To calculate the total capacitance (\(C\)) of capacitors in series:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_x} \]
Where:
Capacitors in series are defined as 2 or more capacitors in a circuit that are connected in series. That is, the current has to flow through the first capacitor before it can get to subsequent capacitors.
Let's assume the following values:
Using the formula:
\[ \frac{1}{C} = \frac{1}{10} + \frac{1}{20} = 0.15 \]
So, the total capacitance is:
\[ C = \frac{1}{0.15} = 6.67 \text{ µF} \]
The total capacitance is 6.67 µF.
Let's assume the following values:
Using the formula:
\[ \frac{1}{C} = \frac{1}{5} + \frac{1}{10} + \frac{1}{15} = 0.3667 \]
So, the total capacitance is:
\[ C = \frac{1}{0.3667} = 2.73 \text{ µF} \]
The total capacitance is 2.73 µF.