# Sinusoidal alternating current/Homework

1. Feb 17, 2015

### ThisVelvetGlove

Hello everyone thanks for giving me your support, sorry if i post in the wrong section.

1. The problem statement, all variables and given/known data

In the picture we can see the graph of sinusoidal i1=i1(t) and i2=i2(t)
ν(as i read, it said only one so that's how i went beyond on resolving ecuations)=?
φ0=?
i1(t1)=?
i2(t2)=?
φ1=?
φ2=?
i hope i translate good, phase shifts φ1221 of i2 behind and before of i1
Given data:
t1=2,5×10-3(s)
t2=5×10-3(s)
T2=0,02
Imax1=1,5(A)
Imax2=1
http://postimg.org/image/jejwzmzn7/ [Broken] (this is the graph)
The attempt at a solution
This is what i did i got stuck at phase shifts.

i(t)=Imaxsin(ωt+φ)

1.i(t1)=Imax1sin(ωt11)
2.i(t2)=Imax2sin(ω22)
ω=2π/T or 2πv
ν=1/T=> ν=1/T2=>ν=1/2×10-2(s)-1
from here => i(t) becomes
1.i(t1)=Imax1sin(2π/T1t11)
2.i(t2)=Imax2sin(2π/T2t22)
from here => φ2 i got it out like this:
Δt.........Δφ
T...........2π
=> Δφ=(Δt×2π)/T =>2 φ20=(t2-t1×2π)/T
=> φ2=(5×10-3-2,5×10-3×2π)/T
=>φ2=(2,5×10-3×2π)/1/2×10-2
=>φ2=5×10-5×2π
so
2.i(t2)=Imax2sin(2π/T2t22)
i(t2)=sin(2π×1/2×10-2×5×10-3+5×10-5×2π)
after this i said that
T1 =t2-t1+T2
=> (2,5×10-3+2×10-2
=>(4,5×10-5)
which got me confused cause if this is different from the other one(T2).... then shouldn't it have to v?

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Last edited by a moderator: May 7, 2017
2. Feb 17, 2015

### lightgrav

If they want 2 distinct φ values, that means one for each curve ... not the difference from one to the other.
(that is, the angle that the sine is offset ... i2 looks like a cosine, so make sure your units are correct)
They both have the same T, and hence the same ω and same ν (or f) .
the initial time offset does not relate to their repeat time periods.