Question

If the two terminals of a battery of $40\text{V}$ are connected across $\text{A}$ and $\text{B}$, the charge flown through the battery will be

Given: $C_1=5\mu F,C_2=2.5\mu F,C_3=6\mu F,C_4=2\mu F$ and $C_5=15\mu F$

• A. $105\mu \text{C}$
• B. $210\mu \text{C}$
• C. $420\mu \text{C}$
• D. $77\mu \text{C}$

Answer 1

The correct option is B $210\mu \text{C}$

Upon redrawing the circuit,


Given: $C_1=5\mu \text{F},C_2=2.5\mu \text{F}{C}_3=6\mu \text{F},C_4=2\mu \text{F}$ and $C_5=15\mu F$

$\frac{C_1}{C_4}=\frac{5}{2}=2.5$

$\frac{C_5}{C_3}=\frac{15}{6}=2.5$

$\Rightarrow$ The given arrangement is balanced wheatstone bridge.

Now, $C_2$ capacitor can be neglected such that $C_1$ and $C_5$ are in series and $C_4$ and $C_3$ are in series, and their combination are in parallel.

$\Rightarrow C_{eq}=\frac{21}{4}\mu F$

$Q=C_{eq}V=(\frac{21}{4}\times 40)\mu C$

$Q=210\mu C$