If an audio amplifier’s power decreases from 20 watts to 9 watts, what is the expected change in decibels?

If an audio amplifier’s power decreases from 20 watts to 9 watts, what is the expected change in decibels?

• A. -3.5 dB
• B. -5.8 dB
• C. -7 dB
• D. -0.35 dB

Answer: (A)

To determine the change in decibels when an audio amplifier's power decreases from 20 watts to 9 watts, the formula used involves the logarithmic relationship between power levels and decibels: \[ \Delta dB = 10 \log_{10}\left(\frac{P2}{P1}\right) \] In this scenario: - \(P1\) (initial power) is 20 watts. - \(P2\) (final power) is 9 watts. Substituting these values into the formula gives: \[ \Delta dB = 10 \log_{10}\left(\frac{9}{20}\right) \] Calculating the fraction: \[ \frac{9}{20} = 0.45 \] Next, computing the logarithm: \[ \log_{10}(0.45) \approx -0.34657 \] Now, multiplying by 10: \[ \Delta dB \approx 10 \times -0.34657 \approx -3.4657 \text{ dB} \] Rounding this value gives approximately -3.5 dB. This indicates that a decrease in power from 20

To determine the change in decibels when an audio amplifier's power decreases from 20 watts to 9 watts, the formula used involves the logarithmic relationship between power levels and decibels:

[

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

]

In this scenario:

• (P1) (initial power) is 20 watts.

• (P2) (final power) is 9 watts.

Substituting these values into the formula gives:

[

\Delta dB = 10 \log_{10}\left(\frac{9}{20}\right)

]

Calculating the fraction:

[

\frac{9}{20} = 0.45

]

Next, computing the logarithm:

[

\log_{10}(0.45) \approx -0.34657

]

Now, multiplying by 10:

[

\Delta dB \approx 10 \times -0.34657 \approx -3.4657 \text{ dB}

]

Rounding this value gives approximately -3.5 dB.

This indicates that a decrease in power from 20