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\[1\mu F\]. \[\therefore \] \[\frac{1}{1}=\frac{1}{2}+\frac{1}{C}\] \[\therefore \] \[\frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2}\] \[C=2\mu F\]You need to login to perform this action.
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