Question

In a Q Meter, the values of the tuning capacitor are \( C_3 \) and \( C_4 \) for resonant frequencies \( f \) and \( 2f \), respectively. The value of distributed capacitance is

• A. \( \frac{(C_3 - C_4)}{2} \)
• B. \( \frac{(C_3 - 2C_4)}{3} \)
• C. \( \frac{(C_3 - 4C_4)}{3} \)
• D. \( \frac{(C_3 - 4C_4)}{2} \)

Hint:

For Q-meter problems involving distributed capacitance, use the resonant frequency formula \( f = \frac{1}{2\pi \sqrt{L(C + C_d)}} \). By squaring and comparing two resonant conditions, you can eliminate the constant terms and solve for the distributed capacitance \( C_d \).

Answer 1

The Correct Option is C

Understanding the Problem:

• A Q meter is used to measure the quality factor (Q) of a circuit.
• At resonance, the inductive reactance ($X_L$) equals the capacitive reactance ($X_C$).
• Distributed capacitance ($C_d$) is the inherent capacitance present in the inductor.

Formulae:

• Resonant frequency ($f$) = $\frac{1}{2\pi\sqrt{L(C + C_d)}}$ where $L$ is inductance, $C$ is tuning capacitance, and $C_d$ is distributed capacitance.

Let's denote:

• $f_1 = f$, $C_1 = C_3$
• $f_2 = 2f$, $C_2 = C_4$
• $L$ = Inductance of the coil
• $C_d$ = Distributed Capacitance

Setting up the equations:

At frequency $f$ ($f_1$): 
$f_1 = \frac{1}{2\pi\sqrt{L(C_3 + C_d)}}$ 
$f = \frac{1}{2\pi\sqrt{L(C_3 + C_d)}}$ 
Squaring both sides: 
$f^2 = \frac{1}{4\pi^2 L (C_3 + C_d)}$      (1)

At frequency $2f$ ($f_2$): 
$f_2 = \frac{1}{2\pi\sqrt{L(C_4 + C_d)}}$ 
$2f = \frac{1}{2\pi\sqrt{L(C_4 + C_d)}}$ 
Squaring both sides: 
$4f^2 = \frac{1}{4\pi^2 L (C_4 + C_d)}$      (2)

Solving for $C_d$:

Divide equation (2) by equation (1):

$\frac{4f^2}{f^2} = \frac{\frac{1}{4\pi^2 L (C_4 + C_d)}}{\frac{1}{4\pi^2 L (C_3 + C_d)}}$

$4 = \frac{C_3 + C_d}{C_4 + C_d}$

$4(C_4 + C_d) = C_3 + C_d$

$4C_4 + 4C_d = C_3 + C_d$

$3C_d = C_3 - 4C_4$

$C_d = \frac{C_3 - 4C_4}{3}$

Therefore, the distributed capacitance is $\frac{C_3 - 4C_4}{3}$.