Why terminal potential difference is less than the emf of the cell?
2 Answers
The terminal potential difference (V) is less than the electromotive force (emf, denoted as \( \mathcal{E} \)) of a cell due to the **internal resistance** of the cell. Here's a detailed explanation of why this happens:
### Key Concepts:
1. **EMF (\( \mathcal{E} \)):**
The emf of a cell is the maximum potential difference between the terminals of the cell when no current is flowing. It's the energy per unit charge provided by the cell to move charge through the entire circuit (both inside the cell and through the external circuit).
2. **Terminal Potential Difference (V):**
The terminal potential difference is the potential difference measured across the terminals of the cell when it is supplying current. This is the effective voltage that can be used by the external circuit.
3. **Internal Resistance (r):**
Every real cell or battery has some internal resistance due to the materials inside the cell. This resistance is denoted by \( r \) and it causes a voltage drop within the cell itself when current flows.
### Why Terminal Potential is Less than EMF:
When a current flows through the circuit, the cell not only has to supply energy to the external circuit but also overcomes its own internal resistance. The total emf of the cell is divided between the external circuit and the internal resistance of the cell.
The relationship can be expressed as:
\[
\mathcal{E} = V + Ir
\]
Where:
- \( \mathcal{E} \) is the emf of the cell.
- \( V \) is the terminal potential difference.
- \( I \) is the current flowing in the circuit.
- \( r \) is the internal resistance of the cell.
From this equation, you can see that:
\[
V = \mathcal{E} - Ir
\]
### Explanation:
- When the cell is supplying current, a part of the emf is "lost" in overcoming the internal resistance of the cell. This "loss" is \( Ir \), where \( I \) is the current and \( r \) is the internal resistance.
- The greater the current \( I \), the larger the voltage drop across the internal resistance \( Ir \), and the smaller the terminal potential difference.
- If no current is flowing (open circuit), \( I = 0 \), and the terminal potential difference equals the emf of the cell. In other words, under no-load conditions, \( V = \mathcal{E} \).
### Conclusion:
The terminal potential difference is less than the emf of the cell whenever current is drawn because part of the emf is used to overcome the internal resistance of the cell. The more current the cell supplies, the greater the internal resistance losses, reducing the terminal voltage further.
### Key Concepts:
1. **EMF (\( \mathcal{E} \)):**
The emf of a cell is the maximum potential difference between the terminals of the cell when no current is flowing. It's the energy per unit charge provided by the cell to move charge through the entire circuit (both inside the cell and through the external circuit).
2. **Terminal Potential Difference (V):**
The terminal potential difference is the potential difference measured across the terminals of the cell when it is supplying current. This is the effective voltage that can be used by the external circuit.
3. **Internal Resistance (r):**
Every real cell or battery has some internal resistance due to the materials inside the cell. This resistance is denoted by \( r \) and it causes a voltage drop within the cell itself when current flows.
### Why Terminal Potential is Less than EMF:
When a current flows through the circuit, the cell not only has to supply energy to the external circuit but also overcomes its own internal resistance. The total emf of the cell is divided between the external circuit and the internal resistance of the cell.
The relationship can be expressed as:
\[
\mathcal{E} = V + Ir
\]
Where:
- \( \mathcal{E} \) is the emf of the cell.
- \( V \) is the terminal potential difference.
- \( I \) is the current flowing in the circuit.
- \( r \) is the internal resistance of the cell.
From this equation, you can see that:
\[
V = \mathcal{E} - Ir
\]
### Explanation:
- When the cell is supplying current, a part of the emf is "lost" in overcoming the internal resistance of the cell. This "loss" is \( Ir \), where \( I \) is the current and \( r \) is the internal resistance.
- The greater the current \( I \), the larger the voltage drop across the internal resistance \( Ir \), and the smaller the terminal potential difference.
- If no current is flowing (open circuit), \( I = 0 \), and the terminal potential difference equals the emf of the cell. In other words, under no-load conditions, \( V = \mathcal{E} \).
