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