What decibel value represents a power increase from 20 watts to 200 watts?

What decibel value represents a power increase from 20 watts to 200 watts?

• A. 6 dB
• B. 10 dB
• C. 3 dB
• D. 12 dB

Answer: (B)

To determine the decibel value that represents a power increase from 20 watts to 200 watts, you can use the formula for calculating power in decibels: \[ dB = 10 \times \log_{10}\left(\frac{P_2}{P_1}\right) \] In this case, \( P_1 \) (initial power) is 20 watts, and \( P_2 \) (final power) is 200 watts. Plugging these values into the formula yields: \[ dB = 10 \times \log_{10}\left(\frac{200}{20}\right) \] Calculating the ratio gives: \[ \frac{200}{20} = 10 \] Now, taking the logarithm: \[ \log_{10}(10) = 1 \] Thus, the calculation becomes: \[ dB = 10 \times 1 = 10 \, dB \] This shows that the increase from 20 watts to 200 watts is represented as a 10 dB increase. Understanding this concept is crucial in ham radio communications, where power levels and their corresponding measurements in decibels are vital for efficient operation and understanding signal strengths.

To determine the decibel value that represents a power increase from 20 watts to 200 watts, you can use the formula for calculating power in decibels:

[ dB = 10 \times \log_{10}\left(\frac{P_2}{P_1}\right) ]

In this case, ( P_1 ) (initial power) is 20 watts, and ( P_2 ) (final power) is 200 watts. Plugging these values into the formula yields:

[ dB = 10 \times \log_{10}\left(\frac{200}{20}\right) ]

Calculating the ratio gives:

[ \frac{200}{20} = 10 ]

Now, taking the logarithm:

[ \log_{10}(10) = 1 ]

Thus, the calculation becomes:

[ dB = 10 \times 1 = 10 , dB ]

This shows that the increase from 20 watts to 200 watts is represented as a 10 dB increase. Understanding this concept is crucial in ham radio communications, where power levels and their corresponding measurements in decibels are vital for efficient operation and understanding signal strengths.