Question

Three capacitors 2, 4, 6 respectively. Find the total capacitance in series and parallel

Answer 1

To find the total capacitance of capacitors in series and parallel, we can use the following formulas:

### 1. Capacitors in Series
For capacitors in series, the total capacitance \( C_{\text{total}} \) is given by the formula:

\[
\frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}
\]

Where \( C_1 = 2 \, \mu F \), \( C_2 = 4 \, \mu F \), and \( C_3 = 6 \, \mu F \).

Calculating:

\[
\frac{1}{C_{\text{total}}} = \frac{1}{2} + \frac{1}{4} + \frac{1}{6}
\]

Finding a common denominator (which is 12):

\[
\frac{1}{C_{\text{total}}} = \frac{6}{12} + \frac{3}{12} + \frac{2}{12} = \frac{11}{12}
\]

Thus:

\[
C_{\text{total}} = \frac{12}{11} \approx 1.09 \, \mu F
\]

### 2. Capacitors in Parallel
For capacitors in parallel, the total capacitance \( C_{\text{total}} \) is simply the sum of the individual capacitances:

\[
C_{\text{total}} = C_1 + C_2 + C_3
\]

Calculating:

\[
C_{\text{total}} = 2 + 4 + 6 = 12 \, \mu F
\]

### Summary
- Total capacitance in **series**: \( \approx 1.09 \, \mu F \)
- Total capacitance in **parallel**: \( 12 \, \mu F \)