Four capacitors of capacitances `2mu F , 3 mu F, 4muF ` and `xmuF` are connected to a battery of emf 6 V and of negligible internal resistance as shown in the figure. If the ratio of the charges on `xmuF ` and `4muF ` capacitances is `(3)/(8)` then the value of x is

Options

A. 2
B. 5
C. 3
D. 8

Answer

B

Explanation

As x and z are in parallel

`implies` Let charges on `2mu F, 3muF,4muF` and `x muF` are `q_(2)q_(3) , q_(4)` and `q_(x)`
`q_(4)= C_(4)V=4muFxx6=24muC`
`implies ` As given `(q_(x))/(q_(4))=(3)/(8) implies q_(x)= 24xx(3)/(8) = 9muC`
Potential difference on `xmu` F
`V_(1)= (q_(x))/(x)= (9)/(x){V = (q)/(C)}`
Same `(9)/(x)` volt will be across `2mu F ` too.
Now remaining `(6-(9)/(x))` volt potential difference will drop across `3muF`.
As `3muF` adn `(x+2)muF` are in series so charge on them will be same `q_(3)=q_(x+2)`
`3(6-(9)/(x))=(x+2) (9)/(x)`
`6x-9=3x+6 implies 3x= 15 implies x=5 `