when two capacitors are connected in series, the effective capacitance is 2.4 micro farad and when connected in parallel, the effective capacitance is 10 micro farad. calculate the individual capacitances.

When two capacitors are connected in series then total capacitance i.e C is
                                                        1/C= 1/C1+1/C2
                                  where,
                                            C1 is the capacitance of 1st capacitor
                                            C2 is the capacitance of 2nd capacitor
                                    Given,
  C = 2.4 micro farad
  1/2.4 = 1/C1+1/C2
    C1× C2   =   2.4 C1 + 2.4 C2............(1)

Now if capacitors are connected in parallel,then total capacitance is i.e. C
  C= C1+C2
  given,  C = 10 micro farad
  10 = C1+ C2...................(2)
  Now, from 1st and 2nd equation,
  C1×C2 = 24
  i.e  C1 = 24/C2 ..............(3)
  now substitute value of C1 in equation 2nd,
  We get,
  C12 -10 C1 +24 = 0
  ⇨      C1  = 6 or 4
  also from 3rd equation,  C2  = 4 or 6
 
 
 

 

 
                           

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c1+c2= 10

also, c1c2/ c1+c2=2.4

or, c1c2= 2.4 c1+ 2.4 c2

thus either c1 is 6 or 4

implies either c2 is 4 or 6 respectively

or,

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