> 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 – LIVE ANSWER TODAY

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:

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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 $2.25, kV$

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