Bad morning.

last night was state of origin (you could probably guess from my last blog post).

went out to the tavern, got plastered with about 80 other people while watching the game. have to say it was a pretty shocker game for both sides, but qld still won.

the try 6 minutes in by the blues was an absolute shocker on qld's part.

haynes passing the ball to idris about 3 meters over his head was abosulutely poor.

brilliant steal by thurston to thaiday.

so i woke up with probably one of my worst hangovers of the year, straight to uni, working on my electronics assignment thats due tomorrow.

pspice is annoying.

anyone want to tell me how to calculate the magnitude of a current as a function of frequency?

Comments

Veho
You need the transfer function of the element you're analysing. I need more info.
 
Cermage
[quote name='Veho' post='2866025' date='May 27 2010, 07:16 PM']You need the transfer function of the element you're analysing. I need more info.[/quote]

entire question is;

Consider the case where the source voltage can have an arbitrary frequency, i.e vs(t) = 10sin(2pi.f.t). Calculate the current, Im, as a function of frequency.

and then some plotting stuff.
 
Veho
It depends on the type of resistance/impedance you connect to the source (and the source's internal impedance but let's assume it's and ideal source). If the resistance is purely ohmic, then the current would be in phase with the voltage, and the amplitude wouldn't be affected by frequency. If the impendance is capacitive or inductive (or a combination), everything depends on the configuration. Again, need moar info. :mellow:
 
Veho
Well the capacitive impedance is 1/(2pifC), and the inductive is (2pifL), they're at 180 degrees, and the total impedance is

Zt2 = (ZL - ZC)2 + R2, and the equation to find the phase differential angle is tg(φ) = (ZL - ZC)/R.

And the current is V/Zt, at an angle of φ. Meaning,

I(t) = (10/Zt)sin(2pift - φ)


EDIT: Sorry, got the L and C the wrong way around.
 
Cermage
thats it? :blink:

man the lecuturer made it much more complicated than it was. thanks veho.
 
ProtoKun7
[quote name='Cermage' post='2866230' date='May 27 2010, 12:13 PM']thats it? :blink:

man the lecuturer made it much more complicated than it was. thanks veho.[/quote]
Some of the other slightly less uneducated members here might take that as quite a relative statement. :rofl:
 

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