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## Homework Statement

A coil is connected to a 60 Hz ac generator with a peak emf equal to 80 V. At this frequency the coil has an impedance of 14 and a reactance of 12 .

a.) What is the peak current in the coil?

answer-5.71 Amps

b.)What is the phase angle between the current and the applied voltage?

answer-58.99 degrees

c.)A capacitor is put into series with the coil and the generator. What capacitance is required so that the current is in phase with the generator emf?

d.) What then is the peak voltage measured across the capacitor?

## Homework Equations

I=V/Z = 80/14=5.17

phase angle [tex]\delta[/tex]=arcsin(X/Z)=arcsin(12/14)=58.99

Resonance, X

_{L}=X

_{C}; [tex]\omega[/tex]

_{res}=1/((LC)^.5)) omega is angular Freq, L is Inductance, C is capacatance.

V

_{L}=I*X

_{C}

## The Attempt at a Solution

confused