Three capacitances, each of 3 μ F , are provided. These cannot be combined to provide the resultant…

1. $2\mu \text{F}$
2. $6\mu \text{F}$
3. $1.5\mu \text{F}$
4. $1\mu \text{F}$

Solution

Equivalent capacitance will be maximum for parallel combination and Minimum for series combination

$\therefore$ Possibilities are

1)           $\text{C}=1\mu \text{F}$

2)           $\text{C}=9\mu \text{F}$

3)         $\frac{1}{\text{C}}=\frac{1}{6}+\frac{1}{3}=\frac{3}{6}$

                                    $\therefore \text{C}=2\mu \text{F}$

4) ​        $\text{C}=\frac{\text{C}}{2}+\text{C}=\frac{3}{2}\times \text{C}$

                                       $=\frac{3}{2}=\text{1.5 }\mu \text{F}$

Asked in: JEE Main 2014 (09 Apr Online)

Practice more Electrostatics questions on Aicharya