Tesla coil model

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I didn't have good intuition for how the parameters of a Tesla coil matter, so I made a toy model and studied how it depends on them. The model is two circuits coupled through the mutual inductance of the primary and secondary windings. Both sides are grounded to the same high-quality ground. (This is probably a poor model for a spark-gap coil, where the driving-circuit capacitance is large; here it's approximated as zero.)

The two coupled differential equations from the circuit are

$$V(t)=L_1\dot{I}_1+kM\dot{I}_2$$

and

$$0=\frac{q_2}{C_2}+L_2\dot{I}_2+R_2 I_2-kM\dot{I}_1$$

where $M=\sqrt{L_1 L_2}$ is the mutual inductance of a perfectly coupled pair of coils. Assuming $V(t)=V_0 e^{i\omega t}$ lets us write

$$V_0=i\omega L_1 I_{1,0}-i\omega kM I_{2,0}$$

and

$$0=\frac{-i I_{2,0}}{\omega C_2}+i\omega L_2 I_{2,0}+R_2 I_{2,0}+i\omega kM I_{1,0}.$$

Doing the algebra, the impedance $Z=V_0/I_{2,0}$ is

$$Z=i\omega\left[\frac{L_1}{kM\omega^2 C_2}+kM-\frac{L_1 L_2}{kM}\right]-\frac{R L_1}{kM}.$$

From the impedance we get the resonant frequency by minimizing $Z$ with respect to $\omega$:

$$\omega=\sqrt{\frac{1}{L_2 C_2\left(1-k^2\right)}}\quad\left[\frac{\text{rad}}{\text{s}}\right]$$

or

$$\boxed{\,f_0=\frac{\omega}{2\pi}=\frac{1}{2\pi\sqrt{L_2 C_2\left(1-k^2\right)}}\ \ [\text{Hz}]\,}$$

So the coupling constant $k$ tunes the resonant frequency of the coil. As Figure 1 shows, as long as $k$ is relatively small it barely affects the resonant frequency; likewise Figure 2 shows the quality factor is essentially unchanged for small $k$.

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