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

Correct Answer: B

Solution :

[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
\[\left. \begin{matrix}    Now\,\,\,\frac{3}{4}V<100 & \Rightarrow V<\frac{400}{3}  \\    \frac{V}{5}<50 & \Rightarrow V<200V  \\    V<400 & \Rightarrow V<400  \\ \end{matrix} \right\}\begin{matrix}    common\,\,solution  \\    V<\frac{400}{3}  \\    V<133\,\,V  \\ \end{matrix}\]