A) 51\[\mu F\]
B) 102 \[\mu F\]
C) 204\[\mu F\]
D) 408 \[\mu F\]
Correct Answer: C
Solution :
In the given circuit, two capacitors of \[20\mu F\]and \[5\mu F\] are connected in series and their resultant is connected in parallel to \[200\mu F\]capacitor. \[\therefore \] \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}-\frac{1}{20}+\frac{1}{5}\] \[\frac{1}{C}=\frac{25}{20\times 5}\] \[{{C}_{s}}=\frac{20\times 5}{25}=4\,\mu F\] Resultant capacitance is \[{{C}_{p}}={{C}_{1}}+{{C}_{2}}\] \[=4+200=204\,\mu F\]You need to login to perform this action.
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