Why the terminal potential difference of a battery decreases when the current drawn from it is increased?
2 Answers
The terminal potential difference of a battery decreases when the current drawn from it is increased due to the internal resistance of the battery. Here’s a detailed explanation:
### Understanding Battery Operation
1. **Ideal Battery vs. Real Battery:**
- **Ideal Battery:** An ideal battery would maintain a constant voltage regardless of the current drawn from it. This is a theoretical concept and doesn't exist in practice.
- **Real Battery:** Real batteries have internal resistance and do not behave like ideal batteries.
2. **Internal Resistance:**
- **Internal Resistance (\(r\))** is the resistance within the battery that causes a voltage drop when current flows through it.
- This internal resistance arises from various factors, such as the electrolyte's resistance, the resistance of the electrodes, and the connections inside the battery.
### Voltage Drop Mechanism
1. **Terminal Voltage (\(V_{\text{term}}\)):**
- The terminal voltage of the battery is the voltage you measure across the battery's terminals.
- It is given by the formula:
\[
V_{\text{term}} = E - I \cdot r
\]
where \(E\) is the electromotive force (EMF) of the battery, \(I\) is the current drawn from the battery, and \(r\) is the internal resistance.
2. **Effect of Increased Current:**
- As the current \(I\) drawn from the battery increases, the term \(I \cdot r\) increases as well (since \(r\) is constant).
- This increased voltage drop across the internal resistance reduces the terminal voltage:
\[
V_{\text{term}} = E - I \cdot r
\]
### Example
Imagine a battery with an EMF of 12V and an internal resistance of 1Ω. If no current is drawn, the terminal voltage is 12V.
- **Current Drawn = 1A:**
\[
V_{\text{term}} = 12V - (1A \times 1Ω) = 11V
\]
- **Current Drawn = 3A:**
\[
V_{\text{term}} = 12V - (3A \times 1Ω) = 9V
\]
As you can see, with higher current, the voltage drop across the internal resistance becomes larger, leading to a lower terminal voltage.
### Summary
The decrease in terminal potential difference with increasing current is due to the internal resistance of the battery. As more current flows, the internal resistance causes a greater voltage drop within the battery, which lowers the voltage measured at the terminals.
### Understanding Battery Operation
1. **Ideal Battery vs. Real Battery:**
- **Ideal Battery:** An ideal battery would maintain a constant voltage regardless of the current drawn from it. This is a theoretical concept and doesn't exist in practice.
- **Real Battery:** Real batteries have internal resistance and do not behave like ideal batteries.
2. **Internal Resistance:**
- **Internal Resistance (\(r\))** is the resistance within the battery that causes a voltage drop when current flows through it.
- This internal resistance arises from various factors, such as the electrolyte's resistance, the resistance of the electrodes, and the connections inside the battery.
### Voltage Drop Mechanism
1. **Terminal Voltage (\(V_{\text{term}}\)):**
- The terminal voltage of the battery is the voltage you measure across the battery's terminals.
- It is given by the formula:
\[
V_{\text{term}} = E - I \cdot r
\]
where \(E\) is the electromotive force (EMF) of the battery, \(I\) is the current drawn from the battery, and \(r\) is the internal resistance.
2. **Effect of Increased Current:**
- As the current \(I\) drawn from the battery increases, the term \(I \cdot r\) increases as well (since \(r\) is constant).
- This increased voltage drop across the internal resistance reduces the terminal voltage:
\[
V_{\text{term}} = E - I \cdot r
\]
### Example
Imagine a battery with an EMF of 12V and an internal resistance of 1Ω. If no current is drawn, the terminal voltage is 12V.
- **Current Drawn = 1A:**
\[
V_{\text{term}} = 12V - (1A \times 1Ω) = 11V
\]
- **Current Drawn = 3A:**
\[
V_{\text{term}} = 12V - (3A \times 1Ω) = 9V
\]
As you can see, with higher current, the voltage drop across the internal resistance becomes larger, leading to a lower terminal voltage.
### Summary
The decrease in terminal potential difference with increasing current is due to the internal resistance of the battery. As more current flows, the internal resistance causes a greater voltage drop within the battery, which lowers the voltage measured at the terminals.
The terminal potential difference of a battery decreases as the current drawn from it increases due to the internal resistance of the battery. Let's break this down in detail.
### Battery Basics
A battery consists of two electrodes and an electrolyte, which together create a chemical reaction that generates electrical energy. This electrical energy is manifested as a voltage (electromotive force, EMF) across the terminals of the battery.
### Terminal Potential Difference and EMF
- **EMF (Electromotive Force):** This is the maximum potential difference a battery can provide when no current is flowing (open circuit condition). It is essentially the voltage generated by the chemical reactions inside the battery.
- **Terminal Potential Difference (V):** This is the voltage measured across the battery terminals when the battery is supplying current to a load.
### Internal Resistance
All batteries have internal resistance, denoted as \( r \). This resistance arises due to the resistance of the electrolyte and the internal connections within the battery.
