The capacitors C, = 11.5 µF, C2 = 31 pf, C3 = 39.5 µf, C4 = 41 pF, C5 = 60 µF, and C, = 13.5 µF are connected as shown below. What is the capacitance of the circuit between points a and b? C2 C3 C5 C, The equivalent capacitance, Ceg x Units uF 121.42 What is the total charge and energy stored in the system if a 18-V battery is connected to points a and b? The charge in the system, Q = Units pC The energy in the system, U = Units mJ

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### Capacitors and Their Arrangements

The circuit depicted involves the following capacitors:
- \( C_1 = 11.5 \, \mu\text{F} \)
- \( C_2 = 31 \, \mu\text{F} \)
- \( C_3 = 39.5 \, \mu\text{F} \)
- \( C_4 = 41 \, \mu\text{F} \)
- \( C_5 = 60 \, \mu\text{F} \)
- \( C_6 = 13.5 \, \mu\text{F} \)

#### Diagram Explanation
- The capacitors \( C_4, C_5, \) and \( C_6 \) are connected in parallel. This parallel combination is in series with three other capacitors \( C_1, C_2, \) and \( C_3 \).

### Calculation of Equivalent Capacitance
1. **Parallel Arrangement**: 
   - For capacitors in parallel, the total capacitance \( C_p \) is the sum:
     \[
     C_p = C_4 + C_5 + C_6 = 41 + 60 + 13.5 = 114.5 \, \mu\text{F}
     \]

2. **Series Arrangement**:
   - The capacitors \( C_1, C_2, C_3, \) and the parallel combination \( C_p \) are in series. The formula for total capacitance \( C_s \) is:
     \[
     \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_p}
     \]
   - Plug in the values:
     \[
     \frac{1}{C_s} = \frac{1}{11.5} + \frac{1}{31} + \frac{1}{39.5} + \frac{1}{114.5}
     \]

The equivalent capacitance \( C_{eq} \) found was incorrectly attempted to be calculated as \( 121.42 \, \mu\text{F} \).

### Additional Calculations with a 18-V Battery
- **Charge in the System**
Transcribed Image Text:### Capacitors and Their Arrangements The circuit depicted involves the following capacitors: - \( C_1 = 11.5 \, \mu\text{F} \) - \( C_2 = 31 \, \mu\text{F} \) - \( C_3 = 39.5 \, \mu\text{F} \) - \( C_4 = 41 \, \mu\text{F} \) - \( C_5 = 60 \, \mu\text{F} \) - \( C_6 = 13.5 \, \mu\text{F} \) #### Diagram Explanation - The capacitors \( C_4, C_5, \) and \( C_6 \) are connected in parallel. This parallel combination is in series with three other capacitors \( C_1, C_2, \) and \( C_3 \). ### Calculation of Equivalent Capacitance 1. **Parallel Arrangement**: - For capacitors in parallel, the total capacitance \( C_p \) is the sum: \[ C_p = C_4 + C_5 + C_6 = 41 + 60 + 13.5 = 114.5 \, \mu\text{F} \] 2. **Series Arrangement**: - The capacitors \( C_1, C_2, C_3, \) and the parallel combination \( C_p \) are in series. The formula for total capacitance \( C_s \) is: \[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_p} \] - Plug in the values: \[ \frac{1}{C_s} = \frac{1}{11.5} + \frac{1}{31} + \frac{1}{39.5} + \frac{1}{114.5} \] The equivalent capacitance \( C_{eq} \) found was incorrectly attempted to be calculated as \( 121.42 \, \mu\text{F} \). ### Additional Calculations with a 18-V Battery - **Charge in the System**
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