The capacitor A shown in Fig. has a capacitance $C_1=3\mu F$ . The dielectric filled in it has a breakdown voltage of 40V and it has a resistance of $3M\Omega$ . The capacitor B has a capacitance of $C_2=2\mu F$ and dielectric in it has a resistance of $2M\Omega$  . Breakdown voltage for B is 50V. The switch is closed at t = 0. The time (in s) at which a capacitor breaks down first will be (Take $\ln \left[2\right]=0.69$ , $\ln \left(3\right)=1.1$ )
 

The final voltage across A will be 

$V_1=\frac{C_2}{C_1+C_2}=\frac{2}{3+2}\times 100=40\mathrm{V}$

across B it will be

$V_2=\frac{C_1}{C_1+C_2}=60V$

Obviously, capacitor B will breakdown first as soon as voltage across it reaches 50 V (at that time voltage across A will be less than 40 V ).

The effective capacitance of the circuit is$t=6\ln 6=10.74s$

$C=\frac{C_1{C}_2}{C_1+C_2}=\frac{3\times 2}{3+2}=\frac{6}{5}\mu \mathrm{F}$

The effective resistance is $R=3+2=5\mathrm{M\Omega }$

 Time constant of the circuit is

$\tau =RC=6\sec$

$q=q_0\left[1-e^{-t/\tau }\right]$

$q_0=CV=\frac{6}{5}\mu \mathrm{F}\times 100\mathrm{V}=120\mu \mathrm{C}$

$q=\left(120\mu C\right)\left(1-e^{-t/\tau }\right)$

Breakdown of B will take place when $Q=2\mu \mathrm{F}\times 50\mathrm{V}=100\mathrm{\mu C}$

$t=6\ln 6=10.74s$