When an electric motor is used with an electric generator, the situation can be viewed as an RLC circuit where the generator is an AC power source. The source produces a sinusoidal AC voltage with an value of V and an angular frequency ω. The work done by the motor is reflected in the resistance R of the circuit. How does changing the angular frequency ω of the generator affect the current flowing in the circuit and the power transferred to the electric motor?
1 Answer
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Let's define inductive reactance $\chi_L=\omega L$ and capacitive reactance $\chi_c=\frac{1}{\omega C}$. Then, using resistance $R$, the impedance of the circuit is $$Z=\sqrt{R^2+(\chi_L-\chi_C)^2}$$ The rms current is $$I=\frac{V}{Z}$$ where $V$ is rms voltage.
The phase between the current and voltage is given by $$\tan\delta=\frac{\chi_L-\chi_C}{R}$$ Then, the power is simply $$P=IV\cos\delta$$ Inserting the initial variables, $$P=\frac{V^2}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}\cdot\sqrt{\frac{\omega L-\frac{1}{\omega C}-R}{R}}$$ The power is the highest when $\delta=0$ which happens when $\chi_L=\chi_C$.
