Which decibel value most closely represents a power decrease from 12 watts to 3 watts?

Which decibel value most closely represents 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 decibel change for a power decrease from 12 watts to 3 watts, the decibel formula for power is used: \[ dB = 10 \cdot \log_{10}\left(\frac{P2}{P1}\right) \] In this scenario, \(P1\) (the initial power) is 12 watts and \(P2\) (the final power) is 3 watts. First, we will take the ratio of these power levels: \[ \frac{P2}{P1} = \frac{3}{12} = \frac{1}{4} \] Now, we can substitute this ratio into the decibel formula: \[ dB = 10 \cdot \log_{10}\left(\frac{1}{4}\right) \] Calculating \(\log_{10}\left(\frac{1}{4}\right)\): \(\frac{1}{4}\) can also be expressed as \(4^{-1}\), which allows us to use properties of logarithms: \[ \log_{10}\left(\frac{1}{4}\right) = -\log_{10}(4) \

To determine the decibel change for a power decrease from 12 watts to 3 watts, the decibel formula for power is used:

[

dB = 10 \cdot \log_{10}\left(\frac{P2}{P1}\right)

]

In this scenario, (P1) (the initial power) is 12 watts and (P2) (the final power) is 3 watts. First, we will take the ratio of these power levels:

[

\frac{P2}{P1} = \frac{3}{12} = \frac{1}{4}

]

Now, we can substitute this ratio into the decibel formula:

[

dB = 10 \cdot \log_{10}\left(\frac{1}{4}\right)

]

Calculating (\log_{10}\left(\frac{1}{4}\right)):

(\frac{1}{4}) can also be expressed as (4^{-1}), which allows us to use properties of logarithms:

[

\log_{10}\left(\frac{1}{4}\right) = -\log_{10}(4)

\