Why the emf of a cell is greater than its terminal voltage?
2 Answers
The relationship between the electromotive force (emf) of a cell and its terminal voltage can be understood through the concepts of internal resistance and the load conditions of the circuit in which the cell operates. Here’s a detailed explanation:
### 1. Definitions
- **Electromotive Force (emf)**: This is the maximum potential difference provided by a cell or battery when no current is flowing. It represents the energy supplied per unit charge and is often denoted by the symbol \( \mathcal{E} \).
- **Terminal Voltage (\( V_T \))**: This is the voltage measured across the terminals of the cell when it is supplying current to a load. It represents the actual voltage available for external use.
### 2. Internal Resistance
Every real cell has some internal resistance (\( r \)). This resistance is due to the materials and electrochemical processes inside the cell. When the cell is delivering current (\( I \)), a voltage drop occurs across this internal resistance, which can be described by Ohm's Law:
\[
V_{\text{drop}} = I \cdot r
\]
### 3. Relationship Between emf and Terminal Voltage
When a load is connected to the cell, the terminal voltage can be expressed in terms of the emf and the internal resistance. The relationship can be described as follows:
\[
V_T = \mathcal{E} - I \cdot r
\]
Where:
- \( V_T \) is the terminal voltage,
- \( \mathcal{E} \) is the emf of the cell,
- \( I \) is the current flowing through the circuit,
- \( r \) is the internal resistance of the cell.
### 4. Explanation of the Difference
From the equation above, it becomes clear why the emf is greater than the terminal voltage when the cell is under load. The key points are:
- **Current Flow**: When current flows from the cell, the internal resistance causes a voltage drop, reducing the voltage available at the terminals.
- **No Load Condition**: If the cell is not supplying current (open circuit), then \( I = 0 \) and the terminal voltage equals the emf:
\[
V_T = \mathcal{E}
\]
- **Loaded Condition**: When the cell is connected to a load and supplying current, the terminal voltage will be less than the emf due to the drop caused by internal resistance.
### 5. Example Calculation
Suppose we have a cell with an emf of \( 12 \, \text{V} \) and an internal resistance of \( 1 \, \Omega \). If the cell is supplying a current of \( 2 \, \text{A} \):
1. Calculate the voltage drop due to internal resistance:
\[
V_{\text{drop}} = I \cdot r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V}
\]
2. Now, calculate the terminal voltage:
\[
V_T = \mathcal{E} - V_{\text{drop}} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V}
\]
In this case, the terminal voltage is \( 10 \, \text{V} \), which is less than the emf of \( 12 \, \text{V} \).
### Conclusion
In summary, the emf of a cell is greater than its terminal voltage when the cell is under load because of the voltage drop that occurs across its internal resistance. This concept is crucial in understanding how batteries and cells function in practical applications, influencing the efficiency and performance of electrical circuits.
### 1. Definitions
- **Electromotive Force (emf)**: This is the maximum potential difference provided by a cell or battery when no current is flowing. It represents the energy supplied per unit charge and is often denoted by the symbol \( \mathcal{E} \).
- **Terminal Voltage (\( V_T \))**: This is the voltage measured across the terminals of the cell when it is supplying current to a load. It represents the actual voltage available for external use.
### 2. Internal Resistance
Every real cell has some internal resistance (\( r \)). This resistance is due to the materials and electrochemical processes inside the cell. When the cell is delivering current (\( I \)), a voltage drop occurs across this internal resistance, which can be described by Ohm's Law:
\[
V_{\text{drop}} = I \cdot r
\]
### 3. Relationship Between emf and Terminal Voltage
When a load is connected to the cell, the terminal voltage can be expressed in terms of the emf and the internal resistance. The relationship can be described as follows:
\[
V_T = \mathcal{E} - I \cdot r
\]
Where:
- \( V_T \) is the terminal voltage,
- \( \mathcal{E} \) is the emf of the cell,
- \( I \) is the current flowing through the circuit,
- \( r \) is the internal resistance of the cell.
### 4. Explanation of the Difference
From the equation above, it becomes clear why the emf is greater than the terminal voltage when the cell is under load. The key points are:
- **Current Flow**: When current flows from the cell, the internal resistance causes a voltage drop, reducing the voltage available at the terminals.
- **No Load Condition**: If the cell is not supplying current (open circuit), then \( I = 0 \) and the terminal voltage equals the emf:
\[
V_T = \mathcal{E}
\]
- **Loaded Condition**: When the cell is connected to a load and supplying current, the terminal voltage will be less than the emf due to the drop caused by internal resistance.
### 5. Example Calculation
Suppose we have a cell with an emf of \( 12 \, \text{V} \) and an internal resistance of \( 1 \, \Omega \). If the cell is supplying a current of \( 2 \, \text{A} \):
1. Calculate the voltage drop due to internal resistance:
\[
V_{\text{drop}} = I \cdot r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V}
\]
2. Now, calculate the terminal voltage:
\[
V_T = \mathcal{E} - V_{\text{drop}} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V}
\]
In this case, the terminal voltage is \( 10 \, \text{V} \), which is less than the emf of \( 12 \, \text{V} \).
