Question

Printr-un conductor străbătut de un curent electric cu intensitatea de 0,32A trece într-un minut un număr de electroni egal cu (e=1,6⋅10−19 C):
A) 1,2⋅ $10^{20}$
B) 3⋅ $10^{20}$
C) 2⋅ $10^{20}$
D) 4⋅ $10^{19}$
E) 4⋅ $10^{20}$
F) 1,2⋅ $10^{19}$

Answer 1

t=1 min=60 s
Q=It=0,32*60=19,2 C
Q=Ne=>N=Q/e=19,2/1,6*10⁻¹⁹=12*10¹⁹ =1,2*10²⁰ electroni 
Varianta corecta este A