The power of a sound from the speaker of a ra | vedclass

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(A) The power level in decibels is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{P_2}{P_1} ight)$.Given initial power $P_1 = 20 	ext{

Vedclass · Accepted Answer

$13$

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(A) The power level in decibels is given by the formula: $\Delta L = 10 \log_{10} \left( \frac{P_2}{P_1} ight)$.Given initial power $P_1 = 20 	ext{ mW}$ and final power $P_2 = 400 	ext{ mW}$.Substituting the values into the formula:$\Delta L = 10 \log_{10} \left( \frac{400}{20} ight) = 10 \log_{10} (20)$.Since $\log_{10} (20) = \log_{10} (2 	imes 10) = \log_{10} (2) + \log_{10} (10)$.Using $\log_{10} (2) \approx 0.301$ and $\log_{10} (10) = 1$,we get:$\Delta L = 10 (0.301 + 1) = 10 (1.301) = 13.01 	ext{ dB}$.Rounding to the nearest integer,the power increase is $13 	ext{ dB}$.

The power of a sound from the speaker of a radio is $20 \text{ mW}$. By turning the knob of the volume control,the power of the sound is increased to $400 \text{ mW}$. The power increase in decibels as compared to the original power is .... $\text{dB}$

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