An audio signal is modulated by a carrier wave of 20 MHz such that the bandwidth required for modulation is 3 kHz. Could this wave be demodulated by a diode detector which has the values of R and C as (i) `R = 1 k Omega, C = 0.01 mu F (ii) R = 10 k Omega, C = 0.01 muF (iii) R = 10k Omega , C = 1 mu muF`.

Options

A. `R=1kOmega,C=0.01muE`
B. `R=10kOmega,C=0.01muF`
C. `R=10kOmega,C=0.1pF`
D. None of these

Explanation

To determine whether the given audio signal modulated by a carrier wave of 20 MHz can be demodulated by a diode detector with specified values of R and C, we need to follow these steps:

### Step 1: Identify the parameters
- **Carrier frequency (f_c)** = 20 MHz = 20 × 10^6 Hz
- **Modulation bandwidth (B)** = 3 kHz = 3 × 10^3 Hz

### Step 2: Calculate the time constant (τ)
The time constant τ is given by the product of R and C:
\[ \tau = R \times C \]

### Step 3: Calculate the cutoff frequency (f_c)
The cutoff frequency (f_c) is given by:
\[ f_c = \frac{1}{2\pi \tau} = \frac{1}{2\pi R C} \]

### Step 4: Check the condition for demodulation
For demodulation to be possible, the following condition must be satisfied:
\[ \tau \ll \frac{1}{B} \]
This implies:
\[ R \times C \ll \frac{1}{B} \]

### Step 5: Calculate the value of \(\frac{1}{B}\)
\[ \frac{1}{B} = \frac{1}{3 \times 10^3} = \frac{1}{3000} \approx 0.000333 \text{ seconds} \]

### Step 6: Evaluate each case
Now we will evaluate the three given sets of R and C values:

#### Case (i): R = 1 kΩ, C = 0.01 µF
1. Convert units:
- R = 1 kΩ = 1000 Ω
- C = 0.01 µF = 0.01 × 10^-6 F = 10^-8 F
2. Calculate τ:
\[ \tau = 1000 \times 10^{-8} = 10^{-5} \text{ seconds} \]
3. Check the condition:
\[ 10^{-5} \ll 0.000333 \text{ (True)} \]
- This case is valid for demodulation.

#### Case (ii): R = 10 kΩ, C = 0.01 µF
1. Convert units:
- R = 10 kΩ = 10,000 Ω
- C = 0.01 µF = 10^-8 F
2. Calculate τ:
\[ \tau = 10,000 \times 10^{-8} = 10^{-4} \text{ seconds} \]
3. Check the condition:
\[ 10^{-4} \ll 0.000333 \text{ (True)} \]
- This case is also valid for demodulation.

#### Case (iii): R = 10 kΩ, C = 1 µF
1. Convert units:
- R = 10 kΩ = 10,000 Ω
- C = 1 µF = 1 × 10^-6 F
2. Calculate τ:
\[ \tau = 10,000 \times 1 \times 10^{-6} = 10^{-2} \text{ seconds} \]
3. Check the condition:
\[ 10^{-2} \not\ll 0.000333 \text{ (False)} \]
- This case is not valid for demodulation.

### Conclusion
- The first two cases (i) and (ii) can demodulate the signal, while case (iii) cannot.