Q141 / 180 · KCET 2017 · Physics
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
capacitances \( 1 \mathrm{pF}, 2 \mathrm{pF} \) and \( 4 \mathrm{pF} \) is
- Correct answerA\( \frac{4}{7} \mathrm{pF} \)
- B\( 1 \mathrm{pF} \)
- C\( \frac{7}{4} \mathrm{pF} \)
- D\( 2 \mathrm{pF} \)
Solution
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
\[ \begin{array}{l} \qquad \begin{array}{l} C_{1}=1 \mathrm{pF} \\ C_{2}=2 \mathrm{pF} \quad C_{3}=4 \mathrm{pF} \\ \frac{1}{C_{e f f}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}} \\ =\frac{1}{1 p F}+\frac{1}{2 p F}+\frac{1}{4 p F}=\frac{(4+2+1)}{4 p F}=\frac{7}{4} \\ \text { Therefore } C_{\text {eff }}=\frac{4}{7} p F \end{array} \end{array} \]
Hence, the correct answer is option: 'A'.