How much charge is stored in the 12.0-uF capacitor?
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Transcribed Image Text:**Question:**
How much charge is stored in the 12.0-µF capacitor?
**Diagram Explanation:**
The diagram illustrates a circuit consisting of a 36 V voltage source and multiple capacitors. Capacitors in the circuit are connected as follows:
- Two capacitors, each with a capacitance of 6.0 µF, are connected in series.
- These are further joined in parallel with a 3.0 µF capacitor.
- This combination is connected in series with a 12.0 µF capacitor.
**Choices:**
- 48 µC
- 259.2 µC
![**Description:**
In the provided circuit diagram, we see a configuration involving a voltage source and capacitors. A 36 V voltage source is connected in series with a capacitor labeled as 12.0 μF. This setup is part of a conceptual physics problem aimed at determining the charge stored in the capacitor.
**Components:**
- **Voltage Source:**
- Value: 36 V
- **Capacitor:**
- Capacitance: 12.0 μF
- **Connection Points:**
- Designated with the label 'd', indicating a point in the circuit.
**Question:**
Calculate the charge stored in the 12.0 μF capacitor.
**Choices:**
- 48 μC
- 259.2 μC
- 144 μC
- 518.4 μC
- 72 μC
- 96 μC
**Solution:**
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
- \( C \) is the capacitance.
- \( V \) is the voltage.
Using the values from the diagram:
\[ Q = 12.0 \, \mu\text{F} \times 36 \, \text{V} \]
To solve for \( Q \):
Convert units if necessary; remember \( 1 \mu\text{F} = 10^{-6} \text{F} \).
Compute the product to obtain the charge in microcoulombs (μC).
Feel free to use this setup as an interactive exercise on an educational platform, allowing users to practice calculations related to capacitors and charge storage.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8daeabfb-08fc-4e08-b5f3-009f3913f10d%2Fe3df4345-d1aa-43f0-8f78-cae5c483b888%2F18j0e6_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Description:**
In the provided circuit diagram, we see a configuration involving a voltage source and capacitors. A 36 V voltage source is connected in series with a capacitor labeled as 12.0 μF. This setup is part of a conceptual physics problem aimed at determining the charge stored in the capacitor.
**Components:**
- **Voltage Source:**
- Value: 36 V
- **Capacitor:**
- Capacitance: 12.0 μF
- **Connection Points:**
- Designated with the label 'd', indicating a point in the circuit.
**Question:**
Calculate the charge stored in the 12.0 μF capacitor.
**Choices:**
- 48 μC
- 259.2 μC
- 144 μC
- 518.4 μC
- 72 μC
- 96 μC
**Solution:**
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
- \( C \) is the capacitance.
- \( V \) is the voltage.
Using the values from the diagram:
\[ Q = 12.0 \, \mu\text{F} \times 36 \, \text{V} \]
To solve for \( Q \):
Convert units if necessary; remember \( 1 \mu\text{F} = 10^{-6} \text{F} \).
Compute the product to obtain the charge in microcoulombs (μC).
Feel free to use this setup as an interactive exercise on an educational platform, allowing users to practice calculations related to capacitors and charge storage.
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