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

KCETKCET 2021Electrostatic Potential and Capacitance

Solution:

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
$N = mn =3 \times 9=27$