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A capacitance of 2 muF is required in an electrical circuit across a potential difference of 1.0 kV A large number of 1 muF capacitors are available which can withstand a potential difference of not more than 300 v. <br> The minimum number of capacitors required to achieve this is

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Answer Text

2432216

Answer :
B
Solution :
Capacity of each condenser `C = 1 muF`. <br> Potential difference it can withstand `= 300V` <br> Required capacitanc e `= 2 muF` across a potantial difference of `1.0 kV i.e., 1000V` <br> Minimum no. of capacitance in series. <br> `x = (1000)/(300) = 3.3 rarr 4` <br> Capacitance of each row, `(1)/(C_(s)) = (1)/(1) + (1)/(1) + (1)/(1) + (1)/(1)` <br> `C_(s) = (1)/(4) muF` <br> If `m` such rows of capacitance are required, then <br> `C_(eq) = 2 = m C_(s) = m xx (1)/(4)` or `m = 8` <br> Hence minimum number of capacitors <br> `= 8xx4 = 32` <br> The arrangement si shown in Fig here, <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PR_XII_V01_C01_E01_350_S01.png" width="80%">

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