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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

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3 Answers 3

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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