A) \[\frac{7}{11}\mu F\]
B) \[\frac{6}{5}\mu F\]
C) \[4\mu F\]
D) \[\frac{7}{10}\mu F\]
Correct Answer: A
Solution :
\[{{C}_{eq}}=0.5\mu F=\frac{\left( 1+\frac{4}{3} \right)C}{\frac{7}{3}+C}=\frac{7C}{\frac{7}{3}+C}=\frac{1}{2}\] |
\[\Rightarrow \] \[\frac{14C}{3}=C+\frac{7}{3}\] |
\[C=\frac{7}{11}.\] |
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