capacitor

Asked by ajaykapil | 30th Oct, 2010, 11:03: AM

Expert Answer:

Dear Student,

 

Figure is for indicative purpose only.

Capacitances and voltage are as given in the query.

V = Q/C

CE=> 1, ED=> 2,  CF=>3,  FD=>4,  FE=> 5   (subscripts)

Because of symmetry, it is clear that:

q1 = q4, and q2=q3.

(We can get above result if we apply Kirchhoff’s Laws to various loops and nodes, and solve the equations.)

 

Applying Kirchhoff’s Law to ACEDB:

10 - q1 - (q2) /3 = 0                            I

 

At point E:

q5 + q1 = q2

q5 = - q1 + q2                                      II

  

Applying Kirchhoff’s Law to EDF:

(q5)/4 + (q2)/3 = q1

q5 = 4q1 - 4/3 (q2)                             . . . III

 

III  -  II  gives

q2 = 15/7 (q1)

  

Substituting in I :

10 - q1 - 5/7 (q1) = 0

Hence

 

q1 = 35/6 = 5.83 microC

q2 = 75/6 = 12.5 mcroC

As per II :

q5 = 40/6 = 6.67 microC

We hope that clarifies your query.
Regards
Team
Topperlearning

Answered by  | 6th Nov, 2010, 11:37: AM

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