KCET · Physics · Capacitance
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
- A 3
- B 9
- C 6
- D 27
Answer & Solution
Correct Answer
(D) 27
Step-by-step Solution
Detailed explanation
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
\(C_{\mathrm{eff}}=\frac{n C}{m}\)
Here, \(m=3, C=2 \mu \mathrm{F}, C_{\text {eff }}=6 \mu \mathrm{F} \)
\( \Rightarrow n \times \frac{2}{3}=6 \)
\( \text {or } n=\frac{18}{2}=9\)
\(\therefore\) 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
\(C_{\mathrm{eff}}=\frac{n C}{m}\)
Here, \(m=3, C=2 \mu \mathrm{F}, C_{\text {eff }}=6 \mu \mathrm{F} \)
\( \Rightarrow n \times \frac{2}{3}=6 \)
\( \text {or } n=\frac{18}{2}=9\)
\(\therefore\) Total number of capacitors required, \(N=m n=3 \times 9=27\).
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