First slide

Capacitance

Question

In the circuit shown in the figure, terminal B is grounded and potential of terminal A is 10 volt. Then charge stored in $3 μ F$ Capacitor is

Moderate

A

$8 μ C$
B

$12 μ C$
C

$16 μ C$
D

$20 μ C$

Solution

(2)
$Sin c e \frac{C_{1}}{C_{3}} = \frac{C_{2}}{C_{4}},$ The given circuit is a balanced wheat stone bridge. So $5 μ F$ can be removed from the circuit.

$\begin{array}{l} N o w C^{'} = \frac{C_{1} . C_{2}}{C_{1} + C_{2}} = \frac{2 \times 3}{2 \times 3} μ F = \frac{6}{5} μ F \\ ∴ C h \arg e i n 3 μ F c a p a c i t o r = C^{'} \times (10 - 0) = \frac{6}{5} \times 10 μ C = 12 μ C \end{array}$

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