A.C. Circuits: Graphs & Voltage/Current Curves

[uzair_ha91]

A.C. Circuits -- Graphs

http://img136.imageshack.us/img136/5782/image64.jpg
In this graph, the curve of current is larger than that of voltage...but in some books the voltage curve is above the current curve (like the one below)...Which one's correct?

 

[waht]

It's arbitrary. You are shown a graph with two curves of different units.

 

[uzair_ha91]

So basically it doesn't matter if you drew both curves of same height, right?

 

[waht]

So basically it doesn't matter if you drew both curves of same height, right?

It doesn't matter, you can scale the y-axis to any size.

 

[Bob S]

For capacitors, The ratio of voltage to current at a frequency f (where f = w/(2 pi) where w is the frequency in radians per second), is

V/I = -j/wC (where the j indicates a 90 degree phase shift). For an inductance:

V/I = +jwL

So for any given C or L, the ratio V/I depends on frequency.

 

[uzair_ha91]

For capacitors, The ratio of voltage to current at a frequency f (where f = w/(2 pi) where w is the frequency in radians per second), is

V/I = -j/wC (where the j indicates a 90 degree phase shift). For an inductance:

V/I = +jwL

So for any given C or L, the ratio V/I depends on frequency.

I'm sorry I don't understand what you are trying to explain..

 

[Bob S]

Hi Usair-

Suppose I have a capacitor C and I put a voltage V on it at frequency f.

The current in it is given by I = (2 pi f) C V. But the phase of the current is shifted by 90 degrees. So we use the short-hand notoation j, so that

I = j (2 pi f C) V

This is because of the equation for charge on a capacitor is

Q = C V

Now suppose V = V₀ sin(wt) where w = 2 pi f

Then I = dQ/dt = C dV/dt = w C V₀ cos(wt)

So there is a 90 degree phase shift (between the sine and cosine), and the amplitude of I depends on frequency f, as well as on C and V₀.

I hope this helps

Bob S