Four capacitors marked with capacitances and breakdown voltages are connected as shown in the figure. The maximum emf of the source, so that no capacitor breaks down is

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Aicharya · Accepted Answer

Resultant capacitance of series combination 1 , $ \begin{aligned} & =\frac{1}{5}+\frac{1}{4}=\frac{9}{20} \\ C_{\mathrm{eq}_1} & =\frac{20}{9}=2.25 \mu \mathrm{F} \end{aligned} $ Resultant capacitance of series combination 2, $ \begin{aligned} & =\frac{1}{3}+\frac{1}{2}=\frac{5}{6} \\ C_{\mathrm{eq}_2} & =\frac{6}{5}=1 \cdot 2 \mu F \end{aligned} $ So, charge on upper branch is $=\frac{20}{9} \mathrm{~V}$ and charge on lower branch is $\frac{6}{5} \mathrm{~V}$.

So, smallest value is 2.25 kV

Four capacitors marked with capacitances and breakdown voltages are connected as shown in the figure. The…

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