Three capacitors of capacitances `2 muF`, `3mu F` and `6muF` are connected in series with a 12 V battery. All the connecting wire are disconnected, the three positive plates are connected together and the three negative plates are connected together. Find the charges on the three capacitors after the reconnection.

Options

A. ` Q_1 = (7)/(11) muC, Q_2 =(10)/(11) muC and Q_3 = (21)/(11) muC`
B. ` Q_1 = (72)/(11) muC, Q_2 =18/(11) muC and Q_3 = (216)/(11) muC`
C. ` Q_1 = (72)/(11) muC, Q_2 =(108)/(11) muC and Q_3 = (216)/(11) muC`
D. ` Q_1 = (2)/(11) muC, Q_2 =(108)/(11) muC and Q_3 = (216)/(11) muC`

Answer

C

Explanation

The equivalent capacitance of the three capacitors joined is series is given by `
`(1)/(C ) = (1)/(2 muF) + (1)/(3 muF)+(1)/(6muF)`
` or, `
`C=1 muF`
`. The charge supplied by the battery `
`= 1 muF X 12V`
` = `
`12muC.`
` As the capacitors are connected in series, `
`12muC`
` charge appears on each of the positive plate and `
`-12muC`
` on each of the negative plates. The charged capacitors are now connected as shown in figure. The `
`36muC`
` charge on the three positive plates now redistribute as `
`Q_1, Q_2 and Q_3`
` on the three connected positive plates. Similarly, `
`-36 muC`
` redistributes as `
`-Q_1, -Q_2 and -Q_3`
`. The three positive plates are now at a common potential and the three nagative plates are also at a common potential. Let the potential difference across each capacitor be V. Then `
`Q_1 = (2muF) V,`
`Q_2 = (3muF) V, and `
`Q_3 = (6muF) V, Also, `
`Q_1+Q_2+Q_3 =36 muC`
`. Solving these equations, `
` Q_1 = (72)/(11) muC, Q_2 =(108)/(11) muC and Q_3 = (216)/(11) muC`
`.