The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude modulated wave is `"x"xx10^(-2)`. The value of x is `"_____"`

Answer

To solve the problem, we need to find the modulation index for an amplitude-modulated wave given the maximum and minimum amplitudes. The maximum amplitude (A_max) is 16 V and the minimum amplitude (A_min) is 8 V.

### Step-by-Step Solution:

1. **Identify the given values:**
- Maximum amplitude, \( A_{max} = 16 \, V \)
- Minimum amplitude, \( A_{min} = 8 \, V \)

2. **Use the formula for modulation index (m):**
The modulation index \( m \) for an amplitude-modulated wave is given by:
\[ m = \frac{A_{max} - A_{min}}{A_{max} + A_{min}} \]

3. **Substitute the values into the formula:**
\[ m = \frac{16 - 8}{16 + 8} \]

4. **Calculate the numerator and denominator:**
- Numerator: \( 16 - 8 = 8 \)
- Denominator: \( 16 + 8 = 24 \)

5. **Calculate the modulation index:**
\[ m = \frac{8}{24} = \frac{1}{3} \]

6. **Convert the modulation index to the required format:**
We know that \( m = 0.33 \) (approximately). According to the problem, the modulation index is given as \( x \times 10^{-2} \).
\[ 0.33 = x \times 10^{-2} \]

7. **Solve for \( x \):**
\[ x = 0.33 \times 10^{2} = 33 \]

### Final Answer:
The value of \( x \) is \( 33 \).

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