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Three capacitors A,B,C are connected in such a way that their equivalent capacitance is equal to the capacitance of B. The capacitances of A and B are 10 muF " and " 30muF respectively and C ne 0. Determine three possible values of C and also show how the capacitors are to be connected in the three cases. |
Answer» SOLUTION :The CAPACITOR B cannot be joined in parallel to any combination of the capacitors A and C because in that case equivalent capacitance will be greater than the capacitance of B. The following THREE combination are POSSIBLE.  i) A and B are connected in series. Now C is connected in parallel with the series combination of A and B. ACCORDING to the question, `(10xx30)/(10+30) +C= 30 " or, "(300)/(40)+C=30` or, `C= 22.5muF` ii) B and C are connected in series. Now A is connected in parallel with the series combination of B and C. According to the question, `(30xxC)/(30+C)+10=30 " or, "(30xxC)/(30+C)=20` or, `C 60muF` iii) A and B are connected in parallel. Now C is connedted in series with the parallel combination of A and B. According to the question, `(40xxC)/(40+C)=30 " or "3(40+C) =4C " or "C= 120muF`. |
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