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