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An electric technicain requires a capacitance fo 2 muF in a circuit across a potential differences of 1 KV. A large number of 1 muF capacitors are available to him, each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires minimum number of capacitors.

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Answer Text

Solution :
Here. Total capacitance, `C = 2 mu F`, <br> Potential difference, `V = 1 kV = 1000 vol t` <br> Capacity of each capacitor, `C_(1) = 1 muF`, <br> Maximum potential difference across each. <br> `V' = 400 vol t` <br> Let n capacitors fo `1 muF` each be connected in series in a row and m such rows be connected in parallel as shown in Fig. <br> As potential difference acoross each row = 1000 volt <br> `:.` Potential difference across each capacitors `= (1000)/(n) = V' = 400` <br> `:. n = (1000)/(400) = 2.5` <br> As n has to be a whole number (not less than 2.5), therefore , `n = 3`. <br> Capacitance of each row of 3 conderes of `11 muF` each, in series `= 1//3`. <br> Total capacitance of m such rows in parallel `= (m)/(3) :. (m)/(3) = 2 (muF) or m = 6` <br> `:.` total number capacitors `= n xx m = 3xx6 = 18` <br> Hence `1 muF` capacitors should be connected in six parallel rows, each row contaning three capacitors in series. <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_XII_V01_C01_S01_592_S01.png" width="80%">

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