In the circuit shown in figure, switch S was closed for a long time. At time t = 0, the switch is opened again. The maximum potential difference across the plates of the capacitor after the switch is opened is 

Maximum current $I_m=\frac{100}{50}2A$

Potential drop across the resistor is 2 $\times$ 50 = 100 V
Initial potential difference across capacitor is zero

$L_{eq}=\left(200+50\right)=250 mH$

Energy stored $=\frac{1}{2}{L}_{eff}{I}_M^2=\frac{1}{2}\times 250\times {10}^{-3}\times 2^2=0.5J$

$When S_1 is opened \frac{1}{2}{L}_{eff}{I}_M^2=\frac{1}{2}C{V}_{max}^2$

$V_{max}=\sqrt{\frac{2\times 0.5}{100\times {10}^{-6}}}=100V$