Internal Resistance of a Battery Formula: EMF, Terminal Voltage, and Current

📚 Understanding Internal Resistance

Every real battery has some internal resistance, which affects the voltage it can actually deliver to a circuit. This resistance is inside the battery itself. The terminal voltage, which is what we measure across the battery's terminals, is less than the electromotive force (EMF) due to the voltage drop across this internal resistance.

🧪 The Formula Unveiled

The relationship between EMF ($\mathcal{E}$), terminal voltage ($V$), current ($I$), and internal resistance ($r$) is given by the following formula:

$V = \mathcal{E} - Ir$

Where:

• ⚡ $\mathcal{E}$ (EMF): The electromotive force is the total voltage produced by the battery when no current is flowing.
• 🔌 $V$ (Terminal Voltage): The voltage available at the battery terminals when a current is flowing.
• 🌊 $I$ (Current): The current flowing through the circuit connected to the battery.
• 🔥 $r$ (Internal Resistance): The resistance within the battery itself.

📜 History and Background

The concept of internal resistance arose from the observation that real batteries don't behave as ideal voltage sources. Early electrical experiments revealed that the voltage provided by a battery decreased as the current drawn from it increased. This led to the postulation of an internal resistance that accounts for these losses within the battery.

💡 Key Principles

• ⚖️ Ohm's Law Connection: The voltage drop across the internal resistance is governed by Ohm's Law ($V = IR$), applied specifically to the internal resistance of the battery.
• ⚡ Energy Dissipation: Internal resistance causes energy dissipation within the battery, usually in the form of heat. This reduces the overall efficiency of the battery.
• 📉 Voltage Drop: As the current increases, the voltage drop ($Ir$) across the internal resistance also increases, leading to a lower terminal voltage.

🌍 Real-World Examples

Example 1: Car Battery

A car battery has an EMF of 12.6 V. When starting the car, a large current of 100 A is drawn, and the terminal voltage drops to 11.6 V. We can calculate the internal resistance:

$11.6 = 12.6 - 100r$

$100r = 1$

$r = 0.01 \Omega$

Example 2: Flashlight Battery

A flashlight battery has an EMF of 1.5 V. When connected to a bulb drawing 0.5 A, the terminal voltage is 1.4 V. The internal resistance is:

$1.4 = 1.5 - 0.5r$

$0.5r = 0.1$

$r = 0.2 \Omega$

📊 Table Summarizing Key Variables

📝 Practice Quiz

• ❓ Question 1: A battery with an EMF of 9V has an internal resistance of 0.5$\Omega$. If a 2$\Omega$ resistor is connected to the battery, what is the current flowing through the circuit?
• 💡 Question 2: A battery's terminal voltage is measured to be 11V when delivering a current of 5A. If its internal resistance is 0.1$\Omega$, what is the EMF of the battery?
• 🤔 Question 3: Explain how internal resistance affects the performance of a battery when used to power a high-current device.
• 🔢 Question 4: A 1.5V battery has an internal resistance of 0.3$\Omega$. What is the terminal voltage when it supplies a current of 0.4A?
• ⚡ Question 5: If the terminal voltage of a battery drops significantly when a load is connected, what does this indicate about its internal resistance?
• 🔋 Question 6: A battery with an EMF of 6V and an internal resistance of 0.2$\Omega$ is connected to a resistor. If the current in the circuit is 2A, what is the value of the external resistance?
• 📈 Question 7: How does the internal resistance of a battery change over time as it discharges, and what effect does this have on its performance?

🔑 Conclusion

Understanding internal resistance is crucial for analyzing and predicting the behavior of real batteries in circuits. It explains why the terminal voltage is always less than the EMF and how the current affects the battery's performance. By considering internal resistance, we can design more efficient and reliable electrical systems.