Question

Three identical capacitors are connected in series. The capacitance of combination is $C$ and break-down voltage is $V$. The net capacitance and break down voltage when these are connected in parallel combination will be:

• A. $\frac{C}{3},3V$
• B. $9C,\frac{V}{3}$
• C. $\frac{C}{3}.V$
• D. $3C,\frac{V}{9}$

Answer 1

The correct option is D $3C,\frac{V}{9}$

In series

$\frac{1}{C^{\prime }}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$

here $C_1=C_2=C_3=C$

Then net capacitance is

$C^{\prime }=\frac{C}{3}$ and V is given

So charge $q=CV$ putting values

$q=\frac{VC}{3}$

Now when capacitors are connected in parallel

$C"=C+C+C=3C$

and charge will remain same

$V^{\prime }=\frac{q}{C"}$

$V^{\prime }=\frac{V}{9}$