A) \[10\,\mu F\]
B) \[15\,\mu F\]
C) \[20\,\mu F\]
D) \[25\,\mu F\]
Correct Answer: B
Solution :
(b) \[15\,\mu F\] Here, \[{{C}_{1}}\] and \[{{C}_{2}}\] are in series, hence their capacitance is given by \[\frac{1}{C'}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{C'}=\frac{1}{20}+\frac{1}{20}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C'=10\,\mu F\] Similarly, \[\frac{1}{C''}=\frac{1}{{{C}_{3}}}+\frac{1}{{{C}_{4}}}\] \[\Rightarrow \,\frac{1}{C''}=\frac{1}{10}+\frac{1}{10}\,\,\Rightarrow C''=5\mu F\] Now, C and C" are in parallel. Hence, resultant capacitance C will be \[C=C'+C''=10+5+15\mu F\]
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