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Capacitance






6. Capacitors in series and parallel combinations



    For prectical applications , two or more capacitors are often used in combination and their total capacitance C must be known.To find total capacitance of the arrangement of capacitor we would use equation
         Q=CV
    (i) Parallel combination of capacitors
  • Figure below shows two capacitors connected in parallel between two points A and B
    Parallel combination of capacitors

  • Right hand side plate of capacitors would be at same common potential VA. Similarly left hand side plates of capacitors would also be at same common potential VB.
  • Thus in this case potential difference VAB=VA-VB would be same for both the capacitors, and charges Q1 and Q2 on both the capacitors are not necessarily equal. So,
         Q1=C1V and Q2=C2V
  • Thus charge stored is divided amongst both the capacitors in direct proportion to their capacitance.
  • Total charge on both the capacitors is,
         Q=Q1+Q2
          =V(C1+C2)
         
    and
         Q/V=C1+C2                              (8)
    So system is equivalent to a single capacitor of capacitance
         C=Q/V
    where,
  • When capacitors are connected in parallel their resultant capacitance C is the sum of their individual capacitances.
  • The value of equivalent capacitance of system is greater then the greatest individual one.
  • If there are number of capacitors connected in parallel then their equivalent capacitance would be
         C=C1+C2+ C3...........               (10)

    (ii) Series combination of capacitors
  • Figure 7 below shows two capacitors connected in series combination between points A and B.
    Series combination of capacitors

  • Both the points A and B are maintained at constant potential difference VAB.
  • In series combination of capacitors right hand plate of first capacitor is connected to left hand plate of next capacitor and combination may be extended foe any number of capacitors.
  • In series combination of capacitors all the capacitors would have same charge.
  • Now potential difference across individual capacitors are given by
         VAR=Q/C1
    and,
         VRB=Q/C2
  • Sum of VAR and VRB would be equal to applied potential difference V so,
         V=VAB=VAR+VRB
          =Q(1/C1 + 1/C2)

    or,

    where

    i.e., resultant capacitance of series combination C=Q/V, is the ratio of charge to total potential difference across the two capacitors connected in series.
  • So, from equation 12 we say that to find resultant capacitance of capacitors connected in series, we need to add reciprocals of their individual capacitances and C is always less then the smallest individual capacitance.
  • Result in equation 12 can be summarized for any number of capacitors i.e.,



7. Energy stored in a capacitor



  • Consider a capacitor of capacitance C, completely uncharged in the begning.
  • Charhing process of capacitor requires expanditure of energy because while charging a capacitor charge is transferred from plate at lower potential to plate at higher potential.
  • Now if we start charging capacitor by transporting a charge dQ from negative plate ti the positive plate then work is done against the potential difference across the plate.
  • If q is the amount of charge on the capacitor at any stage of charging process and φ is the potential difference across the plates of capacitor then magnitude of potential difference is φ=q/C.
  • Now work dW required to transfer dq is
    dW=φdq=qdq/C
  • To charge the capacitor starting from the uncharged state to some final charge Q work required is
    Integrating from 0 to Q
         W=(1/C)∫qdq
         =(Q2)/2C      
    (14a)
         =(CV2)/2
         =QV/2


    Which is the energy stored in the capacitor and can also be written as
         U=(CV2)/2 ---(15)

  • From equation 14c,we see that the total work done is equal to the average potential V/2 during the charging process ,multiplied by the total charge transferred
  • If C is measured in Farads ,Q in coulumbs and V in volts the energy stored would in Joules
  • A parallel plate capacitor of area A and seperation d has capacitance

    C=ε0A/d

  • electric field in the space between the plates is
    E=V/d or V=Ed

    Putting above values of V and C in equation 14b we find
    W=U=(1/2)(ε0A/d)(Ed)2
    =(1/2)ε0E2(Ad)
    =(1/2)ε0E2.V
    ---(16)

  • If u denotes the energy per unit volume or energy density then
    u=(1/2)ε0E2 x volume
  • The result for above equation is generally valid even for electrostatic field that is not constant in space.




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Class 12 Maths Class 12 Physics