Find R for Max Power: 2304000/80 = 28,800

[nhrock3]

find the R so we will have maximal power on it

so now i need to use this formula
$$P_{max}=\frac{1}{8}\frac{|\tilde{V_s}|^2}{R_s}$$
$$|\tilde{V_s}|^2=480^2+1440^2=2304000$$
R_s is the real part of the impedance (80)

$$P_{max}=\frac{1}{8}\frac{2304000}{80}=28800$$

but here it doesn't resemble at all the maximal power formula

they are counting the average power and saying something about rms
??

 

[Valkarie]

No, the formula you have provided is for calculating the maximum power that can be transferred from a source to a load. The maximum power transfer theorem states that the maximum power is transferred when the source and load impedances are equal. In this case, the source impedance is 80 ohms and the load impedance is also 80 ohms. Therefore, the maximum power that can be transferred is 28800 watts.

 

[Mene]

It is important to note that the formula for maximum power, P_max, is different from the formula for average power. The formula for maximum power is P_max = 1/8 * (|V_s|^2 / R_s), where |V_s| is the magnitude of the source voltage and R_s is the real part of the impedance. This formula is used to calculate the maximum amount of power that can be delivered to a load, given the source voltage and impedance. It is not related to the RMS (root mean square) value of the voltage.

In this case, the value of R_s that will result in maximum power is 28,800 Ohms. This means that if a load with an impedance of 28,800 Ohms is connected to the source, it will receive the maximum amount of power possible. It is important to note that this value may not always be the most practical or efficient choice for a given system, as other factors such as cost and efficiency should also be considered.