Question regarding alternating current

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When alternating current (a.c.) flows through a resistor, the average power dissipated remains constant if the peak voltage is unchanged, regardless of frequency changes. The mean power can be calculated using the formula P = (I_rms)^2 R, where I_rms is the root mean square current. Since the root mean square voltage is derived from mean power considerations and is independent of frequency, the average power dissipated remains the same. However, in circuits with inductance, frequency can affect the current due to inductive reactance. Therefore, the assumption holds true only in purely resistive circuits.
Sanosuke Sagara
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When an a.c. flows in a resistor,the mean power dissipated is P.If the frequency of the a.c. is doubled with the same value of peak voltage,what id the average power dissipated now ?

My solution :

P = (I_rms)^2 R

I assume that as the peak voltage remains the same and the mean power value doesn't influenced by the frequency of the a.c. , so the mean power remains the same, that is P.

Is my assumption right ?I hope that someone will help me figure out.Do comment if I have any error in my assumption.Thanks for anybody that spend some time on this question.
 
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The root mean square voltage is derived from mean power considerations. Hence it is independent of the frequency. As long as the peak voltage is the same, the r.m.s. voltage is the same, and hence the mean power output is the same.
 
Sanosuke Sagara said:
I assume that as the peak voltage remains the same and the mean power value doesn't influenced by the frequency of the a.c. , so the mean power remains the same, that is P.

Is my assumption right ?
Your assumption is correct, in a circuit with no inductance. Inductive reactance, which limits current, is frequency dependent.

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