The output current of an 80% mondulating amplitude modulated generator is 1.8A. To what value will the current rise if the generator is additionally modulated by another audiowave of modulation index 0.6?

Options

A. 1.71A
B. 1.81A
C. 1.91A
D. 2.01A

Explanation

To solve the problem, we need to find the new output current of an amplitude modulated generator when it is modulated by an additional audio wave. Here’s the step-by-step solution:

### Step 1: Understand the Given Information
- The initial output current (I) is 1.8 A.
- The first modulation index (μ1) is 80%, which can be converted to a decimal:
\[ μ1 = \frac{80}{100} = 0.8 \]
- The second modulation index (μ2) is given as 0.6.

### Step 2: Calculate the Current Just After the First Modulation
The formula for the output current after the first modulation is:
\[ I_c = \frac{I}{\sqrt{1 + \frac{μ1^2}{2}}} \]
Substituting the values we have:
\[ I_c = \frac{1.8}{\sqrt{1 + \frac{0.8^2}{2}}} \]
Calculating \(0.8^2\):
\[ 0.8^2 = 0.64 \]
Thus, we have:
\[ I_c = \frac{1.8}{\sqrt{1 + \frac{0.64}{2}}} = \frac{1.8}{\sqrt{1 + 0.32}} = \frac{1.8}{\sqrt{1.32}} \]

### Step 3: Calculate the Value of \(I_c\)
Now we need to calculate \(I_c\):
\[ \sqrt{1.32} \approx 1.1487 \]
So,
\[ I_c \approx \frac{1.8}{1.1487} \approx 1.568 \]

### Step 4: Calculate the Total Output Current After Additional Modulation
The formula for the total output current after additional modulation is:
\[ I_{net} = I_c \cdot \sqrt{1 + \frac{μ1^2}{2} + \frac{μ2^2}{2}} \]
Substituting the values:
\[ I_{net} = 1.568 \cdot \sqrt{1 + \frac{0.8^2}{2} + \frac{0.6^2}{2}} \]
Calculating \(0.6^2\):
\[ 0.6^2 = 0.36 \]
Now substituting:
\[ I_{net} = 1.568 \cdot \sqrt{1 + \frac{0.64}{2} + \frac{0.36}{2}} = 1.568 \cdot \sqrt{1 + 0.32 + 0.18} = 1.568 \cdot \sqrt{1.5} \]

### Step 5: Calculate the Value of \(I_{net}\)
Now we calculate \(I_{net}\):
\[ \sqrt{1.5} \approx 1.2247 \]
Thus,
\[ I_{net} \approx 1.568 \cdot 1.2247 \approx 1.92 \]

### Final Result
The new output current after additional modulation is approximately:
\[ I_{net} \approx 1.92 \text{ A} \]