A) \[1.6\,\,\mu F\]
B) \[3.2\,\,\mu F\]
C) \[6.4\,\,\mu F\]
D) \[4.5\,\,\mu F\]
Correct Answer: B
Solution :
Capacitance \[{{C}_{1}}\] and \[{{C}_{2}}\] are connected in parallel. Hence, the equivalent capacitance \[{{C}_{eq}}={{C}_{1}}+{{C}_{2}}=5+10=15\,\,\mu F\] As \[{{C}_{eq}}\] and \[{{C}_{3}}\] are connected in series Hence, effective capacitance between \[P\] and \[Q\] is given by \[\frac{1}{{{C}_{eff}}}=\frac{1}{{{C}_{eq}}}+\frac{1}{{{C}_{3}}}=\frac{1}{15}+\frac{1}{4}=\frac{19}{60}\] \[{{C}_{eff}}=\frac{60}{19}\approx 3.2\,\,\mu F\]
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