### Electrical Circuit Analysis #### Circuit Diagram A simple parallel circuit is shown with a 12V battery connected to two resistors: one with a resistance of 2 ohms and another with a resistance of 3 ohms. #### Questions and Analysis **A) Fill in a VIPR table for this circuit:** | Device | Voltage (V) | Current (I) | Resistance (R) | Power (P) | |--------------|-------------|-------------|----------------|---------------| | 2 ohm bulb | | | 2 ohm | | | 3 ohm bulb | | | 3 ohm | | | Battery | | | | | To complete this table, calculate the voltage, current, and power for each resistor. **B) Is the 2 ohm or 3 ohm bulb brighter? Briefly explain or show calculations to prove your answer.** To determine brightness: - **Power (P)** must be calculated using the formula: \( P = \frac{V^2}{R} \). For the 2-ohm bulb: \( P_{2 ohm} = \frac{12V^2}{2 \Omega} = \frac{144}{2} = 72W \) For the 3-ohm bulb: \( P_{3 ohm} = \frac{12V^2}{3 \Omega} = \frac{144}{3} = 48W \) Thus, the 2-ohm bulb is brighter, as it has a higher power dissipation of 72W compared to 48W for the 3-ohm bulb. **C) If you attached a third lightbulb in parallel to the other two, how would it affect the brightness of the 2 ohm lightbulb?** In a parallel circuit, the voltage across each branch remains unchanged. Adding a third lightbulb in parallel will not affect the voltage across the 2-ohm lightbulb. Consequently, the brightness of the 2-ohm lightbulb will remain the same. --- This analysis helps in understanding the behavior of parallel circuits and the distribution of voltage, current, and power within them. Proper understanding of these concepts is crucial for designing efficient electrical systems and devices.

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### Electrical Circuit Analysis

#### Circuit Diagram
A simple parallel circuit is shown with a 12V battery connected to two resistors: one with a resistance of 2 ohms and another with a resistance of 3 ohms.

#### Questions and Analysis

**A) Fill in a VIPR table for this circuit:**

| Device       | Voltage (V) | Current (I) | Resistance (R) | Power (P)     |
|--------------|-------------|-------------|----------------|---------------|
| 2 ohm bulb   |             |             | 2 ohm          |               |
| 3 ohm bulb   |             |             | 3 ohm          |               |
| Battery      |             |             |                |               |

To complete this table, calculate the voltage, current, and power for each resistor.

**B) Is the 2 ohm or 3 ohm bulb brighter? Briefly explain or show calculations to prove your answer.**

To determine brightness:
- **Power (P)** must be calculated using the formula: \( P = \frac{V^2}{R} \).

For the 2-ohm bulb:
 \( P_{2 ohm} = \frac{12V^2}{2 \Omega} = \frac{144}{2} = 72W \)

For the 3-ohm bulb:
 \( P_{3 ohm} = \frac{12V^2}{3 \Omega} = \frac{144}{3} = 48W \)

Thus, the 2-ohm bulb is brighter, as it has a higher power dissipation of 72W compared to 48W for the 3-ohm bulb.

**C) If you attached a third lightbulb in parallel to the other two, how would it affect the brightness of the 2 ohm lightbulb?**

In a parallel circuit, the voltage across each branch remains unchanged. Adding a third lightbulb in parallel will not affect the voltage across the 2-ohm lightbulb. Consequently, the brightness of the 2-ohm lightbulb will remain the same.

---

This analysis helps in understanding the behavior of parallel circuits and the distribution of voltage, current, and power within them. Proper understanding of these concepts is crucial for designing efficient electrical systems and devices.
Transcribed Image Text:### Electrical Circuit Analysis #### Circuit Diagram A simple parallel circuit is shown with a 12V battery connected to two resistors: one with a resistance of 2 ohms and another with a resistance of 3 ohms. #### Questions and Analysis **A) Fill in a VIPR table for this circuit:** | Device | Voltage (V) | Current (I) | Resistance (R) | Power (P) | |--------------|-------------|-------------|----------------|---------------| | 2 ohm bulb | | | 2 ohm | | | 3 ohm bulb | | | 3 ohm | | | Battery | | | | | To complete this table, calculate the voltage, current, and power for each resistor. **B) Is the 2 ohm or 3 ohm bulb brighter? Briefly explain or show calculations to prove your answer.** To determine brightness: - **Power (P)** must be calculated using the formula: \( P = \frac{V^2}{R} \). For the 2-ohm bulb: \( P_{2 ohm} = \frac{12V^2}{2 \Omega} = \frac{144}{2} = 72W \) For the 3-ohm bulb: \( P_{3 ohm} = \frac{12V^2}{3 \Omega} = \frac{144}{3} = 48W \) Thus, the 2-ohm bulb is brighter, as it has a higher power dissipation of 72W compared to 48W for the 3-ohm bulb. **C) If you attached a third lightbulb in parallel to the other two, how would it affect the brightness of the 2 ohm lightbulb?** In a parallel circuit, the voltage across each branch remains unchanged. Adding a third lightbulb in parallel will not affect the voltage across the 2-ohm lightbulb. Consequently, the brightness of the 2-ohm lightbulb will remain the same. --- This analysis helps in understanding the behavior of parallel circuits and the distribution of voltage, current, and power within them. Proper understanding of these concepts is crucial for designing efficient electrical systems and devices.
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