Question

An RLC circuit such as that of Figure (a) has R = 5.25 Ω, C = 25.1 μF, L = 1.47 H, and εm = 37.9 V. (a) At what angular frequency ωd 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 ωd1 and (d) higher angular frequency ωd2 will the current amplitude be half this maximum value? (e) What is (ωd2 - ωd1)/ωd, the fractional half-width of the resonance curve for this circuit?

R=1092 Current amplitude I Xc> XL X2 > Xc -R= 30 12 -R=1002 0.90 0.95 1.00 1.05 1.10 02/

Answer 1

In an R - L - C circuit, the resonance and maximum current amplitude occurs when

w = 1 / $\sqrt{LC}$

so,

w = 1 / sqrt ( 1.47 * 25.1e-6)

w = 164.6 rad/sec

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(b)

at resonance,

x_L = x_C

so,

Z = R

I = E_m / R

I = 37.9 / 5.25

I = 7.219 A

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(c) to find the lower and higher angular frequencies, we need to have

half current amplitude means E_m / 2R

so,

E_m / z = E_m / 2R

where z = sqrt ( R² + ( x_L - x_C ) ²)

cancel E_m on both sides,

1 / sqrt ( R² + ( x_L - x_C ) ²) = 1 / 2R

where

x_L = wL

and

x_C = 1 / wC

simplifying this, we get a quadratic polynomial equation

(LC) w² +/- ( 1.732 RC) w - 1 = 0

put in the values

3.689e-5 * w² + / - 2.2824e-4 * w - 1 = 0

solve this quadratic equation

w = - 2.2824e-4 + 0.01215 / 2 * 3.689e-5

wd1 = 161.6 rad/s (smaller value)

and

(d)

wd2 = 167.8 rad/sec (larger value)

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(e) fractional width = 167.8 - 161.6 / 164.6 = 0.0373