question_answer

A network of four \[20\mu F\]capacitors is connected to a 600V supply as shown in the figure. The equivalent capacitance of the network is:

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, \[{{C}_{eq}}={{C}_{s}}+{{C}_{4}}=\frac{20}{3}+20=\frac{80}{3}=26.67\mu F\]