First slide

Capacitors and capacitance

Question

Find the equivalent capacitance between A and B

.

Difficult

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
$q_{1} = \frac{2}{5} q, q_{2} = \frac{3}{5} q and q_{3} = - \frac{q}{5}$
$Now, let C_{eq} be the equivalent capacitance between A and B .$
$\begin{array}{l} V_{A} - V_{B} = \frac{q}{C_{eq}} = \frac{q_{1}}{C} + \frac{q_{2}}{2 C} \\ \Rightarrow \frac{q}{C_{eq}} = \frac{2 q}{5 C} + \frac{3 q}{10 C} = \frac{7 q}{10 C} \\ \Rightarrow C_{eq} = \frac{10}{7} C \end{array}$

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