Ohms law understanding

180

04:00:37

Electronics noob here, please don't be toxic. 

Why is this not giving me 18mA on the LED ( 18mA = 5v / 277ohm)?

EDIT: Fixed thanks to the comments :) . 

published 6 years ago

kiani

6 years ago

No you have 2 volt drop across the led.. So put a voltmeter across the 277 ohms ti see what voltage you have across it!? .... Vr = I * r... 3v = I * 277.so. I = 3/277...

Electronics3038

6 years ago

Series circuit current is the same, voltage drop different. ,😉

kiani

6 years ago

I think @elect..3. He just didn't resluse there is a 2v. Volt drop across the Led.. Csuse he is dividing the whole supply voltage by the restlistor vslue. Ti get the current.

DevDorian

6 years ago

For context I am trying to apply what I have learned for choosing a resistor for simply lighting an LED. From a tutorial on Sparkfun. Wanted to see the math in action here to where I know I am giving the LED 18mA. I'm confused now :\ . I expected a voltage drop from the resistor to give me 18mA but instead it changed my 5v to 1.94v. I have more to learn about this.

hurz

6 years ago

but no reason to be confused if you read what @kiani calculated for you. Ask him if you still have a question.

kiani

6 years ago

Think of the Led as another resistor., so the supply voltage (5 volts) divides..some across the resustor (277 ohms) and dome acriss the led.. When we tslk about v=ir. We talk about the voltage directly across the resistor.,, and the current through it....

kiani

6 years ago

Chek on this link..... http://everycircuit.com/circuit/6494515784056832

kiani

6 years ago

Therr was never an 18 mA. Cause you didn't take into account the led resistsnce..

crake

6 years ago

(5-2)V÷0.018A = 166.7ohm

crake

6 years ago

you have to include to "on" voltage of the led in your calculation. 5 volt source minus 2 volt "on" voltage drop across led. using ohms law you can now calculate the resistance required for 18mA to flow through your series circuit.

DevDorian

6 years ago

I see, that tutorial didn't account for the LED drop voltage (https://learn.sparkfun.com/tutorials/voltage-current-resistance-and-ohms-law/an-ohms-law-experiment). I worked out the same formula you posted and now I understand. Thanks :) .

DevDorian

6 years ago

Thank you to everyone

kiani

6 years ago

Lookedcat this spark fun. There are much better sites...much better...

selman

6 years ago

Tips. Voltage is across while current is through, Voltage is the force while current is the movement, Voltage is "potential difference" and always measured as a " difference between two wires, flow of conventional current is oppsite to the flow of electrons, if current is going from more posetive terminal of a component to the negative terminal then that component is consuming power and vice versa...

selman

6 years ago

"Fundamental of Electrical Engineering" by Alexander and Sadiku... great book

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