A circuit has resistors R1 = 353 N and R2 125 2, and two batteries V = 36.0 V and V2 = 9.00 V. The variable resistor R is adjusted until the galvanometer reads 0 A. Calculate the unknown resistance R.
A circuit has resistors R1 = 353 N and R2 125 2, and two batteries V = 36.0 V and V2 = 9.00 V. The variable resistor R is adjusted until the galvanometer reads 0 A. Calculate the unknown resistance R.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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![### Problem Description
A circuit has resistors \( R_1 = 353 \, \Omega \) and \( R_2 = 125 \, \Omega \), and two batteries \( V_1 = 36.0 \, \text{V} \) and \( V_2 = 9.00 \, \text{V} \). The variable resistor \( R \) is adjusted until the galvanometer reads \( 0 \, \text{A} \). Calculate the unknown resistance \( R \).
\[ R = \_\_\_\_\_\_\_\_\_ \Omega \]
### Diagram Description
The diagram depicts a circuit consisting of the following components:
- **Resistor \( R_1 \)**: Positioned at the top left of the circuit and connected in series with battery \( V_1 \).
- **Battery \( V_1 \)**: Located at the bottom left of the circuit, providing \( 36.0 \, \text{V} \).
- **Variable Resistor \( R \)**: Positioned between the connection point of \( R_1 \) and \( R_2 \). It can be adjusted.
- **Resistor \( R_2 \)**: Positioned at the top right of the circuit and connected in series with battery \( V_2 \).
- **Battery \( V_2 \)**: Located at the bottom right of the circuit, providing \( 9.00 \, \text{V} \).
- **Galvanometer \( G \)**: Positioned between \( R_2 \) and \( V_2 \), used to measure the current. When adjusted correctly, it reads \( 0 \, \text{A} \).
### Calculation and Results
To find the unknown resistance \( R \), we need to use the principle that, when the galvanometer reads \( 0 \, \text{A} \), the potential difference across the variable resistor \( R \) is zero. This means the voltage drops across \( R_1 \) and \( R_2 \) must be equal. By using the following equations, we can solve for \( R \):
\[ I_1 = \frac{V_1}{R_1 + R} \]
\[ I_2 = \frac{V_2}{R_2} \]
When the current through the galvanometer is zero (\( I_G = 0 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F37e1db01-06d0-4e20-aae1-26f96b6814c3%2Fff186dc8-e5dd-41b5-96a0-b9fdfd086709%2F6xfsdss_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Description
A circuit has resistors \( R_1 = 353 \, \Omega \) and \( R_2 = 125 \, \Omega \), and two batteries \( V_1 = 36.0 \, \text{V} \) and \( V_2 = 9.00 \, \text{V} \). The variable resistor \( R \) is adjusted until the galvanometer reads \( 0 \, \text{A} \). Calculate the unknown resistance \( R \).
\[ R = \_\_\_\_\_\_\_\_\_ \Omega \]
### Diagram Description
The diagram depicts a circuit consisting of the following components:
- **Resistor \( R_1 \)**: Positioned at the top left of the circuit and connected in series with battery \( V_1 \).
- **Battery \( V_1 \)**: Located at the bottom left of the circuit, providing \( 36.0 \, \text{V} \).
- **Variable Resistor \( R \)**: Positioned between the connection point of \( R_1 \) and \( R_2 \). It can be adjusted.
- **Resistor \( R_2 \)**: Positioned at the top right of the circuit and connected in series with battery \( V_2 \).
- **Battery \( V_2 \)**: Located at the bottom right of the circuit, providing \( 9.00 \, \text{V} \).
- **Galvanometer \( G \)**: Positioned between \( R_2 \) and \( V_2 \), used to measure the current. When adjusted correctly, it reads \( 0 \, \text{A} \).
### Calculation and Results
To find the unknown resistance \( R \), we need to use the principle that, when the galvanometer reads \( 0 \, \text{A} \), the potential difference across the variable resistor \( R \) is zero. This means the voltage drops across \( R_1 \) and \( R_2 \) must be equal. By using the following equations, we can solve for \( R \):
\[ I_1 = \frac{V_1}{R_1 + R} \]
\[ I_2 = \frac{V_2}{R_2} \]
When the current through the galvanometer is zero (\( I_G = 0 \
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