Question

In the circuit shown in figure, each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed.

• A. Positive charge will flow out of the positive terminal of the cell
• B. Positive charge will enter the positive terminal of the cell
• C. The amount of charge flowing through the cell will be $CE$
• D. The amount of charge flowing through the cell will be $\frac{4}{3}CE$

Answer 1

Answer: (A), (C)

Positive charge will flow out of the positive terminal of the cell, The amount of charge flowing through the cell will be CE

Answer 2

Initially, two capacitors (2C equivalent) are in parallel with the cell, storing $2CE$ charge. When the switch closes, a third capacitor connects in parallel, making the total equivalent capacitance $3C$. The final charge stored is $3CE$. The additional charge flowing from the cell is $\Delta Q = 3CE - 2CE = CE$. Since $\Delta Q$ is positive, positive charge flows out of the cell's positive terminal.