A) \[4.7\,\mu F\]
B) \[1.2\,\mu F\]
C) \[3.2\,\mu F\]
D) \[2.2\,\mu F\]
Correct Answer: C
Solution :
Seeing in the given circuit \[{{C}_{1}}\] and \[{{C}_{2}}\] are connected in parallel. Hence, their equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=5+10\] \[=15\,\mu F\] As \[{{C}_{eq}}\] and \[{{C}_{3}}\] are connected in series, Hence, resultant capacitance between P and Q is given by \[\frac{1}{{{C}_{PQ}}}=\frac{1}{{{C}_{eq}}}+\frac{1}{{{C}_{3}}}=\frac{1}{15}+\frac{1}{4}=\frac{19}{60}\] \[\frac{1}{{{C}_{PQ}}}=\frac{60}{19}=3.2\mu F\]
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