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

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Question

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

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) The change in sound level in decibels $(dB)$ 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)$$\Delta L = 10 \log_{10} (20)$Since $\log_{10} (20) = \log_{10} (2 	imes 10) = \log_{10} (2) + \log_{10} (10) \approx 0.301 + 1 = 1.301$.$\Delta L = 10 	imes 1.301 = 13.01 	ext{ dB}$.Rounding to the nearest integer,the power increase is $13 	ext{ dB}$.

The power of 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 ..... $dB$.

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