Question

A constant voltage at a frequency of 1 MHz is applied to an inductor in series with a variable capacitor. When the capacitor is set 500 pF, the current has its maximum value while it is reduced to one-half when the capacitance is 600 pF. Find

(i) the resistance, (ii) the inductance, (iii) the Q-factor of the inductor

Answer 1

Resonance takes palce at 1 MHz, for C = 500 pF.

LC = 1/ω02 = 1/(2π x 106)2 = 10-12/4π2

L = 10− 12/(4 × π2 × 500 × 10− 12) = 1/(4 × π2 × 500)

= 50.72 mH

Z1 = Impedance with 500 pF capacitor = R + j ωL −j1/ωC

= R + j(2π × 106 × 50.72 × 10− 6) − j(1/(2π x 106 x 500 x 10-12))

= R, since resonance occurs.

Z2 = Impedance with 600 pF capacitor

|Z|2 = R + jωL - j(1/(ω x 600 x 10-12) = 2R, since current is halved.

ωL − 1/ωC = √3R

√3R = 2π x 106 x 50.7 x 10-6 - (1/(2π x 106 x 600 x 10-12))

= 2π x 50.72(106/2π600)

= 318.52 − 265.4 = 53.12

R = 30.67 ohms

Q Factor of coil = (ω0L)/R

= 50.72 × 10− 6 × 2π × 106 /30.67 = 50.72 × 2π/30.67 = 10.38