Respuesta :
Answer:
a) 0 A
b)9.6 A
c)4.38 A
Explanation:
Given that
[tex]I_{r.m.s}=3.1\ A[/tex]
[tex]V_{r.m.s}=60\ V[/tex]
We know that
[tex]I_{r.m.s}=\sqrt{\int_{0}^{T}\dfrac{1}{T}I^2(t)\ dt}[/tex]
Lets take current in sinusoidal from ,so the average values of sinusoidal function will be zero.
So the average current = 0 A
[tex]{\int_{0}^{T}\dfrac{1}{T}I^2(t)\ dt}\ is\ the\ average\ of\ the\ square\ of\ the\ current.[/tex]
Average of the square of the current
[tex]Average\ of\ the\ square\ of\ the\ current\ = I_{r.m.s}^2 [/tex]
Average of the square of the current = 9.6 A
The amplitude of current
[tex]I_o=\sqrt 2\ I_{r.m.s}[/tex]
So now by putting the values
[tex]I_o=\sqrt 2\ \times 3.1\ A[/tex]
[tex]I_o=4.38\ A[/tex]
