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An electrician requires a capacitance of $6 \mu \mathrm{F}$ in a circuit across a potential difference of $1.5 \mathrm{kV}$. A large number of $2 \mu \mathrm{F}$ capacitors which can withstand a potential difference of not more than $500 \mathrm{~V}$ are available. The minimum number of capacitors required for the purpose is
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Verified Answer
The correct answer is:
27
Number of capacitors that can be connected in each row,
$\begin{aligned}
m &=\frac{\text { Desired voltage }}{\text { Voltage across each capacitor }} \\
&=\frac{1.5 \mathrm{kV}}{500 \mathrm{~V}}=\frac{1500}{500}=3
\end{aligned}$
Effective capacitance when $m$ capacitors are connected in $n$ rows is given as
$C_{\mathrm{eff}}=\frac{n C}{m}$
$\begin{array}{lc}
\text { Here, } m=3, C=2 \mu \mathrm{F}, C_{\text {eff }}=6 \mu \mathrm{F} \\
\Rightarrow & n \times \frac{2}{3}=6 \\
\text { or } & n=\frac{18}{2}=9
\end{array}$
$\therefore$ Total number of capacitors required, $N=m n=3 \times 9=27$.
$\begin{aligned}
m &=\frac{\text { Desired voltage }}{\text { Voltage across each capacitor }} \\
&=\frac{1.5 \mathrm{kV}}{500 \mathrm{~V}}=\frac{1500}{500}=3
\end{aligned}$
Effective capacitance when $m$ capacitors are connected in $n$ rows is given as
$C_{\mathrm{eff}}=\frac{n C}{m}$
$\begin{array}{lc}
\text { Here, } m=3, C=2 \mu \mathrm{F}, C_{\text {eff }}=6 \mu \mathrm{F} \\
\Rightarrow & n \times \frac{2}{3}=6 \\
\text { or } & n=\frac{18}{2}=9
\end{array}$
$\therefore$ Total number of capacitors required, $N=m n=3 \times 9=27$.
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