Question

A reading of 120 is obtained when standard inductor was connected. The circuit of Q-meter and the variable capacitor is adjusted to a value of 300 pF. A lossless capacitor of unknown value $C_x$ is then connected in parallel with the variable capacitor and the same reading was obtained when the variable capacitor is readjusted to a value of 200 pF. The value of $C_x$ in pF is

• A. 100
• B. 200
• C. 300
• D. 500

Answer 1

The correct option is A 100

Q of the coil =120

$Q=\frac{\omega L}{R}=\frac{1}{\omega RC}$

Case-I:
Q=120 is obtained for C=300 pF

Case-II:
For same reading i.e Q=120
Effective capacitance of the circuit should be 300 pF.

Effective capacitance of parallel capacitor,
$C_{eff}=C+C_x$
$C_x=C_{eff}-C$
$=300-200$
$=100pF$