Series and parallel circuits

Replies

• 

  reachrkata

  There are 2 ways to understand this -
  
  1) Consider the voltage equation of a cap V = 1/C * (integral) i.dt. if there are 2 caps in series you have V1 = 1/C1 * (integral) i.dt and V2 = 1/C2 * (integral) i.dt. Total voltage V = V1+V2 = (1/C1+1/C2) * (integral) i.dt. Now if we have an effective cap Ceff such that the same current i flows through it, V = 1/Ceff * (integral) i.dt. Equating these two you get 1/Ceff = 1/C1 + 1/C2.
  
  2) Adding capacitors in series is like having a single cap with the same plate area as the individual caps and distance between plates as sum of distances of individual caps. That is if C1 = EA/d1, C2 = EA/d2, Ceff = EA/(d1+d2) where E = dielectric constant.
  Solve th above equations and you again get 1/Ceff = 1/C1 + 1/C2
  
  
  For paralle circuits, instead of adding voltages you have to add currents (I = CdV/dt) and you get Ceff = C1+C2.
  
  -Karthik
  
• 

  harish3h

  What is the need for considering effective capacitance/resistance/inductance?
  
  In what way does it help building the circuits..
  
  Why do we calculate it?
• 

  d_vipul

  hey harish3h,
  
  actually the response of the whole circuit is depend upon each parameter of circuit.........
  so while designing it kept into mind......
  
  Like in OSCILLATOR circuits 3 capacitors are used because one capacitor gives you 60 degree phase shift where as a combination of three gives you 180 degree phase shift.....
  
  Hope this helps you......
  
  Regards.
  VIPUL
• 

  Harshad Italiya

  reachrkata

  There are 2 ways to understand this -
  
  1) Consider the voltage equation of a cap V = 1/C * (integral) i.dt. if there are 2 caps in series you have V1 = 1/C1 * (integral) i.dt and V2 = 1/C2 * (integral) i.dt. Total voltage V = V1+V2 = (1/C1+1/C2) * (integral) i.dt. Now if we have an effective cap Ceff such that the same current i flows through it, V = 1/Ceff * (integral) i.dt. Equating these two you get 1/Ceff = 1/C1 + 1/C2.
  
  2) Adding capacitors in series is like having a single cap with the same plate area as the individual caps and distance between plates as sum of distances of individual caps. That is if C1 = EA/d1, C2 = EA/d2, Ceff = EA/(d1+d2) where E = dielectric constant.
  Solve th above equations and you again get 1/Ceff = 1/C1 + 1/C2
  
  
  For paralle circuits, instead of adding voltages you have to add currents (I = CdV/dt) and you get Ceff = C1+C2.
  
  -Karthik
  

  Superb Explanation !
• 

  reachrkata

  Yo thanks !!!
• 

  shreyasm89

  @karthik- best explanation...concise, but complete.😀
• 

  sachleen

  hey. #-Link-Snipped-#
  
  best explanation!!!!!!!!