Driven RLC Resonance and Phasors
Drive a series resistor, inductor and capacitor with an alternating voltage and the circuit answers most strongly at one special frequency. The impedance is $Z = R + i(\omega L - 1/\omega C)$: the inductor and capacitor reactances fight each other, and at $\omega_0 = 1/\sqrt{LC}$ they exactly cancel, leaving only $R$. There the current is largest, $V_0/R$, and it swings in phase with the drive. The phasor diagram tells the whole story: $V_R$ points along the current, $V_L$ leads it by 90 degrees, $V_C$ lags by 90 degrees, and they add tip-to-tail to the source voltage. The sharpness of the resonance is the quality factor $Q = (1/R)\sqrt{L/C}$, which is also how many times larger $V_L$ and $V_C$ grow than the source at resonance (they are equal and opposite, each $Q$ times $V_0$), and it sets the bandwidth $\Delta\omega = \omega_0/Q$. This single peak is how every radio tunes to one station.
WHAT TO TRY
- Sweep the drive frequency $f$ across the peak: the current marker climbs to the top of the resonance curve at $f_0$, and in the phasor diagram the reactive part shrinks to zero and the source lines up with $V_R$.
- Drop the resistance $R$: the peak gets taller and narrower, $Q$ rises, and $V_L$ and $V_C$ balloon far beyond the source voltage at resonance.
- Below $f_0$ the capacitor wins (the phase is negative, the circuit is capacitive); above $f_0$ the inductor wins (the phase is positive).
- Change $L$ or $C$ and watch $f_0 = 1/(2\pi\sqrt{LC})$ slide along the frequency axis.