question_answer

The equivalent capacitance for the network shown in the figure is:

A) \[\frac{1200}{7}pF\]                  

B) \[\frac{1000}{4}pF\]

C) \[\frac{1800}{7}pF\]                  

D) \[\frac{1300}{3}pF\]

Correct Answer: A

Solution :

(a) \[\frac{1200}{7}pF\] Capacitors \[{{C}_{2}}\]and \[{{C}_{3}}\] are connected in series, equivalent capacitance, \[\frac{1}{C'}=\frac{1}{400}+\frac{1}{400}=\frac{2}{400}\] or \[C'=200\,pF\] Capacitors \[{{C}_{1}}\] and \[C'\]are in parallel then their equivalent capacitance,             \[C''=C'+{{C}_{1}}=200+100=300\,pF\] Capacitors \[C''\]and \[{{C}_{4}}\]are connected in series             Equivalent capacitance \[\frac{1}{{{C}_{eq}}}=\frac{1}{C''}+\frac{1}{{{C}_{4}}}=\frac{1}{300}+\frac{1}{400}\] \[\frac{1}{{{C}_{eq}}}=\frac{7}{1200}\] \[\therefore \,\,{{C}_{eq}}=\frac{1200}{7}pF\]