# 3capacitors of capacitances 2 microfarad ,3 microfarad and 6 microfrarad are connected in series with a 12 V battery .all the connecting wires are disconnected .the 3 positiveplates are connected together and the 3 negative plates are connected together.find the charges on the 3 capacitors after the recnnection

### Asked by Subhashree Lenka | 16th May, 2015, 01:18: PM

### The equivalent capacitance of the three capacitors in series is given as
The charge supplied by the battery of 12 V = q × V

= 1 μF × 12 V = 12 μC

Since the capacitors are then connected in series, 12 μC charge appears on each of the positive plates and -12μC charge on each of the negative plates.

The capacitors are connected as shown below:

The total capacitance is:

The 36 μC charge is distributed as Q_{1}, Q_{2} and Q_{3} on the three positive plates and –Q_{1}, -Q_{2} and –Q_{3} on the negative plates.

Also, the three positive terminals are now at the same common potential and the three negative plates are also at the same potential.

Let the potential difference across each capacitor be V.

Then,

Q_{1} = (2μF)V

Q_{2} = (3μF)V

Q_{3} = (6μF)V

Substituting in the equation :

we get:

The charge supplied by the battery of 12 V = q × V

= 1 μF × 12 V = 12 μC

Since the capacitors are then connected in series, 12 μC charge appears on each of the positive plates and -12μC charge on each of the negative plates.

The capacitors are connected as shown below:

The total capacitance is:

The 36 μC charge is distributed as Q_{1}, Q_{2} and Q_{3} on the three positive plates and –Q_{1}, -Q_{2} and –Q_{3} on the negative plates.

Also, the three positive terminals are now at the same common potential and the three negative plates are also at the same potential.

Let the potential difference across each capacitor be V.

Then,

Q_{1} = (2μF)V

Q_{2} = (3μF)V

Q_{3} = (6μF)V

Substituting in the equation :

we get:

### Answered by Yashvanti Jain | 13th Dec, 2017, 06:38: PM

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