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

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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.
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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