Need help with a Resistor calculation please

I have this question and I am wondering if someone can tell me the formula and how they derived the answer (steps) - it is probably simple, but I am learning this as I go :

Electronic

1) A simple circuit consists of a current limiting resistor in series with an LED across a supply voltage of 5VDC. The forward voltage drop is 1.5VDC and the forward current is 10mA. The series resistor would have a value of: a=2E 1k=CE=A9 b=2E 500=CE=A9 c=2E 350=CE=A9 d=2E 100=CE=A9
Reply to
David D
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This is a newsgroup about the repair of electronic equipment.

In what way is your basic question related to repair?

If you'd looked at Mark Zenier's guide to the sci.electronics.* hierarchy: ftp://ftp.eskimo.com/u/m/mzenier/seguide9706.txt you'd likely catch on that this isn't the place to ask such things, while sci.electronics.basics is.

And of course, this is an example of the damage google is doing to the newsgroups. It makes it too easy for people to blunder into the wrong newsgroup.

Michael

Reply to
Michael Black

I regret posting it on here. I see you are passionate about this group, I was hoping I could get an answer quickly and this group has always been good about replying.

Reply to
David D

Electronic

1) A simple circuit consists of a current limiting resistor in series with an LED across a supply voltage of 5VDC. The forward voltage drop is 1.5VDC and the forward current is 10mA. The series resistor would have a value of: a. 1k? b. 500? c. 350? d. 100?

OK you need to conduct 10ma through a resistor with 5 - 1.5 = 3.5 volts across it. Since R = 3.5/.01, you can take it from there. Is this a homework assignment? If so and you can not figure this out, switch majors.

David

Reply to
David

V=3DIR. If you think folk are gonna do your homework for you, get serious.

FWIW the q's a bit ambiguous

NT

Reply to
meow2222

Pay no attention to Michael. There are plenty of people here who would answer your questions and not attempt to net-nanny you.

Reply to
Meat Plow

The formula for calculating the series resistor value is:-

(Vs-Vf) R = ------- I

Where R is the resistance in ohms Vs the supply volts Vf the forward voltage drop I the forward current in amps.

--
*The first rule of holes:  If you are in one, stop digging!

    Dave Plowman        dave@davenoise.co.uk           London SW
                  To e-mail, change noise into sound.
Reply to
Dave Plowman (News)

This is not kosher.

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Do it this way if you MUST post it multiple times.

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Reply to
JeffM

Isn't Michael engaged to Malissa Baldwin?

Peter Wieck Wyncote, PA

Reply to
pfjw

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Wait a minute. I took him to task for posting a basic question in a repair newsgroup. I think the post in .basics was a result of that. Certainly I saw it there after I had posted here (and I check .basics before .repair).

The off-topic posts here are becoming as common as the spam.

(And I gave up on .design months ago)

Michael

Reply to
Michael Black

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Much as you would like it otherwise, life is not black and white. It is full of grey areas, and Usenet newsgroups are one such area. Just because I am a radio ham, it doesn't mean that I have to talk to other radio hams just about radio topics. Sometimes, we talk about motor cycles. Or sewage works. Or the grand Canyon. Such is the way it is also on here. Intelligent like-minded engineers, sometimes talk among themselves, in one thread amongst many, about things of mutual interest, which may or may not be electronics related, and often, even if they are electronic, not related directly to repair. And sometimes, we all help those who either blunder in here with a slightly off-topic question, or come here deliberately because they know that they will get help, as this poster did. Why do you always have such a cry about it ? It is not a big problem on this group is it ? Or perhaps you *do* see it as one ? I just see it as a bunch of Usenet 'friends' acting responsibly, and trying to extend that frienship to others, who might subsequently become regular and valued contributors ...

Honestly, Michael, lighten up a bit.

Arfa

Reply to
Arfa Daily

I second Arfa's comments. If the OP comes back repeatedly, that's the time to make suggestions as to where to go, otherwise, lighten up.

H. R. (Bob) Hofmann

Reply to
hrhofmann

Thirded. A far greater problem has been all the spam emanating from Google. Off topic posts are no big deal, and at least the OP's question was electronics related.

Reply to
JW

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