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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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Answer» 10.5kV  Resultant capacitance of SERIES combination 1, `=(1)/(5) + (1)/(4) = (9)/(20)` `C_(eq_(1)) = (20)/(9) = 2.25muF` Resultant capacitance of series combination 2. `=(1)/(3) + (1)/(2) = (5)/(6)` `C_(eq_(2)) = (6)/(5) = 1.2 MU F` so, charge on upper BRANCH is `= (20)/(9)V` and charge on lower branch is `(6)/(5)V`. 
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