A)
B)
C)
D)
Correct Answer: A
Solution :
\[{{C}_{1}}\]is given by \[\frac{1}{{{C}_{1}}}=\frac{1}{3C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{1}{6}+\frac{4}{2}(\because C=2\mu F)\] \[{{C}_{1}}=\frac{6}{13}\mu F\] \[{{C}_{2}}\]is given by \[\frac{1}{{{C}_{2}}}=\frac{1}{5C}+\frac{2}{C}=\frac{1}{10}+1=\frac{11}{10}\] \[{{C}_{2}}=\frac{10}{10}\mu F\] \[{{C}_{3}}\]is given by \[\frac{1}{{{C}_{3}}}=\frac{1}{4C}+\frac{3}{C}=\frac{1}{8}+\frac{3}{2}=\frac{26}{16}=\frac{13}{8}\] \[{{C}_{3}}=\frac{8}{13}\mu F\] \[{{C}_{4}}\]is given by \[\frac{1}{{{C}_{4}}}=\frac{1}{2C}+\frac{5}{C}=\frac{1}{4}+\frac{5}{2}=\frac{11}{4}\] \[{{C}_{4}}=\frac{4}{11}\mu F\]
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