Question

An electrostatic voltmeter reading upto 2000 V is controlled by a spring with a torsion constant of $5\times {10}^{-6}Nm/rad$ has has a full scale deflection of ${90}^{\circ }$. The capacitance at zero voltage is 15 pF. What is the capacitance when the pointer indicates 2000 V?

• A. 21.16 pF
• B. 15 pF
• C. 6.16 pF
• D. 8.84 pF

Answer 1

The correct option is A 21.16 pF

Deflection,
$\theta ={90}^{\circ }=\frac{\pi }{2}rad$

Final steady deflection,
$\theta =\frac{1}{2}\cdot \frac{V^2}{K}\cdot \frac{dC}{d\theta }$

Rate of change of capacitance,

$\frac{dC}{d\theta }=\frac{2\theta K}{V^2}=\frac{2\times \left(\frac{\pi }{2}\right)\times 5\times {10}^{-6}}{(2000{)}^2}=3.92pF/rad$

Change in capacitance when reading from 0 to 2000 V

$=\frac{dC}{d\theta }=3.92\times \frac{\pi }{2}=6.16pF$

$\therefore$ capacitance when reading 2000 V,

$C=15+6.16=21.16pF$