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

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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.

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The reason for this phenomenon is called resonance, in this case between the capacitor and the inductor. When you apply DC voltage to a capacitor or an inductor they will store energy by forming an electric field. When you take away the voltage, the stored energy will be flowing back into the circuit. As mentioned, both a capacitor and an inductor will do this but they do this in a different way.

Now when you apply AC voltage the voltage will constantly change so the elements constantly store energy and release energy back into the circuit. Because a capacitor and an inductor do this at different times it is possible that they will for a resonating circuit, meaning that the energy of the capacitor flows into the inductor and vice versa. Because of this extra element of energy transfer on top of the supplied AC voltage, the voltage between these 2 elements will be higher than the AC voltage you applied.

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