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KCET · Physics · Alternating Current

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

  1. A \( \frac{4}{7} \mathrm{pF} \)
  2. B \( 1 \mathrm{pF} \)
  3. C \( \frac{7}{4} \mathrm{pF} \)
  4. D \( 2 \mathrm{pF} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( \frac{4}{7} \mathrm{pF} \)

Step-by-step Solution

Detailed explanation

The effective capacitance will be minimum when the threecapacitors are connected in series. The effective capacitance is given by
\(C_{1}=1 \mathrm{pF} \)
\(C_{2}=2 \mathrm{pF} C_{3}=4 \mathrm{pF} \)
\(\frac{1}{C_{e f f}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \)
\(=\frac{1}{1 p F}+\frac{1}{2 p F}+\frac{1}{4 p F}=\frac{(4+2+1)}{4 p F}=\frac{7}{4} \)
Therefore \(\text C_{\text {eff }}=\frac{4}{7} p F\)