 or  # An electrical technician requires a capacitance of 2muF in a circuit across a potential difference of 1kV. A large number of 1muFcapacitors are available to him each of which can withstand a potential difference of not more than 400V. Suggest a possible arrangement that requires the minimum number of capacitors. Apne doubts clear karein ab Whatsapp par bhi. Try it now.
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Solution :
Here total capacitance, `C = 2 mu F` <br> Potential difference V = 1Kv = 1000 volt <br> Capacity of each capacitor `C_(1) = 1 mu F` <br> Maximum potential difference acorss each V = 400 volt <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/VIK_PHY_QB_C05_E05_013_S01.png" width="80%"> <br> Let n capacitors of `1 mu F` each be connected in series in a row and m such rows be connected in parallel as shown in the figure. As potential difference in each row = 1000 volt <br> `:.` Potential difference across each capacitor `= (1000)/(n) = 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 condensors of `1 mu F` <br> Each is series = 1/3 <br> Total capacitance of m such rows in parallel `= (m)/(3)` <br> `:. (m)/(3) = 2 (mu F)` or m = 6 <br> `n xx m = 3 xx 6 = 18`. <br> Hence `1 mu F` capacitors should be connected in six parallel rows, each row containing three capactiros in series. 