2 A cell is yohed with restetance of 2r and in a series a current of O-25 flews Arozegh it- ahen a second resistance of tase 22 is connected n parallel to te ferce, the current m He Cell increases fo O:30.0what oShe internal resistance of He cell (2) EMF
2 A cell is yohed with restetance of 2r and in a series a current of O-25 flews Arozegh it- ahen a second resistance of tase 22 is connected n parallel to te ferce, the current m He Cell increases fo O:30.0what oShe internal resistance of He cell (2) EMF
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Title: Understanding Electrical Circuits: Series and Parallel Resistors**
**Introduction:**
In this module, we will explore the concept of electrical circuits with resistors arranged in series and parallel configurations. We will analyze the behavior of the circuit when additional resistors are added and determine key electrical parameters such as current, internal resistance, and electromotive force (EMF).
**Problem Statement:**
A cell is joined in a series with a resistance of 2Ω and a current of 0.25A flows through it. When a second resistance of 2Ω is connected in parallel to the first, the current in the cell increases to 0.30A. What are:
(i) The internal resistance of the cell?
(ii) The EMF of the cell?
**Explanation of Terms:**
- **Resistance (Ω)**: The opposition that a substance offers to the flow of electric current.
- **Current (A)**: The rate at which charge is flowing.
- **Internal Resistance**: The resistance within the source of electrical energy (the cell).
- **Electromotive Force (EMF)**: The voltage generated by a battery or cell when no current is flowing.
**Calculations and Diagrams:**
1. **Initial Configuration (Series):**
- Given: \( R_1 = 2Ω \), \( I_1 = 0.25A \)
- Circuit Diagram: A single resistor of 2Ω connected to a cell in series.
2. **Modified Configuration (Parallel):**
- Second Resistance \( R_2 = 2Ω \) connected in parallel to \( R_1 \).
- Total Resistance in parallel: \[ R_{total} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \left(\frac{1}{2} + \frac{1}{2}\right)^{-1} = 1Ω \]
- New Current \( I_2 = 0.30A \).
- Circuit Diagram: Two resistors of 2Ω each, connected in parallel to the cell.
3. **Calculating Internal Resistance and EMF:**
- By applying Ohm’s Law (V = IR), we will determine the internal resistance and the EMF of the cell from the above configurations.
**Conclusion:**
Through the given problem, incorporating resistors in different configurations](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0ead8cad-5c25-4a2c-93dd-9a85004ac883%2F95fde5b5-6beb-48ee-aaf6-8e97c3e662d1%2Fejq2p1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Electrical Circuits: Series and Parallel Resistors**
**Introduction:**
In this module, we will explore the concept of electrical circuits with resistors arranged in series and parallel configurations. We will analyze the behavior of the circuit when additional resistors are added and determine key electrical parameters such as current, internal resistance, and electromotive force (EMF).
**Problem Statement:**
A cell is joined in a series with a resistance of 2Ω and a current of 0.25A flows through it. When a second resistance of 2Ω is connected in parallel to the first, the current in the cell increases to 0.30A. What are:
(i) The internal resistance of the cell?
(ii) The EMF of the cell?
**Explanation of Terms:**
- **Resistance (Ω)**: The opposition that a substance offers to the flow of electric current.
- **Current (A)**: The rate at which charge is flowing.
- **Internal Resistance**: The resistance within the source of electrical energy (the cell).
- **Electromotive Force (EMF)**: The voltage generated by a battery or cell when no current is flowing.
**Calculations and Diagrams:**
1. **Initial Configuration (Series):**
- Given: \( R_1 = 2Ω \), \( I_1 = 0.25A \)
- Circuit Diagram: A single resistor of 2Ω connected to a cell in series.
2. **Modified Configuration (Parallel):**
- Second Resistance \( R_2 = 2Ω \) connected in parallel to \( R_1 \).
- Total Resistance in parallel: \[ R_{total} = \left(\frac{1}{R_1} + \frac{1}{R_2}\right)^{-1} = \left(\frac{1}{2} + \frac{1}{2}\right)^{-1} = 1Ω \]
- New Current \( I_2 = 0.30A \).
- Circuit Diagram: Two resistors of 2Ω each, connected in parallel to the cell.
3. **Calculating Internal Resistance and EMF:**
- By applying Ohm’s Law (V = IR), we will determine the internal resistance and the EMF of the cell from the above configurations.
**Conclusion:**
Through the given problem, incorporating resistors in different configurations
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