Question

Five capacitors of $10\mu F$ capacity each are connected to a d.c. potential of $100$ volts as shown in the adjoining figure. The equivalent capacitance between the points $A$ and $B$ will be equal to :

• A. $40\mu F$
• B. $20\mu F$
• C. $30\mu F$
• D. $10\mu F$

Answer 1

The correct option is D $10\mu F$

$\text{Step 1: Balanced Wheat stone bridge}$

Given circuit is an example of the wheat stone bridge.

Here the ratio of upper and lower capacitances on both sides is equal i.e. $\frac{10\mu F}{10\mu F}=\frac{10\mu F}{10\mu F}$, therefore this is a balanced wheat stone bridge.

Hence, the potential difference across the middle capacitor will be zero, so we can eliminate the middle capacitor(Refer figure).

$\text{Step 2: Equivalent capacitance}$

Now solve series and parallel arrangement shown in figure

Firstly solve for series arrangement in upper and lower branches:

$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}=\frac{1}{10}+\frac{1}{10}=\frac{1}{5}$

So, $C=5\mu F$

Now both upper and lower $C$ will be in parallel.

So, $C_{net}=C+C=5+5=10\mu F$

Therefore the equivalent capacitance across AB is $10\mu F$.

Hence, option D is correct.