A) \[40\,\,\mu F\]
B) \[20\,\,\mu F\]
C) \[13.3\,\,\mu F\]
D) \[10\,\,\mu F\]
Correct Answer: C
Solution :
The given circuit consists of three capacitors in series, connected to fourth capacitor in parallel. \[\therefore \] \[\frac{1}{{{C}_{1}}}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}\] \[\Rightarrow \] \[\frac{1}{{{C}_{1}}}=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}=\frac{3}{10}\] \[\Rightarrow \] \[{{C}_{1}}=\frac{10}{3}\mu F\] This is in parallel to \[10\mu F\] capacitor, hence, \[{{C}_{2}}=\frac{10}{3}+10=\frac{40}{3}=13.3\,\,\mu F\]
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