Simply the following expressions using phasors a i1t 20 cos

Simply the following expressions using phasors: (a) i_1(t) = 20 cos (omega t - 30degree) + 16 cos (omega t + 15degree) A (b) i_2(t) = 14 sin(omega t + 45degree) - 17 cos (omega t + 60degree) A (c) i_3(t) = 2 cos(5t) - 7 sin(5t) A

Solution

i1(t)=20.cos(wt-30)+16cos(wt+15)

i1(t)=20<-30+16<15

i1(t)=33.294<-10.135 amps

i1(t)=33.294cos(wt-10.135) amps

i2(t)=14sin(wt+45)-17cos(wt+60) amps

i2(t)=14cos(90-wt-45)-17cos(wt+60)

i2(t)=14cos(45-wt)-17cos(wt+60)

i2(t)=14cos(wt-45)-17coss(wt+60)

i2(t)=14<-45-17<60

i2=24.66<-86.746 amps

i2(t)=24.66cos(wt-86.746) amps

i3(t)=2cos(5t)-7sin(5t) amps

i3(t)=2sin(90-5t)-7sin(5t)

i3(t)=-2sin(5t-90)-7sin(5t)

i3(t)=-2<-90-7<0

i3(t)=7.28<164.054 amps

i3(t)=7.28sin(5t+164.054) amps.