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

In a Q-meter, the resonant frequency of an LC circuit is given by: \[ f = \frac{1}{2\pi \sqrt{L(C + C_d)}} \] where:

• \( L \) is the inductance of the coil,
• \( C \) is the tuning capacitor value,
• \( C_d \) is the distributed capacitance of the coil.

Let: \begin{align*} f &= \text{resonant frequency when the capacitor is } C_3 \\ 2f &= \text{resonant frequency when the capacitor is } C_4 \end{align*}
Step 1: Apply the resonance formula. Using the resonance formula: \[ f = \frac{1}{2\pi \sqrt{L(C_3 + C_d)}} \quad \text{and} \quad 2f = \frac{1}{2\pi \sqrt{L(C_4 + C_d)}} \]
Step 2: Square both equations. \[ f^2 = \frac{1}{4\pi^2 L(C_3 + C_d)} \quad \text{(1)} \] \[ (2f)^2 = 4f^2 = \frac{1}{4\pi^2 L(C_4 + C_d)} \quad \text{(2)} \]
Step 3: Equate and simplify. From (1) and (2), eliminate the common constant: \[ \frac{1}{4\pi^2 L(C_3 + C_d)} = f^2, \quad \frac{1}{4\pi^2 L(C_4 + C_d)} = 4f^2 \] Divide the two equations: \[ \frac{C_3 + C_d}{C_4 + C_d} = 4 \]
Step 4: Solve for \( C_d \). \[ C_3 + C_d = 4(C_4 + C_d) \] \[ C_3 + C_d = 4C_4 + 4C_d \] \[ C_3 - 4C_4 = 3C_d \] \[ C_d = \frac{C_3 - 4C_4}{3} \] \[ \boxed{C_d = \frac{C_3 - 4C_4}{3}} \]