Question

If the equivalent capacitance between A and B is $ \text{1}\text{F,} $ then the value of C will be

• A. $ \text{2}\text{F} $
• B. $ \text{4}\text{F} $
• C. $ \text{3}\text{F} $
• D. $ \text{6}\text{F} $

Answer 1

The Correct Option is A

Capacitors $ {{C}_{3}} $ and $ {{C}_{4}} $ are in parallel, therefore their resultant capacitance, $ C={{C}_{3}}+{{C}_{4}}=1+1=2\mu F $ Now, capacitors $ {{C}_{2}} $ and $ C $ are in series, therefore their resultant capacitance, $ \frac{1}{C\,\,}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2} $ $ C\,\,=1\mu F $ Capacitors $ {{C}_{6}} $ and $ {{C}_{8}} $ are in series, therefore their resultant capacitance, $ \frac{1}{C\,\,\,\,}=\frac{1}{2}+\frac{1}{2}=\frac{2}{2}=1 $ $ C\,\,\,\,=1\mu F $ Now, $ C\,\,\,\,\,\, $ and $ {{C}_{5}} $ are in parallel, therefore their resultant capacitance, $ C\,\,\,\,\,\,=1+1=2\mu F $ Now, $ C\,\,\,\,\,\, $ and $ {{C}_{5}} $ are in series. Therefore, their resultant capacitance, $ \frac{1}{{{(C)}^{5}}}=\frac{1}{2}+\frac{1}{2} $ Now, $ C\,\, $ and $ {{(C)}^{5}} $ are in parallel. Therefore, their resultant capacitance, $ {{(C)}^{6}}=1+1=2\mu F $ Now, $ {{C}_{1}} $ and $ {{(C)}^{6}} $ are in series and their resultant capacitance is given $ 1\mu F $ . $ \therefore $ $ \frac{1}{1}=\frac{1}{2}+\frac{1}{C} $ $ \therefore $ $ \frac{1}{C}=\frac{1}{1}-\frac{1}{2}=\frac{1}{2} $ $ C=2\mu F $