Find the equivalent capacitance between A and B - Infinity Learn by Sri Chaitanya

A
$\frac{10 C}{5}$
B
$\frac{5 C}{7}$
C
$\frac{10 C}{7}$
D
$\frac{7 C}{10}$

Solution:

$q_{1} + q_{2} = q$ …………..(1)

Using KVL,

$- \frac{q_{1}}{C} - \frac{q_{3}}{2 C} + \frac{q_{2}}{2 C} = 0$

$\Rightarrow q_{2} - q_{3} - 2 q_{1} = 0$ ………(2)

Plates inside dotted line form an isolated system.

Hence, $q_{2} + q_{3} - q_{1} = 0$ ………..(3)

Solving these equations, we get

$q_{1} = \frac{2}{5} q, q_{2} = \frac{3}{5} q and q_{3} = - \frac{q}{5}$

$Now, let C_{eq} be the equivalent capacitance between A and B .$

$\begin{array}{l} V_{A} - V_{B} = \frac{q}{C_{eq}} = \frac{q_{1}}{C} + \frac{q_{2}}{2 C} \\ \Rightarrow \frac{q}{C_{eq}} = \frac{2 q}{5 C} + \frac{3 q}{10 C} = \frac{7 q}{10 C} \\ \Rightarrow C_{eq} = \frac{10}{7} C \end{array}$

Solution:

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