# 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