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: Understanding the Concept:
When multiple capacitors are combined in an electrical circuit, their total or equivalent capacitance depends on how they are wired together. There are two primary configuration arrangements: 1. Parallel Configuration: Capacitors are connected side-by-side across the same voltage source. This arrangement maximizes the total surface area available to store electrical charge, resulting in the maximum possible effective capacitance. 2. Series Configuration: Capacitors are connected end-to-end in a single line. This arrangement effectively increases the total separation distance between the outermost plates, which minimizes the total effective capacitance. To achieve the minimum possible equivalent capacitance from a group of capacitors, they must be wired together in a series configuration.
Step 2: Key Formula or Approach:
The reciprocal of the total equivalent capacitance ($C_s$) for a group of capacitors connected in series is equal to the sum of the reciprocals of their individual capacitances: $$ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots $$ Let's identify the given individual values: - $C_1 = 2 \mu\text{F}$ - $C_2 = 4 \mu\text{F}$ - $C_3 = 8 \mu\text{F}$
Step 3: Detailed Explanation:
Let's substitute our specific capacitance values into the series reciprocal formula: $$ \frac{1}{C_s} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} $$ To add these fractions together, find a common denominator for 2, 4, and 8, which is 8. Convert each fraction accordingly: $$ \frac{1}{C_s} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} $$ $$ \frac{1}{C_s} = \frac{4 + 2 + 1}{8} $$ $$ \frac{1}{C_s} = \frac{7}{8} \mu\text{F}^{-1} $$ Now, invert both sides of the equation to find the actual value of the total series capacitance ($C_s$): $$ C_s = \frac{8}{7} \mu\text{F} $$ Perform the division to convert the fraction into a decimal value: $$ C_s \approx 1.1428 \mu\text{F} $$ Rounding this result to two decimal places gives $1.14 \mu\text{F}$. This matches option (C).
Step 4: Final Answer:
The effective minimum capacitance of the configuration is 1.14 $\mu$F.