we square the voltage in the formula
R=E^2/P ? Why can't we just use the actual voltage ?

TIA

Think of these two equations:

P = E*I
and
I = E/R

Let's assume you believe those.

Combine them and you get

P = E * E/R or E^2 / R

So we have to use E^2, not just E.

Mark
 
H

hdjim69

Ahh ! we're breaking out the I into it's component parts. I see.
great, thank you !
 
D

Dorian McIntire

I agree that 3.6 ohms should have been a choice or things should have worked
out so that 5 ohms was an exact answer. As an adjunct electronics instructor
I can tell you that these kind of poorly thought out problems are the best
way to confuse students. You need to build up the students confidence using
ohms law formulas before you present them with problems like this one. The
point may have been to get students to think in generalities but they are
more confusing than helpful.

Dorian
 
J

John Fields

Ahh ! we're breaking out the I into it's component parts. I see.
great, thank you !

---
If you'd be so kind as to leave a little of the post you're replying
to in your reply, then we all could see to whom and to what you're
responding.


Guidelines from:

http://groups.google.com/support/bin/answer.py?answer=12348&topic=250

"Summarize what you're following up.

When you click "Reply" under "show options" to follow up an existing
article,
Google Groups includes the full article in quotes, with the cursor
at the top
of the article. Tempting though it is to just start typing your
message,
please STOP and do two things first. Look at the quoted text and
remove parts
that are irrelevant. Then, go to the BOTTOM of the article and start
typing there.
Doing this makes it much easier for your readers to get through your
post.
They'll have a reminder of the relevant text before your comment,
but won't have to re-read the entire article. And if your reply
appears on a site
before the original article does, they'll get the gist of what
you're talking about."
 
E

ehsjr

John said:
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
 
J

John Fields

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

---
LOL, yup, you're right!

I also missed the bit on the bottom of the page where the bulb makes
reference to being burned out.

Thanks,
 
H

hdjim69

If you'd be so kind as to leave a little of the post you're replying
to in your reply, then we all could see to whom and to what you're
responding.


will do !

j
 
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