Help to understand Capacitor action??

Replies

• 

  Saandeep Sreerambatla

  Do you have anything else in the circuit?
• 

  sayandev

  sorry for late reply.
  no nothing else in the circuit.
• 

  sayandev

  hello?? please anyone help me to sort this out????
• 

  d_vipul

  Hey Friend,
  
  For AC voltages the CAPACITOR acts like a short circuit condition (It's Property)........
  It will directly shorten the AC VOLTAGE SUPPLY and allow to pass maximum current through it..........
  (Practical: If such condition remains for Long time Both Power Supply and Capacitor will burn)
  
  So the output voltage will be around zero.........
  Hope you get this.......If not then me ready to explain you more.......
  Keep on asking........
  
  Regards,
  VIPUL
• 

  sayandev

  Sir, I didn't get that. Actually what's my present concept regarding the capacitor, I am telling. If I am wrong just tell me. actually in this case, means in this circuit at 1st there was a 5V ac voltage source and a capacitor in series. Now during the +ve half cycle of the voltage, the capacitor must be charged upto +5V till 90 degree phase of ac source, and then from 90 to 180 degree phase the capacitor will start discharging till 0V. again during the negative half cycle, the capacitor will start charging upto -5V from 180 to 270 degree phase of ac source and then it will again discharge till 0V during 270 to 360 phase of ac. Now please tell me am I right in the concept??? then I will ask my next query.
• 

  Voltaire

  Hi
  On #-Link-Snipped-# it explains it quite nicely:
  
  
  

  Now suppose we apply a fixed voltage across the plates of our construction, as shown to the left. The battery attempts to push electrons onto the negative plate (blue in the figure), and pull electrons from the positive plate (the red one). Because of the large surface area between the two plates, the battery is actually able to do this. This action in turn produces an electric field between the two plates, and actually distorts the motions of the electrons in the molecules of air in between the two plates. Our construction has been given an electric charge, such that it now holds a voltage equal to the battery voltage. If we were to disconnect the battery, we would find that this structure continues to hold its charge — until something comes along to connect the two plates directly together and allow the structure to discharge itself.
  
  Because this structure has the capacity to hold an electrical charge, it is known as a capacitor. How much of a charge it can hold is determined by the area of the two plates and the distance between them. Large plates close together show a high capacity; smaller plates kept further apart show a lower capacity. Even the cut ends of the wire we described at the top of this page show some capacity to hold a charge, although that capacity is so small as to be negligible for practical purposes.
  
  The electric field between capacitor plates gives this component an interesting and useful property: it resists any change in voltage applied across its terminals. It will draw or release energy in the form of an electric current, thus storing energy in its electric field, in its effort to oppose any change. As a result, the voltage across a capacitor cannot change instantaneously; it must change gradually as it overcomes this property of the capacitor.

  My understanding of the above is that the DC cap will simply hold charge, i.e. no actual current flow, until a load is applied. The AC cap works differently because the alternating current unloads the cap and then you get the phase shift (current moves out of phase with voltage) and capacitive reactance (in Ω).
• 

  Voltaire

  To make this more understandable for d_vipul:
  ..............and...............read............etc.
  then.............understand...............
  ...........
• 

  sayandev

  I got it for dc. now can u please just tell for ac ??????
• 

  Voltaire

  Because A(lternating)C(urrent) alternates from minimum to maximum the amps and volts alternate according to a sine curve like this:
  
  When the power is generated the amps and the volts are in phase. That means that the current and the potential lie on the same curve so that the power is always at a maximum (can you still remember that P = VI).
  When a capacitor or inductor is installed in an AC system it causes the amps and volts to move along different curves. In capacitive circuits the current leads the voltage and in inductive circuits the current lags behind the voltage. Look at this graph
  
  imagine that the red line is voltage and the blue line is current. The power for this current will be less than the one above because VI is out of phase viz. in the top graph at π/2, V = 1 and I = 1 hence VI = 1 but in the bottom graph at π/2 V =1 and I =0 so VI = 0. At π both graphs show zero power i.e. either V or I or both V and I are zero hence VI = 0.
  
