## In This post you will get answer of Q.

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

Solution:

Resultant capacitance of series combination 1 .

$=frac{1}{5}+frac{1}{4}=frac{9}{20}$

$C_{e q_{1}} =frac{20}{9}=2.25, mu F$

Resultant capacitance of series combination $2 .$

$=frac{1}{3}+frac{1}{2}=frac{5}{6}$

$C_{text {eq }_{2}}=frac{6}{5}=1 cdot 2 mu F$

So, charge on upper branch is $=frac{20}{9} V$ and charge on lower branch is $frac{6}{5} V$.

Capacitor | Voltage drop across it | Given breakdown voltage | Maximum value of emf it can bear |

$C_{1}$ | $V=q / C$ $=frac{20}{9} times frac{1}{5}=frac{4}{9} V$ | $1, kg$ | $frac{1 kV }{(4 / 9 V) }=frac{9}{4}$ $=2 cdot 25, kV$ |

$C_{2}$ | $=frac{20}{9} times frac{1}{4}=frac{5}{9} V$ | $2, kg$ | $frac{2 kV }{(5 / 9) V }=frac{18}{5}$ $=3.6, kV$ |

$C_{3}$ | $=frac{6}{5} times frac{1}{2}=frac{3}{5} V$ | $2, kg$ | $frac{2 kV }{(3 / 5) V }=frac{10}{3}$ $=333, kV$ |

$C_{4}$ | $=frac{6}{5} times frac{1}{3}=frac{2}{5} v$ | $1, kg$ | $frac{1 kV }{(2 / 5)}=frac{5}{2}$ $=2.5, kV$ |

So, smallest value is $2.25, kV$