Question

The charge flowing through the cell on closing the key $k$ is equal to

• A. $\frac{4}{3}CV$
• B. $4CV$
• C. $\frac{3}{4}CV$
• D. $\frac{CV}{4}$

Answer 1

The correct option is D $\frac{CV}{4}$

Consider the case of open key

$\frac{1}{C_{eq.}}=\frac{1}{C}+\frac{1}{C+C+C}$

$=\frac{1}{C}+\frac{1}{3C}$

$=\frac{4}{3C}$

$C_{eq.}=\frac{3C}{4}$

Thus, initial charge flown $\left(q_i\right)=\frac{3}{4}CV$

Now, consider the case of closed key.
Net capacitance of this case is $C$.
Charge flown $\left(q_f\right)=CV$
$\therefore \Delta q=q_f-q_i$

$=CV-\frac{3}{4}CV$

$=\frac{1}{4}CV$

Hence, charge passed through cell is $\frac{1}{4}CV$.