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To determine the equivalent capacitance of the given series-parallel combination of capacitors, let's analyze the circuit step-by-step.

### Step-by-Step Solution
1. **Identify Capacitors in Series and Parallel:**
   - From the diagram:
     - \( C_1 = 2 \, \mu\text{F} \)
     - \( C_2 = 3 \, \mu\text{F} \)
     - \( C_3 = 4 \, \mu\text{F} \)
     - \( C_4 = 5 \, \mu\text{F} \)

2. **Combine Capacitors \( C_2 \) and \( C_3 \) in Series:**
   - When capacitors are in series, the equivalent capacitance \( C_{23} \) is given by:
     \[
     \frac{1}{C_{23}} = \frac{1}{C_2} + \frac{1}{C_3}
     \]
     Substituting values:
     \[
     \frac{1}{C_{23}} = \frac{1}{3 \, \mu\text{F}} + \frac{1}{4 \, \mu\text{F}}
     \]
     \[
     \frac{1}{C_{23}} = \frac{4 + 3}{12} = \frac{7}{12}
     \]
     Therefore:
     \[
     C_{23} = \frac{12}{7} \approx 1.71 \, \mu\text{F}
     \]

3. **Combine \( C_{23} \) with \( C_4 \) in Parallel:**
   - When capacitors are in parallel, the equivalent capacitance \( C_{234} \) is:
     \[
     C_{234} = C_{23} + C_4
     \]
     Substituting values:
     \[
     C_{234} = 1.71 \, \mu\text{F} + 5 \, \mu\text{F} = 6.71 \, \mu\text{F}
     \]

4. **Combine \( C_{234} \) with \( C_1 \) in Series:**
   - Now, we combine \( C_{234} \) and \( C_1 \) in series:
     \[
     \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_{234}}
     \]
     Substituting values:
     \[
     \frac{1}{C_{\text{eq}}} = \frac{1}{2 \, \mu\text{F}} + \frac{1}{6.71 \, \mu\text{F}}
     \]
     \[
     \frac{1}{C_{\text{eq}}} = \frac{6.71 + 2}{13.42} = \frac{8.71}{13.42}
     \]
     Therefore:
     \[
     C_{\text{eq}} \approx 1.18 \, \mu\text{F}
     \]

### Answer
The resulting capacitance of the circuit is:
\[
C_{\text{eq}} = 1.18 \, \mu\text{F}
\]

Would you like more details or have any questions?

### Related Questions:
1. What would happen to the equivalent capacitance if \( C_1 \) were increased?
2. How do you calculate equivalent capacitance for purely parallel combinations?
3. What changes if \( C_2 \) and \( C_3 \) are arranged in parallel instead of series?
4. Can you explain why capacitors in series result in a smaller equivalent capacitance?
5. How does adding a capacitor in parallel affect the total capacitance?

### Tip:
For series capacitors, the equivalent capacitance is always less than the smallest individual capacitor in the series.

¿Cuál es la capacidad resultante de la siguiente combinación serie-paralelo que se muestra en la figura?

Learn how to calculate the equivalent capacitance of a complex series-parallel circuit with capacitors of 2 μF, 3 μF, 4 μF, and 5 μF. This example demonstrates key concepts in circuit analysis, including capacitance in series and parallel arrangements. Suitable for students in advanced physics courses.

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