question_answer

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

A)  \[\text{2}\text{F}\]                                       

B)  \[\text{4}\text{F}\]

C)  \[\text{3}\text{F}\]                       

D)         \[\text{6}\text{F}\]

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,                 \[\frac{1}{C\,\,}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}\]                 \[C\,\,=1\mu F\] Capacitors \[{{C}_{6}}\] and \[{{C}_{8}}\] 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 \[{{C}_{5}}\] 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,                 \[\frac{1}{{{(C)}^{5}}}=\frac{1}{2}+\frac{1}{2}\] 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\[1\mu F\]. \[\therefore \]  \[\frac{1}{1}=\frac{1}{2}+\frac{1}{C}\] \[\therefore \]  \[\frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\]                 \[C=2\mu F\]