JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 14)
Three capacitors each of 4 $$\mu $$F are to be connected in such a way that the effective capacitance is 6 $$\mu $$F. This can be done by connecting them :
all in series
two in series and one in parallel
all in parallel
two in parallel and one in series
Explanation
(a) $${1 \over {{C_{eq}}}} = {1 \over 4} + {1 \over 4} + {1 \over 4} = {3 \over 4}$$
$$ \Rightarrow $$ $${C_{eq}} = {4 \over 3}\,\mu F$$
(b) $${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$$
(c) $${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$$
(d) $${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$$
$$ \Rightarrow $$ $${C_{eq}} = {4 \over 3}\,\mu F$$
(b) $${C_{eq}} = {{4 \times 4} \over {4 + 4}} + 4 = 6\mu F$$
(c) $${C_{eq}} = 4 + 4 + 4 = 12\,\mu F$$
(d) $${C_{eq}} = {{\left( {4 + 4} \right) \times 4} \over {\left( {4 + 4} \right) + 4}} = {8 \over 3}\mu F$$
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