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 \[\frac{1}{{{C}_{eq}}}=\frac{1}{C''}+\frac{1}{{{C}_{4}}}=\frac{1}{300}+\frac{1}{400}\] \[\frac{1}{{{C}_{eq}}}=\frac{7}{1200}\] \[\therefore \,\,{{C}_{eq}}=\frac{1200}{7}pF\]You need to login to perform this action.
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