**Calculating Potential Difference Across Terminals of a Battery in a Circuit****Objective:**To find the potential difference across the terminals of the battery in which chemical energy is being converted to electrical potential energy.**Given Data:**- \( \mathcal{E}_1 \) (Electromotive force of the first battery) = 10.9 V- \( R \) (Resistance in the circuit) = 17.3 Ω- \( \mathcal{E}_2 \) (Electromotive force of the second battery) = 4.6 V- \( r \) (Internal resistance of each battery) = 1.0 Ω**Circuit Explanation:**The circuit diagram consists of two batteries and three resistors. The batteries have electromotive forces \( \mathcal{E}_1 \) and \( \mathcal{E}_2 \), and internal resistances both have a value of 1.0 Ω. These are in a configuration where:- The resistor value \( R \) is connected in parallel with respect to the batteries.- Each battery has its internal resistance in series with it.**Detailed Circuit Analysis:**1. Identify battery configuration: - Battery \( \mathcal{E}_1 \) and its internal resistance \( 1.0 \, \Omega \) are connected in series as one loop. - Similarly, battery \( \mathcal{E}_2 \) and its internal resistance \( 1.0 \, \Omega \) are in series as another loop. - The two loops are connected in parallel to a resistor \( R \), which is 17.3 Ω.2. By Kirchhoff's Voltage Law (KVL): To find the overall potential difference across the terminals, we analyze either of the loops. We apply KVL to each circuit: - For Loop 1 (with \( \mathcal{E}_1 \)): \[ V_{\text{loop1}} = \mathcal{E}_1 - I_1 \cdot 1.0 \, \Omega \] - For Loop 2 (with \( \mathcal{E}_2 \)): \[ V_{\text{loop2}} = \mathcal{E}_2 - I_2 \cdot 1.0 \, \Omega \]Where \( I

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In the figure below, find the potential difference across the terminals of the battery in which chemical energy is being converted to electrical potential energy. (Take &₁ = 10.9 V, R = 17.3 , and E₂ = 4.6 V.) E₁ 1.02 1.02 E2 www R

In the figure below, find the potential difference across the terminals of the battery in which chemical energy is being converted to electrical potential energy. (Take &₁ = 10.9 V, R = 17.3 , and E₂ = 4.6 V.) E₁ 1.02 1.02 E2 www R

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter28: Direct-current Circuits

Section: Chapter Questions

Problem 28.2OQ: A battery has some internal resistance. (i) Clan the potential difference across the terminals of...

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Combination of Capacitors

The connection of capacitors can be established in a circuit in two ways.

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**Calculating Potential Difference Across Terminals of a Battery in a Circuit**

**Objective:**
To find the potential difference across the terminals of the battery in which chemical energy is being converted to electrical potential energy.

**Given Data:**
- \( \mathcal{E}_1 \) (Electromotive force of the first battery) = 10.9 V
- \( R \) (Resistance in the circuit) = 17.3 Ω
- \( \mathcal{E}_2 \) (Electromotive force of the second battery) = 4.6 V
- \( r \) (Internal resistance of each battery) = 1.0 Ω

**Circuit Explanation:**
The circuit diagram consists of two batteries and three resistors. The batteries have electromotive forces \( \mathcal{E}_1 \) and \( \mathcal{E}_2 \), and internal resistances both have a value of 1.0 Ω. These are in a configuration where:
- The resistor value \( R \) is connected in parallel with respect to the batteries.
- Each battery has its internal resistance in series with it.

**Detailed Circuit Analysis:**
1. Identify battery configuration:
- Battery \( \mathcal{E}_1 \) and its internal resistance \( 1.0 \, \Omega \) are connected in series as one loop.
- Similarly, battery \( \mathcal{E}_2 \) and its internal resistance \( 1.0 \, \Omega \) are in series as another loop.
- The two loops are connected in parallel to a resistor \( R \), which is 17.3 Ω.

2. By Kirchhoff's Voltage Law (KVL):
To find the overall potential difference across the terminals, we analyze either of the loops. We apply KVL to each circuit:
- For Loop 1 (with \( \mathcal{E}_1 \)):
\[
V_{\text{loop1}} = \mathcal{E}_1 - I_1 \cdot 1.0 \, \Omega
\]

- For Loop 2 (with \( \mathcal{E}_2 \)):
\[
V_{\text{loop2}} = \mathcal{E}_2 - I_2 \cdot 1.0 \, \Omega
\]

Where \( I

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