Question

IP Tuning a Radio A radio tuning circuit contains an RLCRLC circuit with RRR = 5.5 ΩΩ and LLL = 3.0 μHμH .

1.) What capacitance is needed to produce a resonance frequency of 82 MHz ? Please answer in pF units.

2.) If the capacitance is increased above the value found in part A, will the impedance increase, decrease, or stay the same?

3.) Find the impedance of the circuit at resonance.

4.) Find the impedance of the circuit when the capacitance is 1% higher than the value found in part A.

Answer 1

Givens RCL circuit Resistance (R) = 5.51 Inductance (L) = 344= 3x10-6H Capacitance (C) = ? Resonance frequency (f)= 82 MHz = 82x100 Hz We know that 0 (W)= 2.7f Angular frequency = 2x1T X 02 x 106 W = 515.221 x 106 zaalls. Capacitive Reactance (Xc) = I Cدی · Inductive Reactance (XL)= WL lo‘x 3x10-6 515.221 x 10 x XL = 1545-664 si (a) find capacitance () at Resonance frequency- In case of Resonance, XL = Xc WL = -18 = co²L W²L - (515:221x10)% 3x10-6 C = 0.7964x102 1.256 x 1612 C = 1.256 PF ANSWER

(2) If the value of capacitance increases above value find in part (1) Imbedence (2) will be- Z = л Z We know that Impedence of circuit R?+(X--Xc)? , R2 + (WL-be), from here we can observe that, if we increase the value of a then (x2-xc) value increases thats why Impedance (z) also - increase. Because- Resonance forequency if fix and at this frequency X -- Xc= 0 ANSWER Impedance (2) Increases (3) Impedance of cipecuit Of Resonance- Impedance (Z) = R2+(x2-xc)2 , At Resonance, XL= Xc R²to z = N - R 5.502 ANSWER N

(4) find Impedance (z) of the circuit when capacitance 1% increases. In part (1) - Capacitance (C) = 1:256 pf Affer increasing 1% , New Capacitance C' = Cx1-01 = 1.256 X 1.01 Jл ' - 1.268 pP How , Impedance (2) - √ R²+(*₂- Xc)² R2 +(WL-6c.) ? = -6 (5.5)2 + (€ 515.221x 3 11 515.221x1-260x16 30.25+ (1545-664 - 1530.69)? - 30-25+ 224.22 Z = 15.95 r ANSWER

From part(4) we can also understand the answer for part(2).

That is , if we increase the value of capacitance then impedance (Z) also increase.