The Series RLC Circuit — Question 6

Question 6

Two equal-current frequencies locate resonance without a power calibration

A fixed-amplitude source drives a series RLC circuit. Equal current amplitudes are observed at distinct positive frequencies $\omega_a<\omega_b$. Assume R, L, and C are frequency-independent.

1. Prove that resonance is the geometric mean of the two frequencies.

2. Explain algebraically why this pair alone does not determine R.

Question 6 - Solution

Two equal-current frequencies locate resonance without a power calibration

(a) Method and calculation

$$X(\omega)=\omega L-1/(\omega C),\quad X(\omega_a)^2=X(\omega_b)^2.$$$$X(\omega_a)=-X(\omega_b)\Rightarrow L(\omega_a+\omega_b)=\frac1C\left(\frac1{\omega_a}+\frac1{\omega_b}\right).$$

(b) Interpretation and further calculation

$$\boxed{\omega_a\omega_b=1/(LC),\quad\omega_0=\sqrt{\omega_a\omega_b}}.$$$$\boxed{R^2+X_a^2=R^2+X_b^2\ \text{eliminates }R}.$$

Physical check

Net reactance is strictly increasing with frequency, so distinct equal-amplitude points must lie on opposite sides of resonance.