### Relationship Between EMF, Terminal Voltage, and Internal Resistance
When a battery is connected to an external circuit, it provides a current \( I \). The relationship between the EMF (\( E \)), terminal potential difference (\( V \)), and internal resistance (\( r \)) is given by:
\[ V = E - Ir \]
Where:
- \( E \) is the EMF of the battery.
- \( I \) is the current flowing through the battery.
- \( r \) is the internal resistance of the battery.
### Why Terminal Potential Difference Decreases with Increased Current
1. **Internal Voltage Drop:** As current \( I \) flows through the battery, there is a voltage drop across the internal resistance \( r \). This drop is given by \( Ir \), which increases with increasing current. Hence, the voltage drop reduces the terminal potential difference \( V \) compared to the EMF.
2. **Increased Current Effect:** When the current drawn from the battery increases, the voltage drop across the internal resistance \( Ir \) becomes more significant. This means that the terminal potential difference \( V \) will be lower because:
\[ V = E - Ir \]
If \( I \) increases, \( Ir \) becomes larger, causing \( V \) to decrease.
### Example
Consider a battery with an EMF of 12 V and an internal resistance of 1 Ω. If no current is flowing, the terminal voltage is 12 V. If a load draws 2 A of current, the voltage drop due to internal resistance is:
\[ V_{\text{drop}} = I \times r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]
The terminal potential difference is:
\[ V = E - V_{\text{drop}} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V} \]
If the current increases to 4 A:
\[ V_{\text{drop}} = I \times r = 4 \, \text{A} \times 1 \, \Omega = 4 \, \text{V} \]
The terminal potential difference becomes:
\[ V = E - V_{\text{drop}} = 12 \, \text{V} - 4 \, \text{V} = 8 \, \text{V} \]
As you can see, the terminal voltage decreases as the current increases due to the greater voltage drop across the internal resistance.
### Summary
The terminal potential difference of a battery decreases with increased current due to the internal resistance of the battery. As more current flows through the battery, the voltage drop across the internal resistance becomes larger, reducing the potential difference that can be measured across the battery terminals. This internal resistance is a key factor in understanding why a battery's voltage might drop under load.
### Battery Basics
A battery consists of two electrodes and an electrolyte, which together create a chemical reaction that generates electrical energy. This electrical energy is manifested as a voltage (electromotive force, EMF) across the terminals of the battery.
### Terminal Potential Difference and EMF
- **EMF (Electromotive Force):** This is the maximum potential difference a battery can provide when no current is flowing (open circuit condition). It is essentially the voltage generated by the chemical reactions inside the battery.
- **Terminal Potential Difference (V):** This is the voltage measured across the battery terminals when the battery is supplying current to a load.
### Internal Resistance
All batteries have internal resistance, denoted as \( r \). This resistance arises due to the resistance of the electrolyte and the internal connections within the battery.
### Relationship Between EMF, Terminal Voltage, and Internal Resistance
When a battery is connected to an external circuit, it provides a current \( I \). The relationship between the EMF (\( E \)), terminal potential difference (\( V \)), and internal resistance (\( r \)) is given by:
\[ V = E - Ir \]
Where:
- \( E \) is the EMF of the battery.
- \( I \) is the current flowing through the battery.
- \( r \) is the internal resistance of the battery.
### Why Terminal Potential Difference Decreases with Increased Current
1. **Internal Voltage Drop:** As current \( I \) flows through the battery, there is a voltage drop across the internal resistance \( r \). This drop is given by \( Ir \), which increases with increasing current. Hence, the voltage drop reduces the terminal potential difference \( V \) compared to the EMF.
2. **Increased Current Effect:** When the current drawn from the battery increases, the voltage drop across the internal resistance \( Ir \) becomes more significant. This means that the terminal potential difference \( V \) will be lower because:
\[ V = E - Ir \]
If \( I \) increases, \( Ir \) becomes larger, causing \( V \) to decrease.
### Example
Consider a battery with an EMF of 12 V and an internal resistance of 1 Ω. If no current is flowing, the terminal voltage is 12 V. If a load draws 2 A of current, the voltage drop due to internal resistance is:
\[ V_{\text{drop}} = I \times r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]
The terminal potential difference is:
\[ V = E - V_{\text{drop}} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V} \]
If the current increases to 4 A:
\[ V_{\text{drop}} = I \times r = 4 \, \text{A} \times 1 \, \Omega = 4 \, \text{V} \]
The terminal potential difference becomes:
\[ V = E - V_{\text{drop}} = 12 \, \text{V} - 4 \, \text{V} = 8 \, \text{V} \]
As you can see, the terminal voltage decreases as the current increases due to the greater voltage drop across the internal resistance.
### Summary
The terminal potential difference of a battery decreases with increased current due to the internal resistance of the battery. As more current flows through the battery, the voltage drop across the internal resistance becomes larger, reducing the potential difference that can be measured across the battery terminals. This internal resistance is a key factor in understanding why a battery's voltage might drop under load.