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# Three capacitors of equal capacitance whenconnected in series have a net capacitance of G.When connected in parallel they have a capacitance of C2.What is the value of C1/C2?

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Solution

## For series combination, $\frac{1}{{\mathrm{C}}_{1}}=\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}\left[\mathrm{Since}\mathrm{all}\mathrm{capacitors}\mathrm{have}\mathrm{equal}\mathrm{capacitances}\right]\phantom{\rule{0ex}{0ex}}=\frac{3}{\mathrm{C}}\phantom{\rule{0ex}{0ex}}\mathrm{Or}{\mathrm{C}}_{1}=\frac{\mathrm{C}}{3}---\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{parallel}\mathrm{combination},\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{2}=\mathrm{C}+\mathrm{C}+\mathrm{C}=3\mathrm{C}---\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{{\mathrm{C}}_{1}}{{\mathrm{C}}_{2}}=\frac{\mathrm{C}/3}{3\mathrm{C}}=\frac{\mathrm{C}}{3}×\frac{1}{3\mathrm{C}}=\frac{1}{9}$  Suggest Corrections  1      Similar questions  Related Videos   Series Combination
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