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The figure below shows four capacitors with  CA = 4.00 µF,   CB = 8.00 µF,   CC = 5.00 µF,  and  CD = 2.00 µF  connected across points a and b,

The figure below shows four capacitors with  CA = 4.00 µF,   CB = 8.00 µF,   CC = 5.00 µF,  and  CD = 2.00 µF  connected across points a and b,

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The figure below shows four capacitors with 

CA = 4.00 µF,

 

CB = 8.00 µF,

 

CC = 5.00 µF,

 and 

CD = 2.00 µF

 connected across points a and b, which have potential difference 

ΔVab = 12.0 V.
 
(a) What is the equivalent capacitance of the four capacitors?
### Capacitors in Combination **Problem Statement:** The figure below shows four capacitors with \( C_A = 4.00 \, \mu\text{F} \), \( C_B = 8.00 \, \mu\text{F} \), \( C_C = 5.00 \, \mu\text{F} \), and \( C_D = 2.00 \, \mu\text{F} \) connected across points \( a \) and \( b \), which have a potential difference \( \Delta V_{ab} = 12.0 \, \text{V} \). **Diagram Explanation:** A schematic diagram is provided in which four capacitors are connected. The points \(a\) and \(b\) are marked, and the potential difference between them is given as \(12.0 \, \text{V}\). - Capacitor \(C_A\) is connected between point \(a\) and one terminal of \(C_B\). - Capacitor \(C_B\) is then connected to point \(b\). - Capacitors \(C_C\) and \(C_D\) are connected in series between point \(a\) and the junction between \(C_B\) and \(b\), forming a parallel circuit with \(C_A - C_B\). **Question:** (a) What is the equivalent capacitance of the four capacitors? **Answer:** The text box provided shows a calculated value of 2.81 but it is marked with a red cross indicating the value is incorrect. **Solution Steps:** 1. **Combine Capacitors \(C_C\) and \(C_D\) in Series:** \[ \frac{1}{C_{CD}} = \frac{1}{C_C} + \frac{1}{C_D} \] \[ \frac{1}{C_{CD}} = \frac{1}{5.00 \, \mu\text{F}} + \frac{1}{2.00 \, \mu\text{F}} \] \[ \frac{1}{C_{CD}} = 0.2 + 0.5 = 0.7 \] \[ C_{CD} = \frac{1}{0.7} \approx 1.43 \,
Transcribed Image Text:### Capacitors in Combination **Problem Statement:** The figure below shows four capacitors with \( C_A = 4.00 \, \mu\text{F} \), \( C_B = 8.00 \, \mu\text{F} \), \( C_C = 5.00 \, \mu\text{F} \), and \( C_D = 2.00 \, \mu\text{F} \) connected across points \( a \) and \( b \), which have a potential difference \( \Delta V_{ab} = 12.0 \, \text{V} \). **Diagram Explanation:** A schematic diagram is provided in which four capacitors are connected. The points \(a\) and \(b\) are marked, and the potential difference between them is given as \(12.0 \, \text{V}\). - Capacitor \(C_A\) is connected between point \(a\) and one terminal of \(C_B\). - Capacitor \(C_B\) is then connected to point \(b\). - Capacitors \(C_C\) and \(C_D\) are connected in series between point \(a\) and the junction between \(C_B\) and \(b\), forming a parallel circuit with \(C_A - C_B\). **Question:** (a) What is the equivalent capacitance of the four capacitors? **Answer:** The text box provided shows a calculated value of 2.81 but it is marked with a red cross indicating the value is incorrect. **Solution Steps:** 1. **Combine Capacitors \(C_C\) and \(C_D\) in Series:** \[ \frac{1}{C_{CD}} = \frac{1}{C_C} + \frac{1}{C_D} \] \[ \frac{1}{C_{CD}} = \frac{1}{5.00 \, \mu\text{F}} + \frac{1}{2.00 \, \mu\text{F}} \] \[ \frac{1}{C_{CD}} = 0.2 + 0.5 = 0.7 \] \[ C_{CD} = \frac{1}{0.7} \approx 1.43 \,
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