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To solve the problem of finding the charge stored on capacitor \( C_3 \), we can break it down step by step based on the circuit provided.

### Given:
- \( C_1 = 5.15 \, \mu\text{F} \)
- \( C_2 = 11.1 \, \mu\text{F} \)
- \( C_3 = 2.35 \, \mu\text{F} \)
- Potential difference between points \( a \) and \( b \), \( V_{ab} = 63.0 \, \text{V} \)

From the circuit diagram, \( C_1 \), \( C_2 \), and \( C_3 \) are arranged in a specific configuration:
- \( C_1 \) and \( C_3 \) are in series.
- This combination is in parallel with \( C_2 \).

### Step 1: Find the equivalent capacitance of \( C_1 \) and \( C_3 \) (series combination).
For capacitors in series, the equivalent capacitance \( C_{13} \) is given by:
\[
\frac{1}{C_{13}} = \frac{1}{C_1} + \frac{1}{C_3}
\]
Substituting the values:
\[
\frac{1}{C_{13}} = \frac{1}{5.15} + \frac{1}{2.35}
\]
\[
C_{13} = \frac{1}{\left(\frac{1}{5.15} + \frac{1}{2.35}\right)} \approx 1.634 \, \mu\text{F}
\]

### Step 2: Find the total equivalent capacitance \( C_{eq} \) of the circuit.
Now, the equivalent capacitance \( C_{eq} \) of the whole system is the parallel combination of \( C_{13} \) and \( C_2 \):
\[
C_{eq} = C_{13} + C_2 = 1.634 \, \mu\text{F} + 11.1 \, \mu\text{F} = 12.734 \, \mu\text{F}
\]

### Step 3: Calculate the total charge stored in the system.
The total charge \( Q_{total} \) is given by:
\[
Q_{total} = C_{eq} \cdot V_{ab}
\]
Substituting the values:
\[
Q_{total} = 12.734 \times 10^{-6} \, \text{F} \times 63.0 \, \text{V} \approx 0.8022 \, \text{mC} = 802.2 \, \mu\text{C}
\]

### Step 4: Find the voltage across \( C_1 \) and \( C_3 \).
Since \( C_1 \) and \( C_3 \) are in series, they have the same charge. To find the charge on \( C_3 \), we first need to find the voltage across the series combination \( C_{13} \).

The voltage across \( C_2 \) is given by:
\[
V_{C2} = \frac{Q_{total}}{C_2} = \frac{802.2 \, \mu\text{C}}{11.1 \, \mu\text{F}} \approx 72.28 \, \text{V}
\]

Now, the voltage across the series combination \( C_3

A group of capacitors is connected as shown in the figure below with C1 = 5.15 μF, C2 = 11.1 μF, and C3 = 2.35 μF. If the potential difference between points a and b is 63.0 V, what charge is stored on C3?

Explore a detailed step-by-step solution to find the charge stored on a specific capacitor (C3 = 2.35 μF) in a complex network. Using series and parallel capacitance formulas and the potential difference of 63.0 V between points a and b, this solution is perfect for Grades 11-12 students.

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