Question

An alternating circuit that has peak voltage of 38.01 V and a frequency of 60 Hz powers a device with a resistance of 48 ohms.
(a) What is the current at 0.03 s?
(b) How much energy does the device use in one hour?

Answer 1

a)

I = Io sin (wt)

I = (38.01/ 48) sin ( 2 * 3.14* 60* 0.03)

I = 0.155 A

b)

E = Vrms^2* t / R

E = (38.01/ sqrt(2)) * 60*60 / 48

E = 2015.785 J

Comment in case any doubt please rate my answer....