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
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.
The resistor value can now be calculated from R=V/I = 3.5/0.01 = 350 (ohms).
Graham
e. do your own homework.
Hint: V = I * R
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)
You got your answer but did you understand the math and theory that was applied?
Tom
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
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