Question

An electric circuit requires a total capacitance of $2\mu F$ across a potential of $1000V$. Large number of $1\mu F$ capacitances are available each of which would breakdown if the potential is more than $350V$. How many capacitances are required to make the circuit?

• A. 24
• B. 20
• C. 18
• D. 12

Answer 1

Answer: (C)

Here, Capacitance of each capacitor, $C=1\mu F$
Voltage rating of each capacitor $=350V$
Supply voltage $=1000V$
Total capacitance $=2\mu F$
Let $n$ capacitors of $1\mu F$ each be connected in series in a row and $m$ such rows be connected in parallel as shown in the figure.

$\because$ Each capacitor can withstand $350V$.
$\therefore n=\frac{1000}{350}=2.8$
As $n$ cannot be fraction, therefore $n=3$
Capacitance of each row of $3$ capacitors of $1\mu F$ each in series is
$C_s=\frac{1}{3}\mu F$
Total capacitance of $m$ such rows in parallel $=\frac{m}{3}\mu F$
$\therefore \frac{m}{3}\mu F=2\mu F$ or $m=6$
Total number of capacitors $=n\times m=3\times 6=18$