Question regarding alternating current

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SUMMARY

The discussion centers on the behavior of alternating current (a.c.) in resistive circuits when the frequency is doubled while maintaining the same peak voltage. The consensus is that the average power dissipated, calculated using the formula P = (I_rms)^2 R, remains unchanged because the root mean square (r.m.s.) voltage is unaffected by frequency in purely resistive circuits. However, it is noted that this conclusion holds true only in the absence of inductance, as inductive reactance introduces frequency dependence that can alter current flow.

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  • Familiarity with Ohm's Law and power calculations
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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.

AM
 

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