Question

In a given network capacitor $ {{C}_{1}}=10\,\,\mu F,\,\,{{C}_{2}}=\mu F, $ , and $ {{C}_{3}}=4\,\,\mu F $ . The equivalent capacitance between $ PQ $ will be

• A. $ 1.6\,\,\mu F $
• B. $ 3.2\,\,\mu F $
• C. $ 6.4\,\,\mu F $
• D. $ 4.5\,\,\mu F $

Answer 1

Answer: (B)

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 $