# In an RLC series circuit on resonance, how can the voltages over the capacitor and the inductor be larger than the source voltage?

Consider an RLC circuit in series, of the form

If the source drives the circuit in AC at the resonance frequency $\omega =1/\sqrt{LC}$, the peak-to-peak voltages on the capacitor and the inductor, $$V_C=\left|\frac{Z_C}{Z_\mathrm{tot}}\right|V_S=\frac{\frac{1}{\omega C}}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S \quad \text{and}\quad V_L=\left|\frac{Z_L}{Z_\mathrm{tot}}\right|V_S=\frac{\omega L}{\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2}}V_S ,$$ can both be larger than the peak-to-peak voltage $V_S$ of the source.

The math might say one thing, but this is till terribly counterintuitive. How can this be?

• The sum of $V_C$ and $V_L$ can be greater than the supply voltage because the voltages are out of phase and don't simply add together. If you're claiming the voltages are individually larger than the supply voltage I think we'd need to see a circuit diagram showing an example. – John Rennie Dec 8 '15 at 7:58
• I'm sorry i missed the word RESONANCE. At resonance aren't Vl is equal to Vc so don't they cancel each other? – Emma Dec 8 '15 at 8:08
• Hi Emma, I've taken the liberty to beef up your post with some backing to answer @JohnRennie's concerns, trying to keep to your original intent. If I have misinterpreted it please feel free to rollback the changes or clarify what you mean to ask. – Emilio Pisanty Dec 8 '15 at 12:11
• Closely related by the OP: Parallel RLC circuit , how branching currents may each be larger than source current at resonance? – Emilio Pisanty Dec 8 '15 at 12:13