question_answer

Minimum number of 8\[{{m}_{1}}{{m}_{2}}\] and 250 V capacitors are used to make a combination of \[\therefore \] and 1000 V are :

A) 4              

B)                        32                          

C)        8                             

D)         3

Correct Answer: B

Solution :

                Let m rows of n series capacitor be taken then minimum number of capacitors required is \[6.31\times {{10}^{-3}}wb\] Also effective voltage is \[2.5\times {{10}^{-7}}wb\] \[\lambda \]  \[\frac{g}{6}\] Also these four capacitor are connected in series then effective capacitance is \[\frac{{{C}_{p}}}{{{C}_{v}}}=\gamma \] \[\sqrt{6}\]   \[\sqrt{3}\] \[n=\frac{1}{\sin \,C},\]   \[m=\frac{{{m}_{0}}}{\sqrt{1-\frac{{{v}^{2}}}{{{c}^{2}}}}}\] \[\omega \]   \[y=a\sin (\omega t-kx)\] Hence,      \[y=10\sin \pi (0.02x-2t)\]