Questions on AC Power Homework: Ip = V/Z & P =(I^2)R

[influx]

Homework Statement

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I did not embed the image as its quite large

Homework Equations

N/A

The Attempt at a Solution

I'm just confused as to why Ip = V/Z (first red box)? I thought I = V/Z so why is it Ip (the peak value) in this case? As for the second red box, why have they divided it by two? Usually P =( I^2)R no?

Thanks

 

[NascentOxygen]

When a voltage is denoted as, e.g., 8∠40°, there can be some ambiguity as to whether the 8 is peak or RMS, and you must rely on other cues to guide you.

If the signal is written as an explicit time function, e.g., 8.cos(wt-40°) as seen in the lower left of the board, then the 8 there is definitely the peak value. So it seems your teacher was representing the phasor by its peak value when writing it in the concise amplitude∠angle form.

power = I² R where I is the RMS value

If you have the peak value, then divide each by √2, then it's all squared

 

[zoki85]

When a voltage is denoted as, e.g., 8∠40°, there can be some ambiguity as to whether the 8 is peak or RMS, and you must rely on other cues to guide you.

If nothing else is specified than the number is RMS value. That's convention in electrical engineering.

 

[influx]

When a voltage is denoted as, e.g., 8∠40°, there can be some ambiguity as to whether the 8 is peak or RMS, and you must rely on other cues to guide you.

If the signal is written as an explicit time function, e.g., 8.cos(wt-40°) as seen in the lower left of the board, then the 8 there is definitely the peak value. So it seems your teacher was representing the phasor by its peak value when writing it in the concise amplitude∠angle form.

power = I² R where I is the RMS value

If you have the peak value, then divide each by √2, then it's all squared

Does the I in the above equation always = the RMS I value or can it also equal Ip? As in can power ever = (Ip)² R ?

Also, Ip = 1.68<-25.4 A , so why is the angle ignored when substituting into the formula for power? (only the magnitude of 1.68 is subbed in?)

 

[NascentOxygen]

instantaneous power p(t) =i(t)² R
so whatever i(t) you use gives the power at that instant

If you need to determine the peak power in a resistive load, then use the peak current.

Usually we are interested in average power, and for that you use the RMS of the time variable. This means that any AC waveform of RMS value V volts gives the same heating in a resistor as does a DC voltage also of exactly V volts.

Power is a time average over one period, so any phase angle disappears.

 

[zoki85]

Does the I in the above equation always = the RMS I value or can it also equal Ip? As in can power ever = (Ip)² R ?

Equation power = I²R refers to the average power and I is RMS value. IF you know the current has a perfect sine waveform, than you can write power = I_p²R/2 . But if the current wave form is distorded (in most of the cases in reality) than with second equation you make error in calculation. OTOH, if you use RMS value , calculation for any waveform is correct.

Also, Ip = 1.68<-25.4 A , so why is the angle ignored when substituting into the formula for power? (only the magnitude of 1.68 is subbed in?)

Becouse voltage is in phase with current if the load is purely resisive.