Saved Bookmarks
| 1. |
An electric circuit requires a total capacitance of 2mu F across a potential of 1000V. Large number of 1muF capacitances are available each of which would breakdown if the potential is more than 350 V. How many capacitances are required to mae the circuit ? |
|
Answer» 24 Voltage rating of each capacitor = 350 V Supply voltage = 1000 V TOTAL capacitance `= 2mu F` Let n capacitors of `1mu F` each be connected in series in a row and m such rows be connected in parallel as SHOWN in the figure.  `because` Each capacitor can withstand 350 V. `:. n = (1000)/(350) = 2.8` As n cannot be fraction, therefore n = 3 Capacitance of each row of 3 capacitors of `1mu F` each in series is `C_(s) = (1)/(3) mu F` Total capacitance of m such rows in parallel `= (m)/(3) mu F` `:. (m)/(3) mu F = 2mu F` or m = 6 Total NUMBER of capacitors `= n xx m = 3 xx 6 = 18` |
|