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The minimum value of effective capacitance that can be obtained by combining \( 3 \) capacitors of
capacitances \( 1 \mathrm{pF}, 2 \mathrm{pF} \) and \( 4 \mathrm{pF} \) is
Options:
capacitances \( 1 \mathrm{pF}, 2 \mathrm{pF} \) and \( 4 \mathrm{pF} \) is
Solution:
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Verified Answer
The correct answer is:
\( \frac{4}{7} \mathrm{pF} \)
The effective capacitance will be minimum when the threecapacitors are connected in series. The effective capacitance
is given by
\[
\begin{array}{l}
\qquad \begin{array}{l}
C_{1}=1 \mathrm{pF} \\
C_{2}=2 \mathrm{pF} \quad C_{3}=4 \mathrm{pF} \\
\frac{1}{C_{e f f}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\
=\frac{1}{1 p F}+\frac{1}{2 p F}+\frac{1}{4 p F}=\frac{(4+2+1)}{4 p F}=\frac{7}{4} \\
\text { Therefore } C_{\text {eff }}=\frac{4}{7} p F
\end{array}
\end{array}
\]
is given by
\[
\begin{array}{l}
\qquad \begin{array}{l}
C_{1}=1 \mathrm{pF} \\
C_{2}=2 \mathrm{pF} \quad C_{3}=4 \mathrm{pF} \\
\frac{1}{C_{e f f}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\
=\frac{1}{1 p F}+\frac{1}{2 p F}+\frac{1}{4 p F}=\frac{(4+2+1)}{4 p F}=\frac{7}{4} \\
\text { Therefore } C_{\text {eff }}=\frac{4}{7} p F
\end{array}
\end{array}
\]
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