AC Voltage and current, amplitude, rms, peak value

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AC voltage amplitude is equivalent to peak voltage, representing the maximum deviation from the DC value. In current waveforms, peak current is typically used for sinusoidal or complex expressions, while RMS current is primarily for power dissipation calculations. Ohm's law can be applied consistently using either RMS or peak values, but one must remain consistent in their calculations. To derive output DC current from an AC waveform, one can calculate the average value over one cycle, often discernible by inspection. Generally, AC voltage and current values are provided as RMS by convention, but peak values are necessary for specific applications like dielectric strengths or nonlinear components.
Deathfish
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General enquiries (template may not be suitable)

- When the question provides value of AC voltage amplitude, is this the same value of peak voltage?

- Also, in current waveform, when do we use peak current and rms current?

- In ohm's law equations we use rms current, so in the current waveform graph do we need to times square root of 2?

- If provided with current waveform, can we get output dc current from using the same equation for voltage? eg. 2 x Vmax / \pi, Vmax / \pi

Relevant Equations :

V (odc) = 2 x Vmax / \pi, V(odc) = Vmax / \pi,
V (peak) = Vrms x \sqrt{2}
 
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Deathfish said:
General enquiries (template may not be suitable)

- When the question provides value of AC voltage amplitude, is this the same value of peak voltage?
yes, voltage amplitude means maximum deviation from DC value, or half the peak-to-peak
Deathfish said:
- Also, in current waveform, when do we use peak current and rms current?
there isn't a set rule for this unfortunately, in my electrical engineering course it is much easier to use peak values for sinusoidal or complex expressions. only cases I have come across using RMS is calculation of power dissapation
Deathfish said:
- In ohm's law equations we use rms current, so in the waveform graph do we need to times square root of 2?
as long as you are consistent, ie use only rms or only peak values in your calculatoins, ohm's law will hold, just like converting all units to metric units
Deathfish said:
- If provided with current waveform, can we get output dc current from using the same equation for voltage? eg. 2 x Vmax / \pi, Vmax / \pi
i'm not sure what these voltage formalae are used for but to obtain DC current you can use the ac waveform and find its average value over one cycle, more often then not they are quite obvious from inspection
 
Unless otherwise specified, AC voltage and current values are given as RMS. This is convention.

Sometimes it is necessary to know about peak values when specifying things like dielectric strengths or insulation. Also, if nonlinear components are involved (like diodes) and time constants or triggering levels are involved, you may need to work with the actual voltage or current waveforms in their 'peak' form.
 

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