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Say I have an AC circuit with rms voltage $\mathcal{E}_{rms}$, reactance $X$ and resistance $R$. We may say that the average power dissipated in the circuit is as follows:

$$P_{avg} = \boxed{\mathcal{E}_{rms}i_{rms}} = \mathcal{E}_{rms}\left(\frac{\mathcal{E}_{rms}}{Z}\right) = \frac{\mathcal{E}_{rms}^2}{\sqrt{R^2 + X^2}}$$

where $Z$ is the impedance of the circuit. Now, the formula in my textbook hinges on the fact that any power dissipated will be through the joule heating only. Therefore, the average power may be written:

$$P_{avg} = i_{rms}^2R = \left(\frac{\mathcal{E}_{rms}}{Z}\right)^2R = \mathcal{E}_{rms}i_{rms}\frac{R}{Z}$$

This is clearly different from the boxed formula. Why is this? Is $P=VI$ not valid for ac sources? (I don't see any reason why that should be). Does the second formula disregard any energy losses? (through EM radiations, maybe). This is supported by the fact that the second formula reduces to the first when $R=Z$.

Also, this is a bit unrelated but what causes power to be dissipated in the primary coil of a transformer?

In a transformer, the primary coil is essentially a purely inductive ac circuit. How does the energy leave the circuit?

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We may say that the average power dissipated in the circuit is as follows:

But the equation following that isn't true.

In AC circuit analysis, the voltage and current are represented as phasors.

Further, $p=v \cdot i$ is the formula for instantaneous power in the time domain.

The average power, over some time $T$ in the time domain, is given by

$$p_{avg} = \frac{1}{T}\int_0^Tv(t) \cdot i(t)\, dt$$

For AC analysis, we assume sinusoidal excitation at one frequency. In that case, we can 'hide' the time dependence and work with just amplitude and phase, expressed as a complex number, of the sinusoidal voltages and currents.

Then, in this context, we can define the complex power $S$, measured in VA (volt-amps), as follows:

$$S = \tilde V\cdot \tilde I^* = P + jQ $$

Above, $\tilde V$ is the (rms) phasor voltage across and $\tilde I^*$ is the complex conjugate of the phasor current through.

The real part of $S$ is the average power $P$, measured in W (watts), while the imaginary part $Q$ is the reactive power and is measured in VAR (volt-amps reactive).

Let's do the math. Assume a phasor voltage $\tilde V$ across an impedance $Z = R + jX$

The (phasor) Ohm's law yields

$$\tilde I = \frac{V}{Z} = \frac{V}{R + jX}$$

Thus, the complex power is

$$S = \tilde V\cdot \tilde I^* = \frac{\tilde V \cdot \tilde V^*}{R - jX} = \frac{V^2}{R - jX}$$

We can write this in the form of $P + jQ$ as follows

$$S = \frac{V^2}{R - jX}\frac{R + jX}{R + jX} = \frac{V^2}{R^2 + X^2}\left(R + jX \right)$$

Thus, the average power is

$$P = \frac{V^2}{R^2 + X^2}R = \frac{V^2}{Z^2}R = I^2R $$


In a transformer, the primary coil is essentially a purely inductive ac circuit. How does the energy leave the circuit?

That's not true either. For an ideal transformer, the impedance looking into the primary is completely determined by the impedance of the circuit connected to the secondary.

Indeed, if the secondary circuit is open, the primary 'looks like' an open circuit.

If the secondary is terminated with an impedance $Z_s$, the impedance looking into the primary is

$$Z_p = n^2Z_s $$

where $n^2$ is the turns ratio.

For the ideal transformer, all of the energy delivered to the primary flows through to the secondary circuit.

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  • $\begingroup$ If you know the answer to the second part of my question, could you add it? $\endgroup$ – Gerard Jul 15 '14 at 13:48
  • $\begingroup$ @Gerard, see update. $\endgroup$ – Alfred Centauri Jul 15 '14 at 14:26

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