AC circuits need help on c.) and d.)

  • Thread starter hitman0097
  • Start date
  • #1
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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, XL=XC; [tex]\omega[/tex]res=1/((LC)^.5)) omega is angular Freq, L is Inductance, C is capacatance.


VL=I*XC

The Attempt at a Solution


confused
 

Answers and Replies

  • #2
1,860
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What's the impedance of a capacitor, and what do impedances do when they are added in series? What kind of impedance do you want to have a phase angle of zero?
 
  • #3
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Why don't people ever finish these forums, I have been stuck on this problem for an hour just part d.

A coil is connected to a 60 Hz, 100 V ac generator. At this frequency the coil has an impedance of 10 ohm and a reactance of 7.6 ohm.

(a) What is the current in the coil?

10 A

(b) What is the phase angle between the current and the applied voltage? (Calculate the angle in degrees.)

49.46 degrees

(c) What series capacitance is required so that the current and voltage are in phase?

349.02 micro FF

(d) What then is the voltage measured across the capacitor?


The impedence would be a total of XC + XL = Z, but then how is the voltage drop reached?
 

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