Question
A power amplifier has an output of 100 W and an input of 10 W. What is the power gain in decibels?
Group of answer choices
1
5
10
2
Group of answer choices
1
5
10
2
Answers
GPT-4o mini
To calculate the power gain in decibels (dB), you can use the following formula:
\[
\text{Power Gain (dB)} = 10 \times \log_{10} \left( \frac{\text{Pout}}{\text{Pin}} \right)
\]
Where:
- \( \text{Pout} \) is the output power
- \( \text{Pin} \) is the input power
In this case:
- \( \text{Pout} = 100 \, \text{W} \)
- \( \text{Pin} = 10 \, \text{W} \)
Now plug in the values:
\[
\text{Power Gain (dB)} = 10 \times \log_{10} \left( \frac{100}{10} \right)
\]
This simplifies to:
\[
\text{Power Gain (dB)} = 10 \times \log_{10} (10)
\]
Since \( \log_{10} (10) = 1 \):
\[
\text{Power Gain (dB)} = 10 \times 1 = 10 \, \text{dB}
\]
So the power gain in decibels is **10 dB**.
The correct answer is **10**.
\[
\text{Power Gain (dB)} = 10 \times \log_{10} \left( \frac{\text{Pout}}{\text{Pin}} \right)
\]
Where:
- \( \text{Pout} \) is the output power
- \( \text{Pin} \) is the input power
In this case:
- \( \text{Pout} = 100 \, \text{W} \)
- \( \text{Pin} = 10 \, \text{W} \)
Now plug in the values:
\[
\text{Power Gain (dB)} = 10 \times \log_{10} \left( \frac{100}{10} \right)
\]
This simplifies to:
\[
\text{Power Gain (dB)} = 10 \times \log_{10} (10)
\]
Since \( \log_{10} (10) = 1 \):
\[
\text{Power Gain (dB)} = 10 \times 1 = 10 \, \text{dB}
\]
So the power gain in decibels is **10 dB**.
The correct answer is **10**.