To find the resonant frequency of the tuned circuit, we use the formula for the resonant frequency of an LC circuit:

$f = \frac{1}{2 \pi \sqrt{L \cdot C}}$

where:

• $L$ is the inductance in henries (H)
• $C$ is the capacitance in farads (F)

Problem 1: Initial Resonant Frequency Calculation

1. Given Values:

   • Inductance: $L = 40 \, \mu H = 40 \times 10^{-6} \, H$
   • Capacitor: $C_1 = 330 \, pF = 330 \times 10^{-12} \, F$
   • Varactor Capacitance: $C_2 = 50 \, pF = 50 \times 10^{-12} \, F$

2. Total Capacitance Calculation: The total capacitance for capacitors in parallel is the sum of individual capacitances:

   $C_{\text{total}} = C_1 + C_2 = (330 + 50) \times 10^{-12} \, F = 380 \times 10^{-12} \, F$

3. Resonant Frequency Calculation: Now, substitute $L = 40 \times 10^{-6} \, H$ and $C_{\text{total}} = 380 \times 10^{-12} \, F$ into the resonant frequency formula:

   $f = \frac{1}{2 \pi \sqrt{(40 \times 10^{-6}) \cdot (380 \times 10^{-12})}}$

   Calculating the expression under the square root:

   $L \cdot C_{\text{total}} = (40 \times 10^{-6}) \cdot (380 \times 10^{-12}) = 15.2 \times 10^{-15}$

   Taking the square root:

   $\sqrt{15.2 \times 10^{-15}} = 3.9 \times 10^{-8}$

   Thus, the resonant frequency becomes:

   $f = \frac{1}{2 \pi \cdot 3.9 \times 10^{-8}} \approx 4.08 \times 10^{6} \, \text{Hz} = 4.08 \, \text{MHz}$

Problem 2: Change in Varactor Capacitance to 25 pF

1. New Total Capacitance:

   • New Varactor Capacitance: $C_2' = 25 \, pF = 25 \times 10^{-12} \, F$
   • New Total Capacitance:

   $C_{\text{total}}' = C_1 + C_2' = (330 + 25) \times 10^{-12} \, F = 355 \times 10^{-12} \, F$

2. New Resonant Frequency Calculation: Substitute $C_{\text{total}}' = 355 \times 10^{-12} \, F$:

   $f' = \frac{1}{2 \pi \sqrt{(40 \times 10^{-6}) \cdot (355 \times 10^{-12})}}$

   Calculate the expression under the square root:

   $L \cdot C_{\text{total}}' = (40 \times 10^{-6}) \cdot (355 \times 10^{-12}) = 14.2 \times 10^{-15}$

   Taking the square root:

   $\sqrt{14.2 \times 10^{-15}} = 3.77 \times 10^{-8}$

   Thus, the new resonant frequency becomes:

   $f' = \frac{1}{2 \pi \cdot 3.77 \times 10^{-8}} \approx 4.22 \times 10^{6} \, \text{Hz} = 4.22 \, \text{MHz}$

Changes in Frequency:

(a) The frequency increases from 4.08 MHz to 4.22 MHz.

(b) The new resonant frequency is 4.22 MHz.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. What is the effect of inductance on the resonant frequency?
2. How does adding more capacitors in parallel affect the resonant frequency?
3. What are the applications of resonant circuits in oscillators?
4. How does quality factor (Q) relate to the resonant frequency of an LC circuit?
5. What happens if the varactor capacitance is made very small?

Tip: The resonant frequency is inversely proportional to the square root of the product of inductance and capacitance. Thus, reducing the capacitance increases the frequency.