Power Dissipated By Resistor is LRC Circuit

[mj23]

[Note by mentor: this post does not use the homework template because it was originally posted in a non-homework forum.]

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Problem:
What is the average power dissipated by a 25-Ω resistor in an LRC series ac circuit for which the power factor is equal to 0.25 and the maximum voltage of the ac source is 8.0 V?

Answer: .08W

Attempt at Solution:
We know that Pav = (1/2)(V)(I)cos(phi) where cos(phi) is the power factor. So if we substitute I for V/I (ohms law) we get Pav = (1.2)((V^2)/R)cos(phi). Plugging in all the values I get that Pav = .32W which is clearly incorrect...

What did I do wrong?
Thanks for the advice!

 

[ehild]

[Note by mentor: this post does not use the homework template because it was originally posted in a non-homework forum.]

---------------------------------------------

Problem:
What is the average power dissipated by a 25-Ω resistor in an LRC series ac circuit for which the power factor is equal to 0.25 and the maximum voltage of the ac source is 8.0 V?

Answer: .08W

Attempt at Solution:
We know that Pav = (1/2)(V)(I)cos(phi) where cos(phi) is the power factor. So if we substitute I for V/I (ohms law) we get Pav = (1.2)((V^2)/R)cos(phi). Plugging in all the values I get that Pav = .32W which is clearly incorrect...

What did I do wrong?
Thanks for the advice!

Check your formulas. What is Ohm's Law for an AC circuit? How do you get the average power in terms of V_rms and R?

 

[mj23]

ok i understand i made a typo by saying substitute I in for V/I meant V/R as you can see in the next step, but why do we need to use the Vrms in this situation rather than V?

 

[ehild]

ok i understand i made a typo by saying substitute I in for V/I meant V/R as you can see in the next step, but why do we need to use the Vrms in this situation rather than V?

Well, you can use the maximum voltage and current if you like, but I=V/R is valid for the voltage across the resistor which is not the same as the given 8.0 V.