Question

If the two terminals of a battery of $21\text{V}$ is connected across $\text{A}$ and $\text{B}$, the charge flown through the battery will be
$Given:C_1=2\text{pF},C_2=13\text{pF},C_3=6\text{pF}{C}_4=1\text{pF}\text{and}{C}_5=12\text{pF}$

• A. $5.4\times {10}^{-12}\text{C}$
• B. $5.4\times {10}^{-13}\text{C}$
• C. $5.4\times {10}^{-10}\text{C}$
• D. $5.4\times {10}^{-11}\text{C}$

Answer 1

The correct option is D $5.4\times {10}^{-11}\text{C}$

The given circuit can be redrawn as shown


$Given:C_1=2\text{pF},C_2=13\text{pF}{C}_3=6\text{pF},C_4=1\text{pF}\text{and}{C}_5=12\text{pF}$

Here, we get
$\frac{C_1}{C_4}=\frac{2}{1}$

$\frac{C_5}{C_3}=\frac{12}{6}=\frac{2}{1}$

$\therefore \frac{C_1}{C_4}=\frac{C_5}{C_3}$

So, this is a case of balanced Wheatstone bridge.

Now, the circuit reduces to


Connect the battery of $21\text{V}$ (given) between $\text{A}$ and $\text{B}$.

Using the formula of equivalent capacitor, we get

$C^{\prime }=\frac{C_1{C}_5}{C_1+C_5}=\frac{2\times 12}{2+12}=\frac{12}{7}\text{pF}$

$C^{\prime \prime }=\frac{C_4{C}_3}{C_4+C_3}=\frac{1\times 6}{1+6}=\frac{6}{7}\text{pF}$

As $C^{\prime }$ and $C^{\prime \prime }$ are connected in parallel.

So, equivalent capacitance between $\text{A}$ and $\text{B}$ will be

$C_{\text{AB}}=C^{\prime }+C^{\prime \prime }=\frac{18}{7}\text{pF}$

Charge flown through $\text{A}$ and $\text{B}$,

$Q=C_{\text{AB}}V=\frac{18}{7}\times 21\text{pC}$

$\Rightarrow Q=5.4\times {10}^{-11}\text{C}[\because 1\text{pC}={10}^{-12}\text{C}]$

Hence, option $\left(d\right)$ is correct.