pokie_panda said:Homework Statement
Calculate the average power (P) absorbed by the load shown below when its impedance is 56 +j15 ohms 21 A rms?
Answer to three significant figures
Homework Equations
P=Irms^2*R
The Attempt at a Solution
Converting 56 +j15 to polar form yields 57.9741 ohms @ 14.9951 degrees
So 21^2*57.9741 = 25.5kW
Impedance is a measure of the opposition to the flow of electrical current in a circuit. It consists of both resistance and reactance. In power calculations, impedance is used to determine the relationship between voltage, current, and power.
RMS (Root Mean Square) current is the effective or average value of the alternating current over a given time period. It takes into account the fluctuation of the current over time, whereas average current only considers the average value of the current without taking into account the fluctuations.
The formula for calculating average power using impedance and rms current is: P = Irms2 * Z, where P is the average power, Irms is the rms current, and Z is the impedance.
Yes, impedance can affect the accuracy of power calculations because it is a measure of the resistance and reactance in a circuit. If the impedance is high, it can cause the current to be reduced and result in a lower power output. This can lead to inaccuracies in power calculations.
There are a few limitations to using impedance and rms current for power calculations. One limitation is that they are only applicable to AC circuits and cannot be used for DC circuits. Additionally, the accuracy of the calculations can be affected by factors such as temperature, frequency, and circuit complexity.