Question

The plate on the back of a personal computer says that the machine draws 3.1 A from a 120 V, 60 Hz line. For this PC, what are (a) the average current, (b) the average of the square of the current, and (c) the current amplitude.

Answer 1

a) 0 A

b)9.6 A

c)4.38 A

Step-by-step explanation:

Given that

`I_(r.m.s)=3.1\ A`

`V_(r.m.s)=60\ V`

We know that

`I_(r.m.s)=\sqrt{\int_(0)^(T)(1)/(T)I^2(t)\ dt}`

Lets take current in sinusoidal from ,so the average values of sinusoidal function will be zero.

So the average current = 0 A

`{\int_(0)^(T)(1)/(T)I^2(t)\ dt}\ is\ the\ average\ of\ the\ square\ of\ the\ current.`

Average of the square of the current

`Average\ of\ the\ square\ of\ the\ current\ = I_(r.m.s)^2`

Average of the square of the current = 9.6 A

The amplitude of current

`I_o=\sqrt 2\ I_(r.m.s)`

So now by putting the values

`I_o=\sqrt 2\ * 3.1\ A`

`I_o=4.38\ A`