What is the approximate amount of change, measured in decibels (dB), of a power decrease from 12 watts to 3 watts?

What is the approximate amount of change, measured in decibels (dB), of a power decrease from 12 watts to 3 watts?

• A. 1 dB
• B. 3 dB
• C. 6 dB
• D. 9 dB

Answer: (C)

To determine the change in power level from 12 watts to 3 watts in decibels (dB), we can use the formula for converting power ratios to decibels: \[ \text{dB} = 10 \times \log_{10} \left( \frac{P2}{P1} \right) \] In this case, \( P1 \) is 12 watts and \( P2 \) is 3 watts. Plugging the values into the formula gives: \[ \text{dB} = 10 \times \log_{10} \left( \frac{3}{12} \right) = 10 \times \log_{10} \left( 0.25 \right) \] Since \( 0.25 \) can also be expressed as \( \frac{1}{4} \), we know \( \log_{10}(0.25) \) corresponds to \( -0.602 \) (because \( 10^{-0.602} \approx 0.25 \)). Thus, the calculation results in: \[ \text{dB} = 10 \times (-0.602) = -6.02

To determine the change in power level from 12 watts to 3 watts in decibels (dB), we can use the formula for converting power ratios to decibels:

[ \text{dB} = 10 \times \log_{10} \left( \frac{P2}{P1} \right) ]

In this case, ( P1 ) is 12 watts and ( P2 ) is 3 watts. Plugging the values into the formula gives:

[ \text{dB} = 10 \times \log_{10} \left( \frac{3}{12} \right) = 10 \times \log_{10} \left( 0.25 \right) ]

Since ( 0.25 ) can also be expressed as ( \frac{1}{4} ), we know ( \log_{10}(0.25) ) corresponds to ( -0.602 ) (because ( 10^{-0.602} \approx 0.25 )). Thus, the calculation results in:

[ \text{dB} = 10 \times (-0.602) = -6.02