In a diode AM detector with the output circuit consists of R=`1M Omega` and `C=1 pF` would be more suitable for detecting a carrier signal of

Options

A. 1MHz
B. 0.1 MHz
C. 0.5MHz
D. 10MHz

Explanation

To determine the suitable carrier frequency for a diode AM detector with an output circuit consisting of a resistor (R) of 1 MΩ and a capacitor (C) of 1 pF, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Values of R and C**:
- Given R = 1 MΩ = \( 10^6 \) ohms
- Given C = 1 pF = \( 10^{-12} \) farads

2. **Calculate the Product of R and C**:
- Calculate \( R \times C \):
\[ R \times C = (10^6 \, \text{ohms}) \times (10^{-12} \, \text{farads}) = 10^{-6} \, \text{seconds} \]

3. **Determine the Frequency of the Carrier Signal**:
- The frequency of the carrier signal (\( \nu_c \)) can be determined using the formula:
\[ \nu_c = \frac{1}{R \times C} \]
- Substitute the value of \( R \times C \):
\[ \nu_c = \frac{1}{10^{-6}} = 10^6 \, \text{Hertz} \]

4. **Convert the Frequency to Megahertz**:
- Since \( 10^6 \, \text{Hertz} = 1 \, \text{MHz} \), we can conclude that:
\[ \nu_c = 1 \, \text{MHz} \]

5. **Conclusion**:
- Therefore, the diode AM detector with the given values of R and C is suitable for detecting a carrier signal of frequency **1 MHz**.