Question

A $3\mu F$ capacitor is charged to a potential of $300V$ and $2\mu F$ capacitor is charged to $200V$. The capacitors are then connected in parallel with plates of opposite polarities joined together. What amount of charge will flow, when the plates are so connected?

• A. $1300\mu C$
• B. $800\mu C$
• C. $600\mu C$
• D. $300\mu C$

Answer 1

The correct option is A $1300\mu C$

Whenever a capacitor is charged by a well-defined Potential, the charge accumulated by it is given by $Q=CV$

The charge stored in 1st capacitor is $3\mu F\times 300V=3\times {10}^{-6}\times 300=9\times {10}^{-4}coulombs$

The charge stored in 2nd capacitor is $2\mu F\times 200V=2\times {10}^{-6}\times 200=4\times {10}^{-4}coulombs$

When they are connected in parallel with opposite polarities, the charges will get neutralized until the total charge remains the difference between them.

Each capacitor will have the half of the total charge.

$\therefore total$ $charge=(9-4)\times {10}^{-4}=5\times {10}^{-4}$

Thus both the capacitors will have $2.5\times {10}^{-4}coulombs.$

Since current always flows from a higher potential to a lower potential, we can say that charges will flow from the 1st capacitor to the 2nd.

Initially, $4\times {10}^{-4}coulombs$ of charge will flow to neutralize the charge on the 2nd capacitor. The 1st capacitor will now have $5\times {10}^{-4}$ $C$ and 2nd will be chargeless.

Thereafter $2.5\times {10}^{-4}C$ will flow from 1st to 2nd.

Thus total $6.5\times {10}^{-4}$ coulombs of charge will flow from the 1st capacitor to the 2nd one.

Answer:- $650\mu C$