# AC-circuit: Finding the current

Discussion in 'Homework Help' started by Niles, Nov 23, 2008.

1. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
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.

Regards

Niles.

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2. ### mik3 Senior Member

Feb 4, 2008
4,846
65
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.

3. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0

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?

4. ### blazedaces Active Member

Jul 24, 2008
130
0
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

5. ### Niles Thread Starter Active Member

Nov 23, 2008
56
0
I am given some numerical values that I've plugged in.

6. ### blazedaces Active Member

Jul 24, 2008
130
0
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