• # question_answer In figure ${{C}_{1}}=2\mu F,$ ${{C}_{2}}=6\mu F$ & ${{C}_{3}}=3.5\mu F.$ If break down voltages of the individual capacitors are ${{V}_{1}}\text{ }=\text{ }100V,$${{V}_{2}}\text{ }=\text{ 5}0V,$ & ${{V}_{3}}\text{ }=\text{ 40}0V,$What maximum voltage can be placed across points a & b- A) 124 V                B) 133 V C) 100 V                D) 200 V

[B]  Let potential applied between a and b = V using the formula ${{V}_{1}}=\,\,\,\frac{\frac{1}{2}}{\frac{1}{2}+\frac{1}{6}}\,\,\times \,\,V$ ${{V}_{2}}=\,\,\frac{\frac{1}{6}}{\frac{1}{2}+\frac{1}{6}}\,\,\times \,\,V$ ${{V}_{1}}=\frac{6}{8}\,\,\times \,\,V=\frac{3}{4}V$ ${{V}_{2}}=\frac{1}{4}V,$ ${{V}_{3}}=V$ These voltage should be less than corresponding break down voltage $\left. \begin{matrix} Now\,\,\,\frac{3}{4}V<100 & \Rightarrow V<\frac{400}{3} \\ \frac{V}{5}<50 & \Rightarrow V<200V \\ V<400 & \Rightarrow V<400 \\ \end{matrix} \right\}\begin{matrix} common\,\,solution \\ V<\frac{400}{3} \\ V<133\,\,V \\ \end{matrix}$