The voltage source produces a sinusoidal AC voltage. An electrical device with resistance and inductance is connected to the source. How does the actual power of the device depend on the angular frequency of the voltage?
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$\begingroup$ Are the resistor and inductor connected in series? $\endgroup$– Bob DCommented Apr 5, 2022 at 13:29
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3 Answers
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The circuit here is a RL circuit.
Let's define inductive reactance: $$\chi_L=\omega L$$ Then, as shown here, the phase between current and voltage is $$\theta=\arctan\frac{\omega L}{R}=\arctan\frac{\chi_L}{R}$$
The impedance of a RL circuit is $$Z=\sqrt{R^2+\chi_L^2}$$ Using this we can find $$I_0=\frac{V_0}{Z}$$ And the power is $$P=\frac{1}{2}I_0V_0\cos\theta$$ Substituting given quantities, we arrive to $$P=\frac{V^2R}{R^2+\omega^2L^2}$$ where $V$ is the effective voltage.
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In DC circuits the power is defined as voltage times current
$$P_\text{dc} = V_\text{dc} \cdot I_\text{dc}$$
Since voltage and current are not constant in AC systems, the power is also not constant. The instantaneous power is then defined as
$$p(t) = v(t) \cdot i(t)$$
The above equation does not give a single number but a time-dependent waveform. If you want single number, then you need to decide which power component are you actually interested in: apparent, active, or reactive power.
The power in AC circuits can be represented as a vector quantity
$$S = V I^* = P + jQ$$
where $S$ is apparent power, $P$ is active power, $Q$ is reactive power, while $V$ and $I$ are voltage and current phasors which are also vector quantities. Note that in phasor form, $V$ and $I$ denote RMS values of the corresponding voltage and current waveforms.
Therefore, to determine the voltage source power, you need to determine the load current which is
$$I = \frac{V}{Z}$$
where $Z$ is the frequency-dependent load impedance. In your post you mention that the load is a combination of a resistor $R$ and an inductor $L$, but you do not specify if they are connected in series or parallel
$$Z_\text{ser} = R + j \omega L, \qquad Z_\text{par} = \frac{j \omega L R}{R + j \omega L} = \frac{\omega L R}{R^2 + (\omega L)^2} (\omega L + j R)$$
where $\omega = 2 \pi f$ is the source angular frequency. The current can be found as
$$I = \frac{V}{Z} \cdot \frac{Z^*}{Z^*} = \frac{V}{|Z|^2} Z^*$$
The AC power for the two cases is
$$S_\text{ser} = \underbrace{\frac{V^2 \cdot R}{R^2 + (\omega L)^2}}_{P} + j \underbrace{\frac{-V^2 \cdot \omega L}{R^2 + (\omega L)^2}}_{Q}, \qquad S_\text{par} = \underbrace{\frac{V^2}{R}}_{P} + j \underbrace{\frac{-V^2}{\omega L}}_{Q}$$
answered Apr 5, 2022 at 13:37
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Increasing the frequency increases the inductive reactance. If the inductor is in series with the resistor, that means the current is reduced as the frequency increases. Since actual (real) power is $I^{2}R$, the power is reduced.
The inductor complex impedance is, where $f$ is the frequency of the voltage,
$$Z_{L}=j2\pi fL$$
The equivalent series impedance with the resistor is
$$Z_{eq}=R+j2\pi fL$$
Its magnitude is
$$Z=\sqrt {R^{2}+(2\pi fL)^2}$$
The magnitude of the current is, from Ohm's law where $I=V/Z$,
$$I=\frac{V}{\sqrt {R^{2}+(2\pi fL)^2}}$$
Since the power dissipated in the resistor is $I^{2}R$
$$P=\frac{V^{2}R}{R^{2}+(2\pi fL)^2}$$
Note if $f\rightarrow0$
$$P\rightarrow\frac{V^2}{R}=\frac{(IR)^2}{R}=I^{2}R$$
And if $f\rightarrow\infty$
$$P\rightarrow 0$$
Hope this helps
