question_answer

If the equivalent capacitance between A and B is \[1\mu F,\] then the value of C will be

A)  \[2\mu F\]                                        

B)  \[4\mu F\]

C)  \[3\mu F\]                                        

D)  \[6\mu F\]

Correct Answer: A

Solution :

Capacitors \[{{C}_{3}}\]and \[{{C}_{4}}\] are in parallel, therefore their resultant capacitance, \[C'={{C}_{3}}+{{C}_{4}}=1+1=2\mu F\] Now, capacitors \[{{C}_{2}}\] and C' are in series, therefore their resultant capacitance,                 \[\frac{1}{C''}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\] Capacitors \[{{C}_{6}}\] and \[{{C}_{7}}\] are in series, therefore their resultant capacitance,                 \[\frac{1}{C'''}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1\]                 \[C'\,'\,'=1\mu F,\] Now, \[C'\,'\,'\] and \[{{C}_{7}}\] are in parallel, therefore their resultant capacitance,                 \[C'\,'\,'\,'=1+1=2\mu F,\] Now, \[C'\,'\,'\,'\] and \[{{C}_{5}}\] are in series. Therefore, their resultant capacitance,                 \[\frac{1}{{{(C)}^{5,}}}=\frac{1}{2}+\frac{1}{2}\]                 \[{{(C)}^{5,}}=1\mu F,\]                 Now, \[C''\] and \[{{(C)}^{5,}}\]are in parallel. Therefore, their resultant capacitance,                 \[{{(C)}^{6,}}=1+1=2\mu F\] Now, \[{{C}_{1}}\]and \[{{(C)}^{6,}}\] are in series and their resultant capacitance is given \[1\mu F\]. \[\therefore \]  \[\frac{1}{1}=\frac{1}{2}+\frac{1}{C}\] \[\therefore \]  \[\frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\]                 \[C=2\mu F\]