Question

A high Q coil having distributed (self) capacitance is tested with a Q-meter. First resonance at ω₁ = 10⁶ rad/s is obtained with a capacitance of 990 pF. The second resonance at ω₂ = 2 ×10⁶ rad/s is obtained with a 240 pF capacitance. The value of the inductance (in mH) of the coil is (up to one decimal place) ______.

Answer 1

ω₁ = 10⁶ rad/s, ω₂ = 2 × 10⁶ rad/s

C₁ = 990 pF, C₂ = 240 pF

\(n = \frac{{{\omega _2}}}{{{\omega _1}}} = \frac{{2 \times {{10}^6}}}{{{{10}^6}}} = 2\)

Distributed capacitance is given by

\({C_d} = \frac{{{C_1} - {n^2}{C_2}}}{{{n^2} - 1}}\)

\(= \frac{{990 - {2^2} \times 240}}{{{2^2} - 1}} = 10\;pF\)

The inductance of the coil is given by,

\({L_{Coil}} = \frac{1}{{w_1^2\left( {{c_1} + {c_d}} \right)}} = \frac{1}{{w_2^2\left( {{c_2} + {c_d}} \right)}}\)

\(= \frac{1}{{{{\left( {{{10}^6}} \right)}^2}\left( {990 + 10} \right) \times {{10}^{ - 12}}}} = 1mH\)