Question

An electrician requires a capacitance of $6\mu F$ in a circuit across a potential difference of $1.5kV$. A large number of $2\mu F$ capacitors which can withstand a potential difference of not more than $500V$ are available. The minimum number of capacitors required for the purpose is

• A. 3
• B. 9
• C. 6
• D. 27

Answer 1

Answer: (D)

Number of capacitors to be connected in each row,
$m=\frac{1500}{500}=3$
Effective capacitance when connected in $n$ rows with $m$ capacitors in each row is,
$C_{\text{eff }}=n\frac{C}{m}$
$\therefore n\times \frac{2}{3}=6$
$n=\frac{18}{2}=9$
$\Rightarrow n=9$
$\therefore$ Total number of capacitors required
$N=mn=3\times 9=27$