wouldn’t calculating the current from RMS voltage through an inductive impedance mess up the math? I thought you had to use the source’s Peak voltage for that
@@torikenyon First, you wouldn't use a peak value for power unless you also had a correction factor for it. I have seen texts do that but I think it's better to stick with the basic definitions and go from there. Remember, RMS exists specifically so that you can perform power calculations. Second, if you divide an RMS voltage by a resistance, you get an RMS current. Whatever you start with (RMS, peak, p-p), that's what you get at the other end.
After searching a couple hours for an explaination i have found this helpful video thank u sir
Can’t thank you enough sir. Thank you so much
Glad you liked it.
Thankyou sirr
wouldn’t calculating the current from RMS voltage through an inductive impedance mess up the math? I thought you had to use the source’s Peak voltage for that
Or does the calculated current also become an RMS current value?
@@torikenyon First, you wouldn't use a peak value for power unless you also had a correction factor for it. I have seen texts do that but I think it's better to stick with the basic definitions and go from there. Remember, RMS exists specifically so that you can perform power calculations. Second, if you divide an RMS voltage by a resistance, you get an RMS current. Whatever you start with (RMS, peak, p-p), that's what you get at the other end.
@@ElectronicswithProfessorFiore thanks for the response!