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

Step 1: Understand series capacitors.
When capacitors are joined end to end in a single line, they are in series. The combined capacitance is always smaller than the smallest one in the line.

Step 2: Write the series rule.
For series, the reciprocals add: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}. \]

Step 3: Put in the values.
With $2\,\mu F$, $3\,\mu F$, and $6\,\mu F$: \[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}. \]

Step 4: Use a common denominator.
The smallest common denominator is $6$: \[ \frac{1}{C_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6}. \]

Step 5: Add the fractions.
\[ \frac{1}{C_{eq}} = \frac{6}{6} = 1. \]

Step 6: State the answer.
Taking the reciprocal gives the equivalent capacitance. \[ \boxed{1\,\mu F} \]