# Capacitor voltage in LCR circuit greater than the supply - are my phasor calculations correct?

When calculating the voltage across the capacitor in the first diagram, It turned out to be greater than the power supply?

$$Z=R+j(X_L-X_C)$$ $$R=250\Omega$$ $$X_L=2\pi\times60\times0.65=245\Omega$$ $$X_C=\frac{1}{2\pi\times60\times1.5\times10^{-6}}=1768.38\Omega$$ $$Z=250-j1523.336\Omega=1543.71\angle-80.68^{\circ}\Omega$$ $$I=\frac{120\angle0^{\circ}}{1543.71\angle-80.68^{\circ}}=0.07773\angle80.68^{\circ}A$$

When I multiply $$I$$ by $$-jX_C$$

$$V_C=I\times -jX_C=0.07773\angle80.68^{\circ}\times 1768.38\angle-90^{\circ}=0.07773\times1768.38\angle80.68^{\circ}-90^{\circ}=137.45\angle-9.32^{\circ}V$$

Is this correct? And if so, why?

• Resonance. The L and C can resonate and you end up with a higher voltage than the supply. – Tom Carpenter May 4 '16 at 14:02
• So my answer is correct? – pkjag May 4 '16 at 14:05
• Calculate Vr and VL as well, then add all three up, see if it matches the supply. – Tom Carpenter May 4 '16 at 14:05
• I haven't calculated anything, but your workings out make sense. – Tom Carpenter May 4 '16 at 14:06
• For V_R=19.43∠80.68° V and V_L=19.05∠170.68° V – pkjag May 4 '16 at 14:07

Basically what you have is a resonance circuit. The inductor and capacitor oscillate together and as a result you can end up with a higher voltage across one or other element than comes from the supply.

You can check that everything makes sense by summing up all the voltages in the circuit and you should end up with the supply voltage. I checked and if you don't round the numbers in your calculations, the voltages do all sum back up to 120.

A way to immediately see that resonance is happening is to look at the phase angles of the voltages. Notice how the voltage across the inductor has a phase angle which is 180 degrees out of phase from the capacitor. This is due to the fact that current leads voltage in a capacitor by 90 degrees, but lags voltage in an inductor by 90 degress. As a result both elements in a series circuit will have voltages which are in anti-phase.