Chapter 31, Problem 047 An RLC crcuit such as that of Figure a has R-4.75 C = 20

Question

Chapter 31, Problem 047 An RLC crcuit such as that of Figure a has R-4.75 C = 20 3 L = 1.14 H, and 39 v. a At what angular equency u will the current amplitude have its maximum value, as in he resonance curves o Figure b ? b) What is this maximum value? At what c ow e angular requency t and d h a rangu a equency a will the current amplitude be half this maximum value? (e) What is (ad2 - wd/od, the fractional half-width of the resonance curve for this circuit? 8t 1-10 R=30 R> 100 0.900.95 1.00 1.05 1.10

Explanation / Answer

a) resonant frequency, wd = 1/sqrt(L*C)

= 1/sqrt(1.14*20.3*10^-6)

= 208 rad/s

b) Imax = E_max/R

= 39.7/4.75

= 8.36 A

c) let w is the angular frenquency where current becomes half of the maximum.

impedance of the ckt,

z = sqrt(R^2 + (XL - Xc)^2)

= sqrt(R^2 + (w*L - 1/(w*c))^2

= sqrt(4.75^2 + (w*1.14 - 1/(w*20.3*10^-6))^2)

Imax/2 = Emax/z

8.36/2 = 39.7/sqrt(4.75^2 + (w*1.14 - 1/(w*20.3*10^-6))^2)

on solving the above equation we get

==> wd1 = 204 rad/s

d) wd2 = 212 rad/s

e) wd2 - wd1 = 212 - 204

= 8 rad/s

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