Recent content by Themadflower

Okay i have a voltage supply of 5v and a Blue LED. How can i make estimations?

Okay Thats nice but is there a Way to Know the Working current ?

Oh i mean the Series resistance i need for My LED. i don't Know the maximum current

Well. Now i Know, and I am sry, i couldn't copy what I'm given. But i have another question, how can i determine the series resistance of My LED? Only with the Characteristic curve ?

But i don't have a function, only the Graph. Should i try to find one for the LED?

Homework Statement I have a LED Given: u-i-characteristic curve -> thus the wavelengh It's Part of a curcuit. Also (After the LED) There is a resistance of R (known) An a capacitor, (known) Homework Equations If i have a Voltage of U, how can i find the current Running through the LED and...

So dt is twice as high. Here dt2 is 8k

than thank you for your help! now i understood !

yes, does that mean it's not getting hotter at all? or can i simply use dT = I^2 * R and than see how the parameters are affected by cutting the conducter in half? than dt2 would be 8K

it will double

than m2 = m1 /2 . if i take Q=c*m*dT = I^2 *R -> dT = (I^2 *R)/(c*m) and dT2= (I^2 *R/2)/(c*m/2) which is the same as dT ...

R/2? could it be?

well, maybe i would it now is half as long as before so the temperatur rise will also be half dT1? that makes 2 K? and a total temperatur of 26? another idea would be to say, it is first: 20 + 1/5 *20 = 24 and now 24 + 1/5 * 1/2 *24 = 26,4

Hello ! 1. Homework Statement i study for my exams. here is my current problem: a non-specific conductor is connected to an ideal battery (surrounding temperature: 20°C) and reaches a temperature of 24 °C . after cut in half it is again connected to the battery. what temperature does it reach...