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AC Sine wave equation help.

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kiddog

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Hell, I'm having a problem working out a few questions....

v = 120 sin (50πt + 1.1) V

I need to work out frequency, amplitude, relative phase, value of v when t is 20ms, if this voltage is applied to a resistance of 100ohms what would be the equation for the instantaneous current?

I think the frequency is 50 and the amplitude is 120?

Thanks for any help in advance
 

2(pi)f = 50(pi)
f = 25Hz

Instantaneous current = v(t)/R = v(t)/100
 
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    kiddog

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v = 120 sin (50πt + 1.1) , at T= 0, the angle of the sinewave is 1.1/2PI X 360 degs. 20 mS later its the same ( 1/F). Relative phase to what? As I = V/100, I =120/100 sin (50πt + 1.1)
Frank
 
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Thanks a lot for your help, very much appreciated.... regarding relative phase the question just asks... what is the relative phase?

- - - Updated - - -

Am I right in saying the relative phase is 1.1 radians?
 

In general you can write a sinewave as:

v(t)=A•sin(ω•t+φ)

Where A is the amplitude, that means v(t) will go from -A to +A
ω is the angular frequency that can be defined as ω=2•Π•f
φ is the initial phase that is the phase of the wave when t=0

if "f" is in Hz and "t" in seconds, then "ω" will be radians and then also "φ" has to be in radians

It could be with "relative phase" you meant "initial phase".

In your case:
V(0)=120*sin(1.1)=106.94
V(20ms)=120*sin(50•Π•0.02+1.1)=-106.94

this last because f is 25 Hz and not 50 Hz (small oversight of chuckey)
 
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Thanks a lot for the help. makes sense to me now!
 

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