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I am trying the following problem. A resistor with $295\Omega$ and an inductor are connected in series across an AC source that has voltage amplitude of $550V$. The rate at which electrical energy is dissipated in the resistor is $224W$.

What is the impedance of the circuit.

I tried this: $224W=I^2R$, from which $I=.87A$. Then, we have that $V=IZ$ in an AC circuit, so $550V=.87A(Z)$, and from here I get that $Z=631.17\Omega$.

I was wondering what is wrong here. Am I using the correct formulas?

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closed as off-topic by Nathaniel, Chris White, Manishearth Jul 1 '13 at 11:22

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

As per our recommendation on asking homework questions, "It's not enough to just show your work and ask where you went wrong. If you just need someone to check your work, you can always seek out a friend, classmate, or teacher. As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on." – Chris White Jul 1 '13 at 4:53

Power formulas for DC circuits is not correct in AC circuits unless you use root mean squared voltages and/or currents. So $I_{RMS} = 0.87A$ and $V_{RMS} = V_{MAX}/\sqrt{2} = 389V$ and the impedance is $Z = {V_{RMS} \over I_{RMS}} = 447 \Omega $

Alternatively you could have computed maximum current from RMS current and find the impedance with maximum voltage and maximum current.

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Attention that 1.X is the impedance of the inductor

where X is the impedance of the inductor 2.The source is an AC, so you have to calculate the effective voltage it provides, which is to
divide square root of two from the MAXIMUM voltage amplifier. In your formula V=IZ, V should be calculated in this way.

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for solving equations with AC sources one of the way is using the RMS values and then calculating.. the values

but a more appropriate method to solve this question will be with the use of phasors.

you can easily make the phasor diagram of an LC circuit and then determine the values as per need... :)

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I'm not sure this answer helps much. Please also see the page: which help you increasing the quality of your answer. – FraSchelle Jun 29 '13 at 17:20

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