Question

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

• A. 10.5 kV
• B. 5.25 kV
• C. 2.25 kV
• D. 1.25 kV

Answer 1

Answer: (C)

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
$C_1$	$V=q/C$ $=\frac{20}{9}\times \frac{1}{5}=\frac{4}{9}V$	$1kg$	$\frac{1kV}{(4/9V)}=\frac{9}{4}$ $=2\cdot 25kV$
$C_2$	$=\frac{20}{9}\times \frac{1}{4}=\frac{5}{9}V$	$2kg$	$\frac{2kV}{(5/9)V}=\frac{18}{5}$ $=3.6kV$
$C_3$	$=\frac{6}{5}\times \frac{1}{2}=\frac{3}{5}V$	$2kg$	$\frac{2kV}{(3/5)V}=\frac{10}{3}$ $=333kV$
$C_4$	$=\frac{6}{5}\times \frac{1}{3}=\frac{2}{5}v$	$1kg$	$\frac{1kV}{(2/5)}=\frac{5}{2}$ $=2.5kV$

So, smallest value is $2.25kV$