  Now
  You remember from DC that V = IR. In AC circuits with capacitors it works similarly but just a bit different. In AC, in stead of resistance, we talk about reactance and impedance. There are two types (maybe more): capacitive reactance and inductive reactance. These two make up the impedance for the circuit - this is resistance. Impedance means to hold someone or something back - in this case it is the current. Electrical Impedancedefines it as
  

  Electrical impedance, or simply impedance, describes a measure of opposition to alternating current (AC). Electrical impedance extends the concept of resistance to AC circuits, describing not only the relative amplitudes of the voltage and current, but also the relative phases. When the circuit is driven with direct current (DC) there is no distinction between impedance and resistance; the latter can be thought of as impedance with zero phase angle.

  The capacitor takes some time to charge and it also takes some time to discharge i.e. when the polarity of the voltage changes from negative to positive or vice versa. This is the time constant for the capacitor. You can see that the higher the frequency the less time the capacitor has to charge and discharge and the less reactance it will have. The less reactance it has the more power is in the circuit - similar to DC resistance.
  
  Here is an example
  Resistor 220 Ω
  Cap 0.47 mF (micro farads)
  Inductor 4.65E-05 mH (milli henry)
  f 50 Hz
  VRMS 220 V
  
  Calculations:
  ω = 2πf = 314.2
  Reactance:
  Xc = 6772.6Ω
  XL = 0Ω
  Impedance:
  Z = 6776.1Ω
  Total current:
  IRMS 0.0325 A
  
  Voltage drop (each element):
  VR 7.1 V
  VC 219.9 V
  VL 0.0 V
  
  Power factor:
  tanφ -30.78
  φ -88.1 degrees (Capacitive)
  
  Circuit power:
  P 0.23 W
  
  This is all due to the particle/wave duality of the electron. Remember that a current in a magnetic field results in a force and a force in a magnetic field results in a current.
• 

  sayandev

  thanks a lot for all the replies ! got my doubts cleared. thanks
• 

  sayandev

  now going back to my original question
  i.e. In a series ac capacitor circuit containing just one capacitor in series with a 5V ac voltage source. the capacitor is charged to 5V and then the ac voltage source is removed by a dc source of -4V. then what will be the voltage across the capacitor at last????
• 

  Voltaire

  4 Volts - given the question.
  What does it matter that the cap was in an AC circuit?
  What have you understood about all of this? Answer this question: what is τ for a cap? How does a thevinen equivalent work?
• 

  sayandev

  In a thevenin ckt. for capacitors the cap gets charged upto peak value of ac voltage and current will decrease to zero till 90 degree of ac voltage. then as the ac starts decreasing the capacitor discharges and voltage across it decreases and current increasestill 180 degree. again the capacitor stars charging upto negative peak value and the same case repeats.
  
  If this is not the answer then I think I didn't get your question?? sorry !!
  
  if I am wrong then just tell me.
• 

  Voltaire

  sayandev
  You really need to think about it before asking for an answer. What is the question in your book? Is this it or is the question you posted here your understanding of the question in the book?
  1. You say the cap is charged to 5V with AC. What is Ï„ for the cap and f for the current? I.e. is this even possible?
  2. How did you manage to disconnect the cap right when it was 5V?
  3. How many amps can the 5V deliver?
  4. What is the equivalent (thevinen) impedance and how did you calculate this?
  5. You assume that the cap will charge to full voltage. How do you know this?
  DC
  1. What do you know about caps without resistors in DC circuits? How is Ï„ calculated?
  
  Mainly: what does thevinen theory state about any Th%C3%A9Venin's Theoremof voltage sources, current sources, and resistors with two terminals?
• 

  sayandev

  actually I was studying a dual slope digital voltmeter. ther I read that the integrator circuit is charged at first upto the peak and then the switch switches to a reference voltage of opposite polarity. and the capacitor so starts discharging and when it reaches the selected voltage (usually taken ground potential) the stop pulse is generated. and with the help of the counter we determine the unknown voltage. so i asked that question in order to understand this.