Why is the terminal voltage always less than the emf?
2 Answers
To understand why the terminal voltage of a battery is always less than its electromotive force (emf), we need to delve into the concepts of voltage, emf, and internal resistance.
### Key Definitions
1. **Electromotive Force (emf)**: This is the maximum potential difference between the terminals of a battery when no current is flowing. It represents the energy per unit charge that the battery can provide, generated by the chemical reactions inside the battery.
2. **Terminal Voltage**: This is the voltage measured across the terminals of a battery when it is connected to a circuit and current is flowing. It is the actual voltage available to power a load.
3. **Internal Resistance**: All real batteries have some resistance to the flow of electric current, which is referred to as internal resistance (denoted as \( r \)). This resistance arises from the materials used in the battery and the chemical processes occurring within it.
### The Relationship Between emf and Terminal Voltage
When a battery is supplying current to a circuit, the terminal voltage (\( V \)) can be described by the equation:
\[
V = \text{emf} - I \cdot r
\]
Where:
- \( I \) is the current flowing through the battery.
- \( r \) is the internal resistance of the battery.
### Why Terminal Voltage is Less Than Emf
1. **Current Flow**: When the battery is connected to a load, it drives a current (\( I \)) through the external circuit as well as through its own internal resistance (\( r \)). The flow of current results in a voltage drop across the internal resistance, which can be calculated as \( I \cdot r \).
2. **Voltage Drop**: This voltage drop due to internal resistance effectively reduces the voltage available at the terminals of the battery. Thus, the terminal voltage is given by subtracting the voltage drop from the emf:
\[
V = \text{emf} - I \cdot r
\]
Since \( I \cdot r \) is a positive value (as long as \( I \) is not zero), the terminal voltage \( V \) is always less than the emf.
### Example for Clarity
Imagine a battery with an emf of 12 volts and an internal resistance of 1 ohm. If the battery supplies a current of 2 amperes to a circuit:
1. Calculate the voltage drop due to internal resistance:
\[
I \cdot r = 2 \, \text{A} \cdot 1 \, \Omega = 2 \, \text{V}
\]
2. Now, calculate the terminal voltage:
\[
V = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V}
\]
In this example, while the battery can theoretically provide 12 volts, the actual voltage available to the circuit when the battery is under load is only 10 volts.
### Conclusion
In summary, the terminal voltage of a battery is always less than its emf because of the voltage drop caused by internal resistance when the battery is delivering current. This concept is crucial in understanding the performance of batteries in real-world applications, where efficiency and voltage drops can significantly impact the operation of electronic devices and systems.
### Key Definitions
1. **Electromotive Force (emf)**: This is the maximum potential difference between the terminals of a battery when no current is flowing. It represents the energy per unit charge that the battery can provide, generated by the chemical reactions inside the battery.
2. **Terminal Voltage**: This is the voltage measured across the terminals of a battery when it is connected to a circuit and current is flowing. It is the actual voltage available to power a load.
3. **Internal Resistance**: All real batteries have some resistance to the flow of electric current, which is referred to as internal resistance (denoted as \( r \)). This resistance arises from the materials used in the battery and the chemical processes occurring within it.
### The Relationship Between emf and Terminal Voltage
When a battery is supplying current to a circuit, the terminal voltage (\( V \)) can be described by the equation:
\[
V = \text{emf} - I \cdot r
\]
Where:
- \( I \) is the current flowing through the battery.
- \( r \) is the internal resistance of the battery.
### Why Terminal Voltage is Less Than Emf
1. **Current Flow**: When the battery is connected to a load, it drives a current (\( I \)) through the external circuit as well as through its own internal resistance (\( r \)). The flow of current results in a voltage drop across the internal resistance, which can be calculated as \( I \cdot r \).
2. **Voltage Drop**: This voltage drop due to internal resistance effectively reduces the voltage available at the terminals of the battery. Thus, the terminal voltage is given by subtracting the voltage drop from the emf:
\[
V = \text{emf} - I \cdot r
\]
Since \( I \cdot r \) is a positive value (as long as \( I \) is not zero), the terminal voltage \( V \) is always less than the emf.
### Example for Clarity
Imagine a battery with an emf of 12 volts and an internal resistance of 1 ohm. If the battery supplies a current of 2 amperes to a circuit:
1. Calculate the voltage drop due to internal resistance:
\[
I \cdot r = 2 \, \text{A} \cdot 1 \, \Omega = 2 \, \text{V}
\]
2. Now, calculate the terminal voltage:
\[
V = 12 \, \text{V} - 2 \, \text{V} = 10 \, \text{V}
\]
In this example, while the battery can theoretically provide 12 volts, the actual voltage available to the circuit when the battery is under load is only 10 volts.
### Conclusion
In summary, the terminal voltage of a battery is always less than its emf because of the voltage drop caused by internal resistance when the battery is delivering current. This concept is crucial in understanding the performance of batteries in real-world applications, where efficiency and voltage drops can significantly impact the operation of electronic devices and systems.
