Battery R₁ R₂ R3 In the circuit above, a 19 V battery is connected to the circuit where R₁ = 30, R₂ = 38 , and R3 = 35 2. What is the current flowing through R₂ in this circuit? • Do NOT include units • Round to 2 decimal points, include zeros if needed. M w M
Battery R₁ R₂ R3 In the circuit above, a 19 V battery is connected to the circuit where R₁ = 30, R₂ = 38 , and R3 = 35 2. What is the current flowing through R₂ in this circuit? • Do NOT include units • Round to 2 decimal points, include zeros if needed. M w M
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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![### Understanding Circuits and Electrical Resistance
In the diagram provided, a 19 V battery is connected to a parallel circuit with three resistors. The resistances of the resistors are as follows:
- \( R_1 = 30 \, \Omega \)
- \( R_2 = 38 \, \Omega \)
- \( R_3 = 35 \, \Omega \)
Here is an overview and a solution to calculate the current \( I_2 \) flowing through resistor \( R_2 \).
#### Diagram Explanation
The diagram illustrates a simple electrical circuit with a 19 V battery connected in parallel with three resistors labeled \( R_1 \), \( R_2 \), and \( R_3 \). Each resistor has a specified resistance, allowing us to solve for current values using Ohm's Law and principles of parallel circuits.
#### Calculating the Current Flowing Through \( R_2 \)
To find the current flowing through \( R_2 \), we can use Ohm's Law, which states:
\[ I = \frac{V}{R} \]
Given:
- Voltage, \( V = 19 \, V \)
- Resistance of \( R_2 \), \( R_2 = 38 \, \Omega \)
Plugging these values into Ohm's Law:
\[ I_2 = \frac{19}{38} = 0.50 \]
Thus, the current flowing through \( R_2 \) is:
\[ I_2 = 0.50 \]
### Conclusion
For this particular parallel circuit, the current flowing through the resistor \( R_2 \) with a resistance of 38 ohms, when connected to a 19 V battery, is 0.50 A.
#### Note:
- Do **NOT** include units in your final answer.
- Round to 2 decimal points, include zeros if needed.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb910418a-e376-48de-ad2c-cad9a3f8e771%2Ff087611d-5fa2-4acc-9d80-9a2313c34dfd%2F5s9gqwc_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Circuits and Electrical Resistance
In the diagram provided, a 19 V battery is connected to a parallel circuit with three resistors. The resistances of the resistors are as follows:
- \( R_1 = 30 \, \Omega \)
- \( R_2 = 38 \, \Omega \)
- \( R_3 = 35 \, \Omega \)
Here is an overview and a solution to calculate the current \( I_2 \) flowing through resistor \( R_2 \).
#### Diagram Explanation
The diagram illustrates a simple electrical circuit with a 19 V battery connected in parallel with three resistors labeled \( R_1 \), \( R_2 \), and \( R_3 \). Each resistor has a specified resistance, allowing us to solve for current values using Ohm's Law and principles of parallel circuits.
#### Calculating the Current Flowing Through \( R_2 \)
To find the current flowing through \( R_2 \), we can use Ohm's Law, which states:
\[ I = \frac{V}{R} \]
Given:
- Voltage, \( V = 19 \, V \)
- Resistance of \( R_2 \), \( R_2 = 38 \, \Omega \)
Plugging these values into Ohm's Law:
\[ I_2 = \frac{19}{38} = 0.50 \]
Thus, the current flowing through \( R_2 \) is:
\[ I_2 = 0.50 \]
### Conclusion
For this particular parallel circuit, the current flowing through the resistor \( R_2 \) with a resistance of 38 ohms, when connected to a 19 V battery, is 0.50 A.
#### Note:
- Do **NOT** include units in your final answer.
- Round to 2 decimal points, include zeros if needed.
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