### Conclusion:
The terminal potential difference is less than the emf of the cell whenever current is drawn because part of the emf is used to overcome the internal resistance of the cell. The more current the cell supplies, the greater the internal resistance losses, reducing the terminal voltage further.
The terminal potential difference (V_t) of a cell is typically less than the electromotive force (EMF) of the cell due to internal resistance. Here's a detailed explanation:
### 1. **Electromotive Force (EMF)**
The EMF of a cell, denoted as \( E \), is the maximum potential difference between the two terminals of the cell when no current is flowing. It's a measure of the cell's ability to drive an electric current through an external circuit.
### 2. **Internal Resistance**
Every real cell has some internal resistance, denoted as \( r \). This internal resistance arises from the cell's materials and construction, which oppose the flow of electric current.
### 3. **Terminal Potential Difference (V_t)**
The terminal potential difference, \( V_t \), is the actual voltage available at the terminals of the cell when it is connected to an external circuit and current is flowing.
### 4. **Relationship Between EMF and Terminal Potential Difference**
When a cell is delivering current \( I \) to an external load with resistance \( R \), the voltage drop across the internal resistance of the cell is \( I \times r \). This voltage drop reduces the potential difference available at the terminals. The relationship can be expressed as:
\[ V_t = E - I \times r \]
Here’s a step-by-step breakdown:
- **EMF (E):** This is the maximum voltage that the cell can provide.
- **Internal Resistance (r):** The internal resistance causes a voltage drop within the cell as current flows.
- **Current (I):** The current flowing through the cell creates a voltage drop across the internal resistance.
### Example
Suppose you have a cell with an EMF of 12V and an internal resistance of 1Ω. If the cell is connected to an external load, and the current flowing through the cell is 2A, then:
- The voltage drop across the internal resistance is \( I \times r = 2 \text{ A} \times 1 \text{ Ω} = 2 \text{ V} \).
- Therefore, the terminal potential difference \( V_t \) would be \( E - I \times r = 12 \text{ V} - 2 \text{ V} = 10 \text{ V} \).
### Summary
The terminal potential difference is always less than the EMF of the cell when current flows because some of the EMF is used to overcome the internal resistance of the cell. This results in a voltage drop within the cell itself, reducing the voltage available at the terminals.
### 1. **Electromotive Force (EMF)**
The EMF of a cell, denoted as \( E \), is the maximum potential difference between the two terminals of the cell when no current is flowing. It's a measure of the cell's ability to drive an electric current through an external circuit.
### 2. **Internal Resistance**
Every real cell has some internal resistance, denoted as \( r \). This internal resistance arises from the cell's materials and construction, which oppose the flow of electric current.
### 3. **Terminal Potential Difference (V_t)**
The terminal potential difference, \( V_t \), is the actual voltage available at the terminals of the cell when it is connected to an external circuit and current is flowing.
### 4. **Relationship Between EMF and Terminal Potential Difference**
When a cell is delivering current \( I \) to an external load with resistance \( R \), the voltage drop across the internal resistance of the cell is \( I \times r \). This voltage drop reduces the potential difference available at the terminals. The relationship can be expressed as:
\[ V_t = E - I \times r \]
Here’s a step-by-step breakdown:
- **EMF (E):** This is the maximum voltage that the cell can provide.
- **Internal Resistance (r):** The internal resistance causes a voltage drop within the cell as current flows.
- **Current (I):** The current flowing through the cell creates a voltage drop across the internal resistance.
### Example
Suppose you have a cell with an EMF of 12V and an internal resistance of 1Ω. If the cell is connected to an external load, and the current flowing through the cell is 2A, then:
- The voltage drop across the internal resistance is \( I \times r = 2 \text{ A} \times 1 \text{ Ω} = 2 \text{ V} \).
- Therefore, the terminal potential difference \( V_t \) would be \( E - I \times r = 12 \text{ V} - 2 \text{ V} = 10 \text{ V} \).
### Summary
The terminal potential difference is always less than the EMF of the cell when current flows because some of the EMF is used to overcome the internal resistance of the cell. This results in a voltage drop within the cell itself, reducing the voltage available at the terminals.