A) \[51\,\mu F\]
B) \[102\,\mu F\]
C) \[204\,\mu F\]
D) \[408\,\mu F\]
Correct Answer: C
Solution :
Capacitors\[20\,\mu F\]and\[5\,\mu F\]are connected in series combination hence, their equivalent capacitance is given by \[{{C}_{S}}=\frac{20\times 5}{20+5}\] \[=\frac{100}{25}\] \[=4\mu F\] Again\[{{C}_{s}}=4\mu F\]and\[200\mu F\]are connected in parallel hence, their equivalent capacitance is \[{{C}_{p}}=4+200=204\mu F\]
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