Question

Three capacitors of capacitances \(2\,\mu F\), \(3\,\mu F\), and \(6\,\mu F\) are connected in series. The equivalent capacitance is:

• A. \( 11\,\mu F \)
• B. \( 1\,\mu F \)
• C. \( 0.5\,\mu F \)
• D. \( 2\,\mu F \)

Hint:

In series: \[ C_{eq} < \text{smallest capacitor} \]

Answer 1

The Correct Option is B

Concept: For capacitors in series: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

Step 1: Substitute values \[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]

Step 2: Take LCM (6) \[ \frac{1}{C_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} \] \[ \frac{1}{C_{eq}} = \frac{6}{6} = 1 \]

Step 3: Final result \[ C_{eq} = 1\,\mu F \]