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

David De wrote:

> 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. 1kΩ
> b. 500Ω
> c. 350Ω
> d. 100Ω

To answer this you need to know the voltage across the resistor. From the
equation R=V/I if you know any two, you can calculate the third.

The voltage across the resistor will be the supply voltage less the LED's
forward voltage = 5 - 1.5V = 3.5V

See Kirchoff's Laws for an analysis of this.
http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

The resistor value can now be calculated from R=V/I = 3.5/0.01 = 350 (ohms).

Graham

thank you for all the help.

On Aug 14, 11:09 am, "Anonymous." <m...@privacy.net> wrote:
> "David De" <netr...@yahoo.com> wrote in message
>
> news:1187103198.637442.98290@w3g2000hsg.googlegroups.com...
> 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. 1k?
> b. 500?
> c. 350?
> d. 100?
>
> -----ooooo-----
>
> The answer is c.350. I did it in my head...it's the sort of mental
> arithmetic that we were taught to do in Junior (Pre age 11) school
> in Britland.
>
> This is the way it works.....If there is 1.5V drop across the LED, then
> there must be 5 -1.5 = 3.5V across the resistor and at 10mA
> with R=V/I that is a resistor of...
>
> 3.5/0.01 = 350.
>
> (Where 10mA is 0.01A)

"David De" <netrate@yahoo.com> wrote in message
news:1187105701.207646.279370@g4g2000hsf.googlegroups.com...
> thank you for all the help.
>

You got your answer but did you understand the math and theory that was
applied?

Tom

David De wrote:

> 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. 1kΩ
> b. 500Ω
> c. 350Ω
> d. 100Ω
>
I don't understand what the crap above is?, maybe my reader didn't
decode it properly?
But for the most part.
R = (5-1.5)/0.010 = 350

and so on and so on..

Ok, Now do the rest of your home work.


--
"I'm never wrong, once i thought i was, but was mistaken"
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5