AC-circuit: Finding the current

Thread Starter

Niles

Joined Nov 23, 2008
56
Hi all.

Please take a look at the attachted circuit.

I have found the following expression for the complex current in the resistor (the hat indicates that it is complex):

\(\widehat_{I (t)} = \frac{{\omega ^2 LC {\widehat{U_0} }}}{{\omega ^2 LRC + i\omega L - R}}\exp ( - i\omega t),\)

where U_0 is the amplitude of U(t), and the hat indicates that it is complex.

Now I wish to find the real current, and I want to write it as:
\(I(t) = I_0\cos(\omega t + \phi).\)

But how do I do this? I have spent like 2 hours trying, but I don't
know how to rewrite the complex amplitude of the current to have a
phase.


Thanks in advance.



Regards

Niles.
 

Attachments

mik3

Joined Feb 4, 2008
4,843
A(cos(a)+isin(a))=Aexp(ia),

where
a is the angle in rads and
A is the amplitude

use this to convert your formula to a cos+isin version.

Or you can re-analyze the circuit but use jωL for the impedance of the inductor and 1/jωC for the impedance of the capacitor and analyze the circuit using the voltage and current divider equations, like you do with resistances.
 

Thread Starter

Niles

Joined Nov 23, 2008
56
Thanks for replying.

I have found the complex current, and it is now on the form:

I_complex (t) = 0.0015 * exp[-i(ωt-φ)],

where φ is the phase, and φ = 5.65.

Does this mean that the real current is given as:

I_real(t) = 0.0015 * cos[ωt+(-φ)],

where φ = 5.65?

Thanks in advance.
 

blazedaces

Joined Jul 24, 2008
130
How are you arriving at these constant values? Your circuit has only variables in it... so shouldn't your final answer contain only variables as well?

-blazed
 

blazedaces

Joined Jul 24, 2008
130
Alright. Then I just have one comment, just in case you forgot: you have an imaginary number in the denominator of your fraction. Just make sure to remove it by multiplying the top and bottom by the conjugate...

Cheers,

-blazed
 
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