Find the equivalent capacitance of the grouping between points a and b

General
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find the equivalent capacitance of the grouping between points a and b.

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Find the equivalent capacitance of the grouping between points a and b

Answer:
To find the equivalent capacitance of a grouping of capacitors between points (a) and (b), we need to analyze the configuration of the capacitors. Capacitors can be connected in series, parallel, or a combination of both. Here’s how to approach each scenario:

1. Capacitors in Series:

When capacitors are connected in series, the reciprocal of the total (equivalent) capacitance (C_{\text{eq}}) is the sum of the reciprocals of the individual capacitances (C_1, C_2, …, C_n):

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots + \frac{1}{C_n}

2. Capacitors in Parallel:

When capacitors are connected in parallel, the total (equivalent) capacitance (C_{\text{eq}}) is the sum of the individual capacitances:

C_{\text{eq}} = C_1 + C_2 + \cdots + C_n

3. Combination of Series and Parallel:

In more complex circuits, capacitors might be arranged in a combination of series and parallel. In such cases, you need to:

  1. Identify and simplify series and parallel groups step by step.
  2. Replace each simplified group with its equivalent capacitance.
  3. Continue simplifying until you reduce the circuit to a single equivalent capacitance between points (a) and (b).

Example Problem:

Consider a circuit with the following capacitors:

  • C_1 = 2 \, \mu\text{F}
  • C_2 = 3 \, \mu\text{F}
  • C_3 = 6 \, \mu\text{F}

Step-by-Step Solution:

  1. Identify Series and Parallel Groups:

    Suppose (C_1) and (C_2) are in series, and their combination is in parallel with (C_3).

  2. Calculate Series Combination:

    For (C_1) and (C_2) in series:

    \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}

    Therefore:

    C_{\text{series}} = \frac{6}{5} \, \mu\text{F} = 1.2 \, \mu\text{F}

  3. Calculate Parallel Combination:

    Now, (C_{\text{series}}) is in parallel with (C_3):

    C_{\text{eq}} = C_{\text{series}} + C_3 = 1.2 \, \mu\text{F} + 6 \, \mu\text{F} = 7.2 \, \mu\text{F}

Conclusion:

The equivalent capacitance of the grouping between points (a) and (b) is 7.2 \, \mu\text{F}.

By following these steps, you can determine the equivalent capacitance for any configuration of capacitors. Make sure to carefully identify series and parallel connections and simplify them step by step.