What decibel value most closely represents a power increase from 5 watts to 10 watts?

What decibel value most closely represents a power increase from 5 watts to 10 watts?

• A. 6 dB
• B. 10 dB
• C. 3 dB
• D. 1 dB

Answer: (C)

To determine the decibel value that represents a power increase from 5 watts to 10 watts, it's important to recall the formula used to calculate the decibel change in power levels. The formula for power in decibels (dB) is: \[ \text{dB} = 10 \times \log_{10}\left(\frac{P2}{P1}\right) \] where \( P1 \) is the initial power and \( P2 \) is the final power. In this case, \( P1 = 5 \) watts and \( P2 = 10 \) watts. Substituting these values into the formula gives: \[ \text{dB} = 10 \times \log_{10}\left(\frac{10 \text{ watts}}{5 \text{ watts}}\right) = 10 \times \log_{10}(2) \] The logarithm of 2 is approximately 0.301, so: \[ \text{dB} = 10 \times 0.301 = 3.01 \text{ dB} \] Thus, the power increase from 5 watts to 10 watts corresponds very closely to an increase

To determine the decibel value that represents a power increase from 5 watts to 10 watts, it's important to recall the formula used to calculate the decibel change in power levels. The formula for power in decibels (dB) is:

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

where ( P1 ) is the initial power and ( P2 ) is the final power. In this case, ( P1 = 5 ) watts and ( P2 = 10 ) watts.

Substituting these values into the formula gives:

[ \text{dB} = 10 \times \log_{10}\left(\frac{10 \text{ watts}}{5 \text{ watts}}\right) = 10 \times \log_{10}(2) ]

The logarithm of 2 is approximately 0.301, so:

[ \text{dB} = 10 \times 0.301 = 3.01 \text{ dB} ]

Thus, the power increase from 5 watts to 10 watts corresponds very closely to an increase