# Homework Help: Alternating current about resonance width

1. Nov 1, 2006

Since the question quite long and contains many formulas,
so i take a photo instead.

my question is that for part b
since $Z= \sqrt{R^2+(X_L-X_C)^2}$

can i subst $X_L = (W_0+\triangle{\omega})L$

and $X_C = \frac{1}{(w_0+\triangle{\omega})L}$

into the $Z= \sqrt{R^2+(X_L-X_C)^2}$

if it does, can u show me some of steps ?
I can derive the result from the problem in part b

By the way, what do the amplitude of current is half the resonance value?
is it $I=\frac{\omega}{2}$ ?

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Last edited: Nov 1, 2006
2. Nov 2, 2006

### Andrew Mason

I think you meant $X_C = \frac{1}{(w_0+\triangle{\omega})C}$

Yes. Work out the expression for Z by expanding:

$$Z = \sqrt{R^2+\left((W_0+\Delta{\omega})L -\frac{1}{(w_0+\Delta{\omega})C} \right)^2}$$

and ignoring the higher order terms of $\Delta\omega$
I am not sure I understand your question here. The amplitude of the current is V/Z. Z is not a linear function of $\omega$

AM

Last edited: Nov 2, 2006