### Conclusion
In summary, the emf of a cell is greater than its terminal voltage when the cell is under load because of the voltage drop that occurs across its internal resistance. This concept is crucial in understanding how batteries and cells function in practical applications, influencing the efficiency and performance of electrical circuits.
The electromotive force (EMF) of a cell is a measure of the maximum potential difference the cell can produce between its terminals when no current is flowing. The terminal voltage, on the other hand, is the actual voltage across the cell's terminals when it is supplying current to an external circuit.
The difference between the EMF and the terminal voltage can be explained by considering the internal resistance of the cell. Here’s a detailed explanation:
### 1. **Internal Resistance**
Every real cell has some internal resistance, denoted as \( r \). This resistance comes from the materials within the cell, such as the electrolyte and the electrodes. When a current \( I \) flows through the cell, this internal resistance causes a voltage drop inside the cell.
### 2. **Voltage Drop Due to Internal Resistance**
The terminal voltage \( V_{\text{terminal}} \) of the cell can be calculated using the following relationship:
\[ V_{\text{terminal}} = \text{EMF} - I \cdot r \]
Here’s a breakdown of this equation:
- **EMF** (\( \mathcal{E} \)): This is the ideal voltage the cell can provide when no current is flowing.
- **Current** (\( I \)): The amount of current flowing through the circuit.
- **Internal Resistance** (\( r \)): The resistance inside the cell that opposes the flow of current.
When current flows through the cell, it encounters the internal resistance, which causes a voltage drop (\( I \cdot r \)) inside the cell. As a result, the voltage available at the terminals (i.e., the terminal voltage) is less than the EMF by the amount of this internal voltage drop.
### 3. **Example**
Consider a simple example with a battery:
- **EMF of the battery**: 12 V
- **Internal resistance**: 1 Ω
- **Current flowing**: 2 A
The voltage drop due to internal resistance is:
\[ \text{Voltage Drop} = I \cdot r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]
Thus, the terminal voltage is:
\[ V_{\text{terminal}} = \text{EMF} - \text{Voltage Drop} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V} \]
### 4. **Practical Implications**
In practical applications, the internal resistance of a cell can impact performance. For high-current applications, the internal resistance can cause significant voltage drops, reducing the efficiency of the power delivered by the cell. For this reason, it's important to account for internal resistance when designing and using batteries and other cells.
In summary, the EMF of a cell is greater than its terminal voltage because the internal resistance of the cell causes a voltage drop when current flows. This drop reduces the actual voltage available at the cell's terminals compared to the theoretical maximum EMF.
The difference between the EMF and the terminal voltage can be explained by considering the internal resistance of the cell. Here’s a detailed explanation:
### 1. **Internal Resistance**
Every real cell has some internal resistance, denoted as \( r \). This resistance comes from the materials within the cell, such as the electrolyte and the electrodes. When a current \( I \) flows through the cell, this internal resistance causes a voltage drop inside the cell.
### 2. **Voltage Drop Due to Internal Resistance**
The terminal voltage \( V_{\text{terminal}} \) of the cell can be calculated using the following relationship:
\[ V_{\text{terminal}} = \text{EMF} - I \cdot r \]
Here’s a breakdown of this equation:
- **EMF** (\( \mathcal{E} \)): This is the ideal voltage the cell can provide when no current is flowing.
- **Current** (\( I \)): The amount of current flowing through the circuit.
- **Internal Resistance** (\( r \)): The resistance inside the cell that opposes the flow of current.
When current flows through the cell, it encounters the internal resistance, which causes a voltage drop (\( I \cdot r \)) inside the cell. As a result, the voltage available at the terminals (i.e., the terminal voltage) is less than the EMF by the amount of this internal voltage drop.
### 3. **Example**
Consider a simple example with a battery:
- **EMF of the battery**: 12 V
- **Internal resistance**: 1 Ω
- **Current flowing**: 2 A
The voltage drop due to internal resistance is:
\[ \text{Voltage Drop} = I \cdot r = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]
Thus, the terminal voltage is:
\[ V_{\text{terminal}} = \text{EMF} - \text{Voltage Drop} = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V} \]
### 4. **Practical Implications**
In practical applications, the internal resistance of a cell can impact performance. For high-current applications, the internal resistance can cause significant voltage drops, reducing the efficiency of the power delivered by the cell. For this reason, it's important to account for internal resistance when designing and using batteries and other cells.
In summary, the EMF of a cell is greater than its terminal voltage because the internal resistance of the cell causes a voltage drop when current flows. This drop reduces the actual voltage available at the cell's terminals compared to the theoretical maximum EMF.