Question

In a variable capacitance transducer the diaphragm are 20mm in diameter & 4 mm apart if a pressure produces an average deflection of 0.25 mm. The capacitance before application of force is 400pF. The value of capacitance (in pF) after the application of force is

• A. 427
• B. 327
• C. 350
• D. 450

Answer 1

The correct option is A 427

$C_1=400pF$ (before force applied)

$C=\frac{{\epsilon }_0{\epsilon }_{r}A}{x}....\left(1\right)$

$x_1=4mm$

$\text{due to pressure average deflection of}\Delta x=0.25mm$

$x_2=x_1-\Delta x=4-0.25=3.75mm....\left(2\right)$

$C_1=\frac{{\epsilon }_o{\epsilon }_{r}A}{x}$

$400\times {10}^{-12}=\frac{{\epsilon }_o{\epsilon }_{r}A}{4mm}$

${\epsilon }_o{\epsilon }_{r}A=1.6\times {10}^{-12}....\left(3\right)$

$C_2=\frac{{\epsilon }_o{\epsilon }_{r}A}{x_2}=\frac{1.6\times {10}^{-12}}{3.75\times {10}^{-3}}=427\left(pF\right)$

$\text{(by using (2) \& (3) )}$