Answer :
Answer:
d)50.4 C
Explanation:
We are given that
[tex]I(t)=6A+(4.8 A/s)t[/tex]
Where t (in s)
We have to find the charge pass a cross section of the wire in the time period t=0 s and t=3.5 s
We know that
[tex]dq=Idt[/tex]
Integrating on both sides
[tex]q=\int_{0}_{3.5} (6+4.8 t)dt[/tex]
[tex]q=[6t+2.4t^2]^{3.5}_{0}[/tex]
[tex]q=6(3.5)+2.4(3.5)^2=50.4 C[/tex]
Hence, option d is true.