An RLC circuit such as that of Figure (a) has R = 4.75 , C = 20.0 F, L = 1.10 H,

Question

An RLC circuit such as that of Figure (a) has R = 4.75 , C = 20.0 F, L = 1.10 H, and m = 25.1 V. (a) At what angular frequency d will the current amplitude have its maximum value, as in the resonance curves of Figure (b)? (b) What is this maximum value? At what (c) lower angular frequency d1 and (d) higher angular frequency d2 will the current amplitude be half this maximum value? (e) What is (d2 - d1)/d, the fractional half-width of the resonance curve for this circuit?

Explanation / Answer

Xc=1/WC

Xl=WL

impedence of ciruit =R+(Xc-Xl)j

Current in max when impedence is min

=> Xc=Xl

=> Wd=1/sqrt(LC) = 213.20

b)

Current, Max = 25.1/R = 5.28 A

c)

For half its magnitude :

impedence=twice at peak

=> impedence= sqrt(R^2+(Xc-Xl)^2)

=> R^2+(wL-1/wC)^2 = 2*4.75^2

22.56 + (w*1.10 - 50000/w)^2 = 45.125

(w*1.10 - 50000/w)^2 = 22.56

=>

w*1.10 - 50000/w = 4.75

this gives w = 215.37 , -211

c)

lower freq = -211

d)

higher = 215.37

e)

(215.37 - 211)/213.20 = 0.0204

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