Question

A pressure measuring instrument uses a capacitive transducer having a spacing of 4mm between its diaphragh. A pressure of $600kN/m^2$ produces an average deflection of 0.3mm of the diaphragm of the transducer. The transducer which has a capacitance of 300 pF before application of pressure & is connected in an oscillator circuit having a frequency of 100 kHz. The change in frequency (in kHz) of the oscillator after the pressure is applied to the transducer is

• A. 3.8
• B. 96.177
• C. 2.8
• D. 9.8

Answer 1

The correct option is A 3.8

$C_1=300pF$

$\Delta R=\text{average deflection = 0.3 ~mm}$

$f_1=100kHz$

$x_1=4mm$

$C_1=\frac{\epsilon A}{x}$

$\epsilon A=300\times {10}^{-12}\times 4\times {10}^{-3}=1.2\times {10}^{-12}$

$x_2=x_1-\Delta x=4mm-0.3mm=3.7mm$

$\frac{f_1}{f_2}=\sqrt{\frac{C_2}{C_1}}$

$C_2=\frac{\epsilon A}{x_2}=\frac{1.2\times {10}^{-12}}{3.7\times {10}^{-3}}$

$=324.32\left(pF\right)$

$\frac{f_1}{f_2}=\sqrt{\frac{C_2}{C_1}}$
$f_2=f_1\sqrt{\frac{C_1}{C_2}}$

$=100\times {10}^3\times \sqrt{\frac{300\times {10}^{-12}}{324.32\times {10}^{-12}}}$

$f_2=96.177kHz$

$\Delta f=f_1-f_2=100kHz-96.177kHz$

$\Delta f=3.823kHz$