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An electrical technician requires a capacitance of 2 muF in a circuit across a potential difference of 1 kV. A large number of 1 mu F 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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Solution :
Total required capacitance , C=`2muF` <br> potential difference, V=1kv =1000 v <br> Capacitance of each capacitor , `C_(1)=1 muF` <br> each capacitor can withstand a potential difference `V_(1)=400 V` <br> Suppose a number of capacitors are connected in series and these series circuits are connected in parallel (row) to each other. the potential difference across each row must be 1000 V and potential difference across each capacitor must be 400 V. hence,number of capacitors in each row is given as <br> `1000/400=2.5` <br> Hence, there are three capacitors in each row. <br> Capacitance of each row `=1/(1+1+1)=1/3 muF` <br> Let there are n rows each hacing three capacitors, which are connected in parallel. <br> Hence, equivalent capacitance of the circuit is given as <br> `1/3+1/3+1/3+....n ` terms <br> `=n/3` <br> However, capacitance of the circuit is given as 2`muF` <br> `:. n/3=2` <br> `n=6` <br> Hence, 6 rows of three capacitors are present in the circuit. a minimum of 6 x 3 i.e., 18 capacitors are required for the given arragements.

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