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An electric technicain requires a capacitance fo 2 muF in a circuit across a potentialdifferences of 1 KV. A large numberof 1 muF capacitors are available to him, each of which can withstanda potentialdifference of not more than 400V. Suggesta possiblearrangement that requiresminimum numberof capacitors. |
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Answer» SOLUTION :Here. Total capacitance,`C = 2 mu F`, Potentialdifference, `V = 1 kV = 1000 vol t` Capacity of eachcapacitor, `C_(1) = 1 muF`, Maximum potentialdifferenceacross each. `V' = 400 vol t` Let n capacitorsfo `1 muF` each be connectedin series in a row and m suchrows be connectedin parallel as shown in FIG. As potentialdifferenceacorosseach row = 1000 volt `:.` Potential DIFFERENCE across each capacitors`= (1000)/(n) = V' = 400` `:. n = (1000)/(400) = 2.5` As n has to be a whole NUMBER(not less than 2.5), therefore , `n = 3`. Capacitanceof eachrow of 3 conderes of `11 muF` each, in series `= 1//3`. Total capacitance of m such rows in parallel `= (m)/(3) :. (m)/(3) = 2 (muF) or m = 6` `:.` total number capacitors `= n xx m = 3xx6 = 18` HENCE `1 muF` capacitors should beconnectedin six parallel rows, each row contaning three capacitors in series. 
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