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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Capacitance Problem: Equivalent Capacitance Calculation**
**Objective**: Find the equivalent capacitance \( C_{\text{eq}} \) of the combination of capacitors shown in the figure.
**Capacitor Values**:
- \( C_1 = 0.775 \, \mu\text{F} \)
- \( C_2 = 10.0 \, \mu\text{F} \)
- \( C_3 = 2.25 \, \mu\text{F} \)
**Diagram Explanation**:
- The diagram illustrates three capacitors. Capacitors \( C_1 \) and \( C_2 \) are in parallel on the left side. Capacitor \( C_3 \) is in series with the combination of \( C_1 \) and \( C_2 \).
**Calculation**:
1. **Parallel Combination**: The equivalent capacitance for capacitors in parallel is calculated by adding their capacitances:
\[
C_{\text{parallel}} = C_1 + C_2
\]
Substituting the given values:
\[
C_{\text{parallel}} = 0.775 \, \mu\text{F} + 10.0 \, \mu\text{F} = 10.775 \, \mu\text{F}
\]
2. **Series Combination**: The overall equivalent capacitance \( C_{\text{eq}} \) for the series combination with \( C_3 \) is found using the formula for capacitors in series:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{C_{\text{parallel}}} + \frac{1}{C_3}
\]
Substituting the values:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{10.775 \, \mu\text{F}} + \frac{1}{2.25 \, \mu\text{F}}
\]
3. **Solve for \( C_{\text{eq}} \)**:
Calculate the reciprocal of the sum to find the equivalent capacitance.
**Note**: The final value should be computed using the relevant formulas to ensure understanding of series and parallel combinations.
**Source**: OpenStax College Physics](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff07b499f-ca10-4d91-9b90-60cf46092d14%2F30c8e931-50db-436d-b168-d92219104356%2Fyfmg0tc_processed.png&w=3840&q=75)
Transcribed Image Text:**Capacitance Problem: Equivalent Capacitance Calculation**
**Objective**: Find the equivalent capacitance \( C_{\text{eq}} \) of the combination of capacitors shown in the figure.
**Capacitor Values**:
- \( C_1 = 0.775 \, \mu\text{F} \)
- \( C_2 = 10.0 \, \mu\text{F} \)
- \( C_3 = 2.25 \, \mu\text{F} \)
**Diagram Explanation**:
- The diagram illustrates three capacitors. Capacitors \( C_1 \) and \( C_2 \) are in parallel on the left side. Capacitor \( C_3 \) is in series with the combination of \( C_1 \) and \( C_2 \).
**Calculation**:
1. **Parallel Combination**: The equivalent capacitance for capacitors in parallel is calculated by adding their capacitances:
\[
C_{\text{parallel}} = C_1 + C_2
\]
Substituting the given values:
\[
C_{\text{parallel}} = 0.775 \, \mu\text{F} + 10.0 \, \mu\text{F} = 10.775 \, \mu\text{F}
\]
2. **Series Combination**: The overall equivalent capacitance \( C_{\text{eq}} \) for the series combination with \( C_3 \) is found using the formula for capacitors in series:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{C_{\text{parallel}}} + \frac{1}{C_3}
\]
Substituting the values:
\[
\frac{1}{C_{\text{eq}}} = \frac{1}{10.775 \, \mu\text{F}} + \frac{1}{2.25 \, \mu\text{F}}
\]
3. **Solve for \( C_{\text{eq}} \)**:
Calculate the reciprocal of the sum to find the equivalent capacitance.
**Note**: The final value should be computed using the relevant formulas to ensure understanding of series and parallel combinations.
**Source**: OpenStax College Physics
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