A diode detector is used to detect and amplitude modulated wave of `60%` modulation by using a condenser of capacity 250 pF in parallel with a load resistance `100 kOmega`. Find the maximum modulated frequency which could be detected by it

Options

A. 10.62 kHz
B. 5.31 MHz
C. 5.31 kHz
D. 10.62 MHz

Answer

To solve the problem, we need to find the maximum modulated frequency \( f_c \) that can be detected by the diode detector. We will use the formula:

\[ f_c = \frac{1}{2 \pi \mu R C} \]

where:
- \( \mu \) is the modulation index (given as 0.6 for 60% modulation),
- \( R \) is the load resistance,
- \( C \) is the capacitance.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Load Resistance \( R = 100 \, k\Omega = 100 \times 10^3 \, \Omega \)
- Capacitance \( C = 250 \, pF = 250 \times 10^{-12} \, F \)
- Modulation Index \( \mu = 0.6 \)

2. **Substitute the Values into the Formula:**
We substitute the values of \( \mu \), \( R \), and \( C \) into the formula for \( f_c \):

\[ f_c = \frac{1}{2 \pi (0.6) (100 \times 10^3) (250 \times 10^{-12})} \]

3. **Calculate the Denominator:**
First, calculate the product in the denominator:

\[ 2 \pi (0.6) (100 \times 10^3) (250 \times 10^{-12}) \]

- Calculate \( 2 \pi \approx 6.2832 \)
- Calculate \( 0.6 \times 100 \times 10^3 = 60 \times 10^3 = 60000 \)
- Calculate \( 60000 \times 250 \times 10^{-12} = 15000 \times 10^{-9} = 15 \times 10^{-6} \)

Now, multiply by \( 2 \pi \):

\[ 6.2832 \times 15 \times 10^{-6} \approx 9.4248 \times 10^{-5} \]

4. **Calculate \( f_c \):**
Now, take the reciprocal to find \( f_c \):

\[ f_c = \frac{1}{9.4248 \times 10^{-5}} \approx 10600 \, Hz \]

Converting to kilohertz:

\[ f_c \approx 10.62 \, kHz \]

5. **Final Answer:**
The maximum modulated frequency that can be detected is approximately:

\[ f_c \approx 10.62 \, kHz \]

### Summary:
The maximum modulated frequency which could be detected by the diode detector is \( 10.62 \, kHz \).