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
AP EAMCETAP EAMCET 2018
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 $2.25\, kV$
| Capacitor | Voltage drop across it | Given breakdown voltage | Maximum value of emf it can bear |
| $C_{1}$ | $V=q / C$ $=\frac{20}{9} \times \frac{1}{5}=\frac{4}{9} V$ | $1\, kg$ | $\frac{1 kV }{(4 / 9 V) }=\frac{9}{4}$ $=2 \cdot 25\, kV$ |
| $C_{2}$ | $=\frac{20}{9} \times \frac{1}{4}=\frac{5}{9} V$ | $2\, kg$ | $\frac{2 kV }{(5 / 9) V }=\frac{18}{5}$ $=3.6\, kV$ |
| $C_{3}$ | $=\frac{6}{5} \times \frac{1}{2}=\frac{3}{5} V$ | $2\, kg$ | $\frac{2 kV }{(3 / 5) V }=\frac{10}{3}$ $=333\, kV$ |
| $C_{4}$ | $=\frac{6}{5} \times \frac{1}{3}=\frac{2}{5} v$ | $1\, kg$ | $\frac{1 kV }{(2 / 5)}=\frac{5}{2}$ $=2.5\, kV$ |