Question

Three capacitors have capacitances \(2\,\mu F\), \(4\,\mu F\) and \(8\,\mu F\). When they are connected in a way to give minimum capacitance, their effective capacitance is:

• A. \(8\,\mu F\)
• B. \(14\,\mu F\)
• C. \(1.14\,\mu F\)
• D. \(11.4\,\mu F\)

Hint:

For capacitors: Maximum capacitance: \[ C_{\max}=C_1+C_2+C_3+\cdots \] Minimum capacitance: \[ \frac{1}{C_{\min}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} +\cdots \] Hence, minimum capacitance is always obtained when all capacitors are connected in series.

Answer 1

The Correct Option is C

Step 1: Determine the arrangement for minimum capacitance. Minimum capacitance is obtained when all capacitors are connected in series. Therefore, \[ \frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \]

Step 2: Calculate the reciprocal. \[ \frac{1}{C_{\text{eq}}} = \frac{4+2+1}{8} \] \[ = \frac{7}{8} \]

Step 3: Find the equivalent capacitance. \[ C_{\text{eq}} = \frac{8}{7} \] \[ = 1.14\,\mu F \] Therefore, the effective capacitance is \[ \boxed{1.14\,\mu F} \]