The terminal voltage of a battery or power supply is always less than its electromotive force (EMF) due to the internal resistance of the battery. Let’s break down why this happens:
### **1. Understanding EMF and Terminal Voltage**
- **Electromotive Force (EMF):** This is the maximum potential difference between the terminals of a battery when no current is flowing through the circuit. It's essentially the voltage that the battery can provide if it were ideal, meaning there’s no internal resistance.
- **Terminal Voltage:** This is the actual voltage you measure across the terminals of the battery when it is delivering current to an external circuit.
### **2. The Role of Internal Resistance**
All real batteries have some degree of internal resistance, denoted as \( r \). This internal resistance causes a drop in the voltage when current flows through the battery. The relationship between EMF (\( \mathcal{E} \)), terminal voltage (\( V \)), and internal resistance (\( r \)) can be expressed by the following equation:
\[ V = \mathcal{E} - I \cdot r \]
where:
- \( V \) is the terminal voltage.
- \( I \) is the current flowing through the circuit.
- \( r \) is the internal resistance of the battery.
### **3. Why Terminal Voltage is Less Than EMF**
When the battery is supplying current to an external circuit, a voltage drop occurs due to its internal resistance. This is analogous to having a resistor inside the battery that consumes part of the voltage generated by the battery's EMF. The more current the battery supplies, the greater the voltage drop across this internal resistance.
Here’s a step-by-step breakdown:
1. **No Current Condition:** When no current flows (open circuit), there is no voltage drop across the internal resistance. The terminal voltage equals the EMF.
2. **Current Flow Condition:** When current flows, the internal resistance causes a voltage drop according to Ohm’s Law (\( V = I \cdot r \)). This drop reduces the terminal voltage.
For example, if a battery has an EMF of 12 volts and an internal resistance of 1 ohm, and it is supplying a current of 2 amps, the voltage drop across the internal resistance is:
\[ \text{Voltage Drop} = I \cdot r = 2 \, \text{amps} \times 1 \, \text{ohm} = 2 \, \text{volts} \]
Thus, the terminal voltage would be:
\[ V = \mathcal{E} - \text{Voltage Drop} = 12 \, \text{volts} - 2 \, \text{volts} = 10 \, \text{volts} \]
3. **Higher Current, Greater Drop:** As the current increases, the voltage drop increases, further decreasing the terminal voltage.
### **4. Summary**
The terminal voltage is always less than the EMF because of the voltage drop across the internal resistance of the battery. When a battery supplies current, this internal resistance causes a reduction in the voltage that can be measured at the terminals. The greater the current drawn from the battery, the larger the voltage drop, and thus the more pronounced the difference between EMF and terminal voltage.
### **1. Understanding EMF and Terminal Voltage**
- **Electromotive Force (EMF):** This is the maximum potential difference between the terminals of a battery when no current is flowing through the circuit. It's essentially the voltage that the battery can provide if it were ideal, meaning there’s no internal resistance.
- **Terminal Voltage:** This is the actual voltage you measure across the terminals of the battery when it is delivering current to an external circuit.
### **2. The Role of Internal Resistance**
All real batteries have some degree of internal resistance, denoted as \( r \). This internal resistance causes a drop in the voltage when current flows through the battery. The relationship between EMF (\( \mathcal{E} \)), terminal voltage (\( V \)), and internal resistance (\( r \)) can be expressed by the following equation:
\[ V = \mathcal{E} - I \cdot r \]
where:
- \( V \) is the terminal voltage.
- \( I \) is the current flowing through the circuit.
- \( r \) is the internal resistance of the battery.
### **3. Why Terminal Voltage is Less Than EMF**
When the battery is supplying current to an external circuit, a voltage drop occurs due to its internal resistance. This is analogous to having a resistor inside the battery that consumes part of the voltage generated by the battery's EMF. The more current the battery supplies, the greater the voltage drop across this internal resistance.
Here’s a step-by-step breakdown:
1. **No Current Condition:** When no current flows (open circuit), there is no voltage drop across the internal resistance. The terminal voltage equals the EMF.
2. **Current Flow Condition:** When current flows, the internal resistance causes a voltage drop according to Ohm’s Law (\( V = I \cdot r \)). This drop reduces the terminal voltage.
For example, if a battery has an EMF of 12 volts and an internal resistance of 1 ohm, and it is supplying a current of 2 amps, the voltage drop across the internal resistance is:
\[ \text{Voltage Drop} = I \cdot r = 2 \, \text{amps} \times 1 \, \text{ohm} = 2 \, \text{volts} \]
Thus, the terminal voltage would be:
\[ V = \mathcal{E} - \text{Voltage Drop} = 12 \, \text{volts} - 2 \, \text{volts} = 10 \, \text{volts} \]
3. **Higher Current, Greater Drop:** As the current increases, the voltage drop increases, further decreasing the terminal voltage.
### **4. Summary**
The terminal voltage is always less than the EMF because of the voltage drop across the internal resistance of the battery. When a battery supplies current, this internal resistance causes a reduction in the voltage that can be measured at the terminals. The greater the current drawn from the battery, the larger the voltage drop, and thus the more pronounced the difference between EMF and terminal voltage.