Two Inductors

Let $\omega$ be the general symbol for the angular frequency. The impedance of the parallel $LC$ branch is:

$Z_p = \frac{j \omega L (1/j \omega C)}{ j \omega L + 1/j \omega C} = \frac{j \omega L}{1 - \omega^2 L C}$

For minimum lamp power, we want the parallel branch impedance to be infinite:

$1 - \omega^2 L C = 0 \\ \omega = \frac{1}{\sqrt{LC}}$

For maximum power, we want the total circuit impedance to be resistive (just the lamp). This means that the parallel branch must resonate with the other series inductor.

$\frac{j \omega L}{1 - \omega^2 L C} = - j \omega L \\ 1 - \omega^2 L C = -1 \\ \omega = \frac{\sqrt{2}}{\sqrt{LC}}$

Interestingly, there is also another frequency at which this happens. At $\omega = 0$ (DC), the two inductors act as short circuits, and the circuit impedance consists solely of the lamp impedance.

The graph below is a plot of lamp power vs. source angular frequency, for some randomly chosen circuit parameter values.

Assume that the lamp is purely resistive and has resistance $R$ , which remains constant with temperature. The brightness of the lamp is then proportional to the power it dissipates. Since the lamp's resistance is constant, power is proportional to $|I|^2$ or square of the amplitude of current passing through the lamp. Since this current $I$ is also the current through the circuit, then we have $|I|=\dfrac {|V|}{|Z|}$ where $V$ is the voltage of the variable frequency voltage source and $Z$ is the resultant impedance of the circuit. Then the brightness of the lamp is proportional to:

$\begin{aligned} |I|^2 & = \frac {|V|^2}{|Z|^2} = \frac {|V|^2}{\left|R+j\omega L + \frac 1{j \omega C}||j\omega L\right|^2} = \frac {|V|^2}{\left|R+j\omega L + \frac 1{j\omega C + \frac 1{j\omega L}}\right|^2} = \frac {|V|^2}{\left|R+j \left(\omega L + \frac {\omega L}{1-\omega^2LC}\right)\right|^2} \\ & = \frac {|V|^2}{R^2 + \left(\omega L + \frac {\omega L}{1-\omega^2LC}\right)^2} \end{aligned}$

We note that $|I|^2$ is minimum, that is the lamp is dimmest, when $|X| = \omega L + \dfrac {\omega L}{1-\omega^2LC}$ is maximum. That is when $1-\omega^2 LC = 0$ , then $|X| \to \infty$ , or $\omega = \dfrac 1{\sqrt{LC}}$ . And $|I|^2$ is maximum or the lamp is brightest, when $|X|$ is minimum or $|X| = 0$ for $\omega \ne 0$ (not DC), $\implies \omega_1 L + \dfrac {\omega_1 L}{1-\omega_1^2LC} = 0$ or $1-\omega_1^2LC = -1 \implies \omega_1^2 LC = 2 \implies \omega_1 = \sqrt{\dfrac 2{LC}} = \sqrt 2 \omega$ . Therefore $\alpha = \sqrt 2 \approx \boxed{1.41}$ .

The following is a plot of $|I(\omega)|^2$ against $\omega$ for $|V|=R=C=L=1$ in value. Note that $\omega = \dfrac 1{\sqrt{LC}} = 1$ and $\omega_1 = \sqrt 2$ .