To determine the equivalent capacitance of the given combination of capacitors, the configuration of the capacitors needs to be clarified (whether they are in series, parallel, or a combination). However, since I cannot see the image you referenced (pic-Q8-Ex2), I will demonstrate how to calculate the equivalent capacitance for both series and parallel configurations, and then you can apply these formulas based on your specific circuit arrangement.
1. Capacitors in Series:
The formula for the equivalent capacitance \(C_{eq}\) of capacitors in series \(C_1, C_2, C_3, \ldots, C_n\) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \ldots + \frac{1}{C_n} \]
2. Capacitors in Parallel:
The formula for the equivalent capacitance \(C_{eq}\) of capacitors in parallel \(C_1, C_2, C_3, \ldots, C_n\) is given by:
\[ C_{eq} = C_1 + C_2 + C_3 + \ldots + C_n \]
Example Calculation:
Assuming you have three capacitors: \(C_1 = 45 \mu F\), \(C_2 = 38 \mu F\), \(C_3 = 52 \mu F\). You can perform the calculations as follows.
If in series:
\[ \frac{1}{C_{eq}} = \frac{1}{45} + \frac{1}{38} + \frac{1}{52} \]
Calculating each term in \( \mu F\):
- Calculate \( \frac{1}{45} \approx 0.0222 \)
- Calculate \( \frac{1}{38} \approx 0.0263 \)
- Calculate \( \frac{1}{52} \approx 0.0192 \)
Adding those,
\[ \frac{1}{C_{eq}} = 0.0222 + 0.0263 + 0.0192 \approx 0.0677 \] Therefore, \( C_{eq} \approx \frac{1}{0.0677} \approx 14.77 \mu F\).
If in parallel:
\[ C_{eq} = 45 + 38 + 52 = 135 \mu F \]
Now, please apply these formulas based on the specific configuration you have in your image. If you can provide the layout (e.g., which capacitors are in series and which are in parallel), I can help you get a more precise result for your situation.
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