Chapter 12, Problem 71GP

To determine

To Find: The power output, if a stereo is rated at 150 W output at 1000 Hz and the power output drops by 10 dB at 15 kHz.

Answer to Problem 71GP

The power output drop is ${\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}=15\text{ W}$ .

Explanation of Solution

Given:

Power output drops (P_out) = 10 dB

Power input (Pin) = 150 Hz

Formula used:

Write the expression for intensity of sound is

  $\mathrm{d}\mathrm{B}=10\log \left(\frac{{\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}}{{\mathrm{P}}_{\mathrm{i}\mathrm{n}}}\right)$

Here, ${\mathrm{P}}_{\text{out}}$ is power output drops.

  ${\mathrm{P}}_{\text{in}}$ is the power input.

Calculation:

The Intensity of sound is given by,

  $\mathrm{d}\mathrm{B}=10\log \left(\frac{{\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}}{{\mathrm{P}}_{\mathrm{i}\mathrm{n}}}\right)$

  $-10=10\log \left(\frac{{\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}}{{\mathrm{P}}_{\mathrm{i}\mathrm{n}}}\right)$

Equation further simplified as

  $-1=\log \left(\frac{{\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}}{{\mathrm{P}}_{\mathrm{i}\mathrm{n}}}\right)$

Dividing by 10 on both sides

  $\frac{1}{10}=\left(\frac{{\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}}{{\mathrm{P}}_{\mathrm{i}\mathrm{n}}}\right)$

  ${\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}={\mathrm{p}}_{\mathrm{i}\mathrm{n}}\frac{1}{10}$

  ${\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}=150\times \frac{1}{10}\text{ W}$

  ${\mathrm{P}}_{\mathrm{o}\mathrm{u}\mathrm{t}}=15\text{ W}$

The power output is 15 W.

Conclusion:

The power output is 15 W, if a stereo is rated at 150 W output at 1000 Hz and the power output drops by 10 dB at 15 kHz.