Question

Three capacitors of equal capacitance, when connected in series have net capacitance \(C_1\) and when connected in parallel have net capacitance \(C_2\). The ratio of \(C_1\) and \(C_2\) is

• A. \(\frac{1}{3}\)
• B. 3
• C. \(\frac{1}{9}\)
• D. 9

Hint:

Series gives \(C/3\) and parallel gives \(3C\).

Answer 1

The Correct Option is C

Step 1: Method: Use a Number:
Pick \(C = 6\ \mu F\) for each capacitor to make the arithmetic easy.

Step 2: Series:
\[ \frac{1}{C_1} = \frac{1}{6}+\frac{1}{6}+\frac{1}{6} = \frac{1}{2} \] So \(C_1 = 2\ \mu F\).

Step 3: Parallel:
\(C_2 = 6+6+6 = 18\ \mu F\).

Step 4: Ratio:
\[ \frac{C_1}{C_2} = \frac{2}{18} = \frac{1}{9} \] The result does not depend on the value we picked. It equals \(\frac{1}{3}\times\frac{1}{3}\).

Final Answer:
\[\boxed{\frac{C_1}{C_2}=\frac{1}{9}}\]