Each capacitor in the combination shown in the figure below (C V 20.0 με 20.0 με с HH μr) has a breakdown voltage of 13.0 V. What is the breakdown voltage of the combination? 20.0 με 20.0 με HI
Each capacitor in the combination shown in the figure below (C V 20.0 με 20.0 με с HH μr) has a breakdown voltage of 13.0 V. What is the breakdown voltage of the combination? 20.0 με 20.0 με HI
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![### Capacitor Combination Breakdown Voltage Calculation
**Problem Statement:**
Each capacitor in the combination shown in the figure below (C = 17.0 μF) has a breakdown voltage of 13.0 V. What is the breakdown voltage of the combination?
**Diagram Explanation:**
The circuit consists of two pairs of capacitors. Each pair includes two capacitors, each with a capacitance of 20.0 μF, connected in parallel. These parallel pairs are then connected in series with a capacitor of 17.0 μF, and there is a reference label for points \( a \) and \( b \).
**Diagram Details:**
- Two capacitors, each of 20.0 μF, connected in parallel between points \( a \) and \( C \).
- A capacitor of 17.0 μF connected in series between points \( C \) and \( b \).
- Two capacitors, each of 20.0 μF, connected in parallel between points \( C \) and \( b \).
**Breakdown Voltage Calculation:**
1. **Parallel Capacitors:**
- When capacitors are connected in parallel, their capacitances add up.
- For each pair of 20.0 μF capacitors in parallel:
\[ C_{\text{parallel}} = 20.0 \,\mu\text{F} + 20.0 \,\mu\text{F} = 40.0 \,\mu\text{F} \]
2. **Series Capacitors:**
- When capacitors are connected in series, the total capacitance (C_total) can be found using the formula:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Applying to our circuit:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{40.0 \,\mu\text{F}} + \frac{1}{17.0 \,\mu\text{F}} + \frac{1}{40.0 \,\mu\text{F}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{1}{40} + \frac{1}{17} + \frac{1}{40} \]
3](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F75edd112-ce06-4202-807b-82434248c0a2%2F5cb9f879-9c0a-450e-96f3-0d947ddcd657%2Fhn2uutg_processed.png&w=3840&q=75)
Transcribed Image Text:### Capacitor Combination Breakdown Voltage Calculation
**Problem Statement:**
Each capacitor in the combination shown in the figure below (C = 17.0 μF) has a breakdown voltage of 13.0 V. What is the breakdown voltage of the combination?
**Diagram Explanation:**
The circuit consists of two pairs of capacitors. Each pair includes two capacitors, each with a capacitance of 20.0 μF, connected in parallel. These parallel pairs are then connected in series with a capacitor of 17.0 μF, and there is a reference label for points \( a \) and \( b \).
**Diagram Details:**
- Two capacitors, each of 20.0 μF, connected in parallel between points \( a \) and \( C \).
- A capacitor of 17.0 μF connected in series between points \( C \) and \( b \).
- Two capacitors, each of 20.0 μF, connected in parallel between points \( C \) and \( b \).
**Breakdown Voltage Calculation:**
1. **Parallel Capacitors:**
- When capacitors are connected in parallel, their capacitances add up.
- For each pair of 20.0 μF capacitors in parallel:
\[ C_{\text{parallel}} = 20.0 \,\mu\text{F} + 20.0 \,\mu\text{F} = 40.0 \,\mu\text{F} \]
2. **Series Capacitors:**
- When capacitors are connected in series, the total capacitance (C_total) can be found using the formula:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
- Applying to our circuit:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{40.0 \,\mu\text{F}} + \frac{1}{17.0 \,\mu\text{F}} + \frac{1}{40.0 \,\mu\text{F}} \]
\[ \frac{1}{C_{\text{total}}} = \frac{1}{40} + \frac{1}{17} + \frac{1}{40} \]
3
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