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: Understanding the Question:
Three identical capacitors, each of capacitance C, are joined in series to give \(C_1\) and in parallel to give \(C_2\). We need \(C_1 : C_2\).

Step 2: Series Combination:
In series, the reciprocals add: \[ \frac{1}{C_1} = \frac{1}{C}+\frac{1}{C}+\frac{1}{C} = \frac{3}{C} \] So \(C_1 = C/3\).

Step 3: Parallel Combination:
In parallel, the capacitances add: \[ C_2 = C + C + C = 3C \]

Step 4: Take the Ratio:
\[ \frac{C_1}{C_2} = \frac{C/3}{3C} = \frac{1}{9} \]

Step 5: Checking the Options:
Option 1 (1/3) would come from forgetting the factor 3 in the parallel value. Options 2 and 4 are larger than 1, but the series value is smaller than the parallel one. Option 3, 1/9, is right.

Final Answer:
The ratio \(C_1 : C_2\) is 1/9, which is option 3. \[ \boxed{\frac{1}{9}} \]