Three capacitors `C_(1) , C_(2) and C_(3) ` are connected as show in the figure to a battery of V volt. If the capacitor `C_(3)` breaks down electrically, the change in total charge on the combination of capacitors is

Options

A. `(C_(1) +C_(2)) V [ 1- (C_(3)/(C_(1) +C_(2) +C_(3)))]`
B. `(C_(1) +C_(2)) V [ 1- ((C_(1)+C_(2))/(C_(1) +C_(2) +C_(3)))]`
C. `(C_(1) +C_(2)) V [ 1+ (C_(3)/(C_(1) +C_(2) +C_(3)))]`
D. `(C_(1) +C_(2)) V [ 1+ (C_(2)/(C_(1) +C_(2) +C_(3)))]`

Answer

A

Explanation

Since , `C_(1) and C_(2)` are parallel, so their equivalent capacitance will be `(C_(1) +C_(2)) " Now " (C_(1) +C_(2)) and C_(3)` are in series, so the eat equvalent capacitance of circuit.
` 1/C = 1/C_(3) + 1/(C_(1)+ C_(2)) = (C_(1)+C_(2) +C_(3))/((C_(1) + C_(2))C_(3))`
` C_(1) = ((C_(1) +C_(2))C_(3))/(C_(1) + C_(2)+ C_(3))`
since V is the voltage of the battery so change on this system .
q = CV
` Rightarrow q = ((C_(1)+C_(2)C_(3)V)/(C_(1)+C_(2)+C_(3)))`
If the capacitor ` C_(3) ` breaks down, then total equivalent equivalent capacitance
` C= C_(1) + C_(2)`
Newm chage stored
q = C' V
q' = `(C_(1) +C_(2)) V`
Change in total charge
` Deltaq = q'-q " " [ therefore q gt q ]`
`(C_(1)+C_(2))V-((C_(1)+C_(2))C_(3)V)/(C_(1)+C_(2)+C_(3))`
`= C_(1)+C_(2)V[ 1- C_(3)/(C_(1)+C_(2)+C_(3))]`
` Deltaq = (C_(1)+C_(2))V [ 1- C_(3)/(C_(1) + C_(2)+C_(3))]`