 or  # An electric technician requires a capacitance of 2muF in a circuit across a potential difference of 1kV. A large number of 1muF capacitors 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 :
Let a combination of capacitors be arranged in which m rows with n capacitors each are connected in parallel. <br> `:. 400n = 1000" "[:.1kV= 1000 V]` <br> or, `n = (1000)/(400) = 2.5 approx 3` <br> The equivalent capacitance of each row <br> `C = (1)/((1)/(1)+(1)/(1)+(1)/(1))muF = (1)/(3)muF` <br> The equivalent capacitance of m such connected in parallel shoukd be `2muF`. <br> But `C_(eq) = mC :. m = (C_(eq))/(C )= (2)/(1/3)=6` <br> Total number of capacitors `= 3xx6 = 18` <br> They should be arranged in 6 rows having 3 in each row and connected in parallel. 