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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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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%">