The average power dissipated in a stereo speaker is 54 W. Assuming that the speaker can be treated as a 4.0- resistance, find the peak value of the ac voltage applied to the speaker.

Is 4.0- supposed to represent 4.0 ohms?

If so,
Average power = Vrms^2/R = 54 W
= 1/2 * peak power
Vrms^2 = 54*4 = 216 volt^2
Vrms = 14.7 volt

Vrms is called the "root mean square" voltage, the usual value cited for alternating currents.

The peak voltage is sqrt2 times higher than Vrms, or 20.8 V

I multipl1ed 14.7 (the root mean squere voltage, Vrms) by the square root of 2. That results in the peak voltage, for a sine wave variation. A single-pitch sound is a sine wave, and that is what a speaker would receive.

still don't understand where the 20.8V comes into it how u get that?

To find the peak value of the AC voltage applied to the speaker, we can use Ohm's Law and the formula for power.

Ohm's Law states that the current flowing through a resistance is equal to the voltage across it divided by the resistance:

I = V / R

We know the resistance of the speaker is 4.0 Ω, and we need to find the voltage. Rearranging the formula, we can express the voltage in terms of current and resistance:

V = I * R

Now we can use the formula for power:

P = V^2 / R

We know the power dissipated in the speaker is 54 W. Rearranging the formula to solve for V^2:

V^2 = P * R

Substituting the given values:

V^2 = 54 W * 4.0 Ω
V^2 = 216 VΩ

Taking the square root of both sides, we can find the peak value of the AC voltage:

V = √(216 VΩ)
V ≈ 14.7 V

Therefore, the peak value of the AC voltage applied to the speaker is approximately 14.7 V.