- #1
pyroknife
- 611
- 4
I attached the link to the problem. Shouldn't the Vlmax be 330V? I tried entering that and it wasn't accepting my answer.
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The discussion revolves around calculating peak EMF in an AC circuit, specifically questioning whether the peak voltage (Vlmax) should be 330V. Participants clarify that when working with AC sources, the RMS voltage is typically used, and the relationship V_peak = √2 * V_rms is emphasized. The original poster confuses the peak voltage definition with the EMF equation, leading to misunderstandings about the circuit's behavior. It is noted that in a series circuit with a resistor and inductor, the peak voltage across components must be less than the supply's peak due to reactance. Understanding these principles is crucial for accurate calculations in AC circuits.
- #2
sandy.bridge
- 797
- 1
Where is the link?
- #3
- #4
sandy.bridge
- 797
- 1
When dealing with ac sources you want to be dealing with the rms voltage.
V_{peak}=\sqrt{2}V_{rms}
V_{peak}=\sqrt{2}V_{rms}
- #5
pyroknife
- 611
- 4
sandy.bridge said:
Well that's the 2nd part of part b where it asks for the rms Voltage. The 1st part is asking for the peak. I thought the peak was defined by emf=emfpeak * cos(wt+weird symbol) and in this case wouldn't emf peak just be 330V?
- #6
gneill
Mentor
- 20,989
- 2,934
You have a series circuit (resistor and inductor), so it's a voltage divider. The peak voltage across either component must be less than the voltage supply's peak. You'll have to deal with the reactance of the inductor (or its impedance if you're familiar with that).
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