Recent content by curiosissimo

Ok I got it, thanks :)

Thank you all

Ah I got it. I wasn't considering that the current was entering the circuit "outiside". What about there's no current entering from outside? I mean, what if I have to find the total current in the circuit with the same problem data (except for i=20)?

I mean, the generators are oriented in the same way, so I1 and I2 go in the sale direction

Why I1+I2 = 0 if I2 is clockwise? I miss this point

I get the solution if the verse of I2 is counterclockwise, but why? It should go clockwise following the negative pole.

I1 flows clockwise as well as I2. In fact I2 can't go counterclockwise from O to N, because after O we have V2 and at first there is the positive pole. But I2 follows the negative one.

The main problem is the verse of the 2 currents ## i1 ## and ## i2 ## . I think they both go clockwise, so in the node A we should have ##i1+i2=20A##. Now let's apply the second law (going clockwise). So ##ΔV1-R1*i3 - R2*i4 + ΔV2=0##, where ##i3## and ##i4## are the currents which pass through...

Of course! What a silly mistake! Thank you very much!

So for the Gaussian theorem we know that $$ \frac{Q}{e} = \vec E \cdot \vec S $$ Q's value is known so we don't need to express it as $$Q=(4/3)\pi*(R_2 ^3-R_1 ^3)*d$$ where d is the density of the charge in the volume. I've expressed the surface $$S=4\pi*x^2$$ where x is the distance of a point...

Hello everyone! I'm curiosissimo, I'm 18 and I'm an Italian high school student. In 2020/2021 I'll attend my last high school class. I'm here because some physics problems are weird and if I make a mistake, I usually waste a lot of time finding out it. But here I'll understand everything better...