Why terminal potential of a cell (or a battery) is always less than its emf ?
2 Answers
The terminal potential difference (or voltage) of a cell or battery is always less than its electromotive force (emf) due to the internal resistance of the cell or battery. Here’s a detailed explanation of why this happens:
### 1. **Understanding EMF and Terminal Potential**
- **Electromotive Force (EMF):** The EMF of a cell is the maximum potential difference between its terminals when no current is flowing. It represents the energy per unit charge provided by the cell’s chemical reaction to move charges through the circuit.
- **Terminal Potential Difference:** This is the actual potential difference between the terminals of the cell when a current is flowing through the circuit. It is measured across the terminals of the cell during operation.
### 2. **Internal Resistance and Its Effect**
A real cell or battery has an internal resistance (\( r \)) which affects the terminal potential difference. When a current (\( I \)) flows through the cell, there is a voltage drop across this internal resistance due to Ohm’s law:
\[ V_{\text{drop}} = I \times r \]
This internal resistance results in a loss of voltage inside the cell, which affects the terminal potential difference.
### 3. **Relationship Between EMF, Terminal Potential, and Internal Resistance**
When a current flows through the cell, the terminal potential difference (\( V_{\text{terminal}} \)) is given by:
\[ V_{\text{terminal}} = \text{EMF} - (I \times r) \]
Here’s why:
- **Internal Voltage Drop:** As current flows, the internal resistance causes a voltage drop inside the cell. This reduces the potential difference available at the terminals.
- **Practical Example:** Suppose a battery has an EMF of 12V and an internal resistance of 1 ohm. If the battery is supplying a current of 2A, the voltage drop across the internal resistance is \( 2 \text{A} \times 1 \text{Ω} = 2 \text{V} \). Therefore, the terminal potential difference is:
\[ V_{\text{terminal}} = 12 \text{V} - 2 \text{V} = 10 \text{V} \]
### 4. **Conclusion**
The terminal potential of a cell is always less than its EMF because some of the EMF is used up overcoming the internal resistance of the cell. The greater the internal resistance or the higher the current, the greater the voltage drop inside the cell and the lower the terminal potential difference.
### 1. **Understanding EMF and Terminal Potential**
- **Electromotive Force (EMF):** The EMF of a cell is the maximum potential difference between its terminals when no current is flowing. It represents the energy per unit charge provided by the cell’s chemical reaction to move charges through the circuit.
- **Terminal Potential Difference:** This is the actual potential difference between the terminals of the cell when a current is flowing through the circuit. It is measured across the terminals of the cell during operation.
### 2. **Internal Resistance and Its Effect**
A real cell or battery has an internal resistance (\( r \)) which affects the terminal potential difference. When a current (\( I \)) flows through the cell, there is a voltage drop across this internal resistance due to Ohm’s law:
\[ V_{\text{drop}} = I \times r \]
This internal resistance results in a loss of voltage inside the cell, which affects the terminal potential difference.
### 3. **Relationship Between EMF, Terminal Potential, and Internal Resistance**
When a current flows through the cell, the terminal potential difference (\( V_{\text{terminal}} \)) is given by:
\[ V_{\text{terminal}} = \text{EMF} - (I \times r) \]
Here’s why:
- **Internal Voltage Drop:** As current flows, the internal resistance causes a voltage drop inside the cell. This reduces the potential difference available at the terminals.
- **Practical Example:** Suppose a battery has an EMF of 12V and an internal resistance of 1 ohm. If the battery is supplying a current of 2A, the voltage drop across the internal resistance is \( 2 \text{A} \times 1 \text{Ω} = 2 \text{V} \). Therefore, the terminal potential difference is:
\[ V_{\text{terminal}} = 12 \text{V} - 2 \text{V} = 10 \text{V} \]
### 4. **Conclusion**
The terminal potential of a cell is always less than its EMF because some of the EMF is used up overcoming the internal resistance of the cell. The greater the internal resistance or the higher the current, the greater the voltage drop inside the cell and the lower the terminal potential difference.
The terminal potential of a cell (or battery) is always less than its electromotive force (emf) due to the internal resistance of the cell. Here's a more detailed explanation:
1. **Electromotive Force (emf)**: This is the maximum potential difference a cell can provide when no current is flowing. It represents the total energy per unit charge available from the chemical reactions inside the cell.
2. **Internal Resistance**: Cells and batteries have some internal resistance, which arises due to the materials and construction of the cell. When a current flows through the cell, there is a voltage drop across this internal resistance.
3. **Terminal Potential**: When a current flows, the terminal potential (the voltage you measure across the cell terminals) is reduced by the voltage drop caused by the internal resistance. This is given by the formula:
\[
V = \text{emf} - I \cdot r
\]
where \( V \) is the terminal potential, \( I \) is the current, and \( r \) is the internal resistance of the cell.
So, the terminal potential is less than the emf by the amount of voltage drop due to the internal resistance when current is flowing. If there were no internal resistance (an ideal cell), the terminal potential would be equal to the emf.
1. **Electromotive Force (emf)**: This is the maximum potential difference a cell can provide when no current is flowing. It represents the total energy per unit charge available from the chemical reactions inside the cell.
2. **Internal Resistance**: Cells and batteries have some internal resistance, which arises due to the materials and construction of the cell. When a current flows through the cell, there is a voltage drop across this internal resistance.
3. **Terminal Potential**: When a current flows, the terminal potential (the voltage you measure across the cell terminals) is reduced by the voltage drop caused by the internal resistance. This is given by the formula:
\[
V = \text{emf} - I \cdot r
\]
where \( V \) is the terminal potential, \( I \) is the current, and \( r \) is the internal resistance of the cell.
So, the terminal potential is less than the emf by the amount of voltage drop due to the internal resistance when current is flowing. If there were no internal resistance (an ideal cell), the terminal potential would be equal to the emf.