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, \[y=10\sin \pi (0.02x-2t)\]You need to login to perform this action.
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