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An electrical technician requires a capacitance of 2 uF in a circuit across a potential difference of 1 kV. A large number of 1 uF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors. |
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Answer» Solution :As potential difference required by the technician=1 kV = 1000 V and potential difference rating of each `1 MUF` capacitor `= 400 V` `:.` Minimum number of capacitros to be joined in SERIES `n= 1000/400 = 2.5` It means that atleast 3 capacitors should be joined in a series row, whose COMBINED capacitance will `be C/3 = (1muF)/3 = 1/3 muF` As technician requires a capacitance of `2 muF` , hence number of capacitor rows to be joined in parallel should be `m = (2muF)/(1/3muF)= 6` Minimum number of capacitors NEEDED by the technician `=nm = 3xx 6 =18`. The technician should arrange these 18 capacitors in 6 parallel rows, each row consisting of 3 capacitors in series. |
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