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

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

• A. 10 dB
• B. 12 dB
• C. 18 dB
• D. 28 dB

Answer: (A)

To accurately calculate the decibel increase associated with a change in power from 20 watts to 200 watts, we can use the formula for power levels in decibels: \[ \text{dB} = 10 \times \log_{10} \left(\frac{P_2}{P_1}\right) \] where \( P_2 \) is the final power and \( P_1 \) is the initial power. In this case, \( P_2 \) is 200 watts and \( P_1 \) is 20 watts. Plugging the values into the formula gives us: \[ \text{dB} = 10 \times \log_{10} \left(\frac{200}{20}\right) \] \[ \frac{200}{20} = 10 \] So we simplify the logarithm: \[ \text{dB} = 10 \times \log_{10}(10) \] Since the logarithm of 10 is 1: \[ \text{dB} = 10 \times 1 = 10 \text{ dB} \] This indicates that increasing the power from

To accurately calculate the decibel increase associated with a change in power from 20 watts to 200 watts, we can use the formula for power levels in decibels:

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

where ( P_2 ) is the final power and ( P_1 ) is the initial power.

In this case, ( P_2 ) is 200 watts and ( P_1 ) is 20 watts. Plugging the values into the formula gives us:

[

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

]

[

\frac{200}{20} = 10

]

So we simplify the logarithm:

[

\text{dB} = 10 \times \log_{10}(10)

]

Since the logarithm of 10 is 1:

[

\text{dB} = 10 \times 1 = 10 \text{ dB}

]

This indicates that increasing the power from