A) \[30.26\mu F\]
B) \[20\mu F\]
C) \[26.67\mu F\]
D) \[10\mu F\]
Correct Answer: C
Solution :
(c) \[26.67\mu F\] From the figure, \[{{C}_{1}},{{C}_{2}},{{C}_{3}}\]are connected in series. \[\therefore \,\,\,\,\,\,\,\frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] \[=\frac{1}{20}+\frac{1}{20}+\frac{1}{20}=\frac{3}{20}\mu F\] Or \[{{C}_{s}}=\frac{20}{3}\mu F\] Now \[{{C}_{s}}\]is in parallel with \[{{C}_{4}}\] \[\therefore \]Equivalent capacitance, \[{{C}_{eq}}={{C}_{s}}+{{C}_{4}}=\frac{20}{3}+20=\frac{80}{3}=26.67\mu F\]You need to login to perform this action.
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