# Voltage across capacitor/inductor in RLC circuit

It is said that voltage across capacitor/inductor is higher than the applied AC voltage. How?

Consider a series LCR circuit with an a.c. voltage, $V=2\:\small V$ applied. Let the resistance be $R=1\:\Omega$, and let the reactances be $X_L=X_C=10\:\Omega$, which means that the complex reactances are $j10$, and $-j10$.
The series circuit has an impedance: $$Z=R+jX_L-jX_C=1+j10-j10=1+j0$$
That is, $Z$ is a resistance of $1\:\Omega$ and the current through the R, L and C is: $$I=\frac{V}{Z}=\frac{V}{R}=\frac{2}{1}=2 A$$ Hence, the voltages across $L$ and $C$ will be: $j I X_L= j20\: V$, and $-j I X_C= -j20\: V$, respectively.
Therefore the capacitor and inductor each have voltages of $20\:V$, but these are $180°$ out of phase so they cancel each other when looking at the total voltage across the series configuration.