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

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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 $2.25, kV$