Alternating current about resonance width

[adrian116]

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} ?

 

[Andrew Mason]

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}

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

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

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