The carrier freqeuncy of a station is 40 MHz. A resistor of 10k `Omega` and capacitor of CpF are available in the detector circuit. The possible value of C will be

Options

A. 12
B. 8.2
C. 5.6
D. All of these

Explanation

To solve the problem, we need to determine the possible value of the capacitor \( C \) in the detector circuit, given the carrier frequency and the resistance. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify Given Values
- Carrier frequency \( f_c = 40 \) MHz
- Resistance \( R = 10 \) kΩ

### Step 2: Convert Units
Convert the carrier frequency from MHz to Hz:
\[ f_c = 40 \text{ MHz} = 40 \times 10^6 \text{ Hz} = 4 \times 10^7 \text{ Hz} \]
Convert the resistance from kΩ to Ω:
\[ R = 10 \text{ kΩ} = 10 \times 10^3 \text{ Ω} = 10^4 \text{ Ω} \]

### Step 3: Calculate the Time Period
The time period \( T \) of the carrier frequency is the reciprocal of the frequency:
\[ T = \frac{1}{f_c} = \frac{1}{4 \times 10^7} = 2.5 \times 10^{-8} \text{ seconds} \]

### Step 4: Determine the Time Constant
The time constant \( \tau \) for an RC circuit is given by:
\[ \tau = R \times C \]
Where \( C \) is in farads.

### Step 5: Set Up the Condition
For the detection circuit to work effectively, the time period \( T \) should be less than the time constant \( \tau \):
\[ T < \tau \]
This implies:
\[ 2.5 \times 10^{-8} < R \times C \]

### Step 6: Substitute the Values
Substituting the value of \( R \):
\[ 2.5 \times 10^{-8} < 10^4 \times C \]
Rearranging gives:
\[ C > \frac{2.5 \times 10^{-8}}{10^4} = 2.5 \times 10^{-12} \text{ F} = 2.5 \text{ pF} \]

### Step 7: Evaluate Possible Values of \( C \)
Now, we will evaluate the provided options for \( C \) (12 pF, 8.2 pF, and 5.6 pF):
1. **For 12 pF:**
\[ \tau = 10^4 \times 12 \times 10^{-12} = 1.2 \times 10^{-7} \text{ seconds} \]
\( T < \tau \) is satisfied.

2. **For 8.2 pF:**
\[ \tau = 10^4 \times 8.2 \times 10^{-12} = 8.2 \times 10^{-8} \text{ seconds} \]
\( T < \tau \) is satisfied.

3. **For 5.6 pF:**
\[ \tau = 10^4 \times 5.6 \times 10^{-12} = 5.6 \times 10^{-8} \text{ seconds} \]
\( T < \tau \) is satisfied.

### Conclusion
All options provided (12 pF, 8.2 pF, and 5.6 pF) satisfy the condition \( T < \tau \). Therefore, the possible values of \( C \) can be any of the given options.