Consider the connections shown in (a) Find the capacitance between the points A and B. (b) find the charge on the three capacitors. (c ) Taking the potential at the point B to be zero, find the potential at the point D.

Options

A. `(12 mu F)`, `48 muC` & 96 micro coulomb `16 V`
B. `(6 mu F)`, 48 micro coulomb & 96 micro coulomb , `16 V`
C. `(6 mu F)`, 48 micro coulomb & 96 micro coulomb, `12 V`
D. `(6 mu F)` 48 micro coulomb & 96 micro coulomb, `10 V`

Answer

B

Explanation

The `
`12 muF`
` and `
`6 muF`
` capacitors are joined in series. The equivalent of these two will have a capacitance given by `
`(1)/(C ) = (1)/(12 mu F) + (1)/(6 mu F)`
`, or, `
`C = 4 muF`
`. The combination of these two capacitors is joined in parallel with the `
`2mu F`
` capacitor. Thus, the equivalent capacitance between A and B is `
`4muF + 2muF = 6 muF`
`. (b) The charge supplied by the battery is `
`Q = CV =6muF X24V = 144muC`
`. The potential difference across the `
`2muF`
` capacitor is 24V . The charge on this capacitor is, therefore, `
`2muFX24V = 48muC`
`. The charge on the `
`12muF`
` and `
`6muF`
` capacitor is, therefore, `
`144muc - 48mu C= 96muC`
`. (c ) The potential difference across the `
`6 muF `
` capacitor is (96 muC)/(6 muF) = 16V`
`. As the potential at the point B is taken to be zero, the potential at the point D is 16 V.