Question

Three capacitors of capacitance 6 µF are available. The minimum and maximum capacitances obtained are

• A. 3 μF, 12 μF
• B. 2 μF, 12 μF
• C. 2 µF, 18µF
• D. 6 μF, 18 μF

Hint:

Remember that the rules for combining capacitors are the opposite of those for resistors. Capacitors in \textbf{parallel} add up like resistors in \textbf{series} (\(C_{eq} = C_1 + C_2 + \dots\)). Capacitors in \textbf{series} use the reciprocal formula, like resistors in \textbf{parallel} (\(1/C_{eq} = 1/C_1 + 1/C_2 + \dots\)).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To obtain the maximum equivalent capacitance from a set of capacitors, they should be connected in parallel. To obtain the minimum equivalent capacitance, they should be connected in series.
Step 2: Key Formula or Approach:
Given three capacitors, each with capacitance \( C = 6 \, \mu F \).
For Parallel Combination (Maximum Capacitance): The equivalent capacitance \(C_{\text{p}}\) is the sum of individual capacitances. \[ C_{\text{max}} = C_1 + C_2 + C_3 \] For Series Combination (Minimum Capacitance): The reciprocal of the equivalent capacitance \(C_{\text{s}}\) is the sum of the reciprocals of individual capacitances. \[ \frac{1}{C_{\text{min}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] Step 3: Detailed Explanation:
Calculation for Maximum Capacitance (Parallel):
Connect all three 6 µF capacitors in parallel. \[ C_{\text{max}} = 6 \, \mu F + 6 \, \mu F + 6 \, \mu F = 18 \, \mu F \] Calculation for Minimum Capacitance (Series):
Connect all three 6 µF capacitors in series. \[ \frac{1}{C_{\text{min}}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{1+1+1}{6} = \frac{3}{6} = \frac{1}{2} \] To find \(C_{\text{min}}\), we take the reciprocal: \[ C_{\text{min}} = 2 \, \mu F \] Step 4: Final Answer:
The minimum capacitance obtained is 2 µF, and the maximum capacitance obtained is 18 µF. Therefore, option (C) is the correct choice.