AP Physics 2: EMF and Internal Resistance

To truly master circuit analysis, you must move beyond the ideal components of introductory physics. Real-world batteries are not perfect voltage sources; they have inherent limitations that cause measurable effects in a circuit. Understanding electromotive force (EMF) and internal resistance is crucial for explaining why a battery heats up during use, why a car headlight dims when the starter motor engages, and for designing reliable electronic systems. This concept bridges the gap between simplified theory and practical, observable circuit behavior.

From Ideal to Real: Modeling a Physical Battery

In an ideal model, a battery is a perfect source of constant voltage, maintaining its potential difference regardless of how much current it supplies. However, a physical battery is a chemical device. As chemical reactions generate electrical energy, they also encounter opposition within the battery's materials—its electrodes and electrolyte. This opposition is modeled as internal resistance ($r$), a small resistor placed in series inside the battery itself.

Therefore, a real battery is best represented as a two-component combination: an ideal EMF source ($\epsilon$), which provides the battery's maximum possible voltage when no current flows, and its internal resistance ($r$). The EMF ($\epsilon$) is the energy per unit charge converted from chemical to electrical form, measured in volts. It is the voltage you would measure across the battery's terminals with an ideal voltmeter (one that draws negligible current).

The Fundamental Equation: $\epsilon =V_{\text{terminal}}+Ir$

This equation is the cornerstone for analyzing circuits with non-ideal batteries. It states that the battery's EMF ($\epsilon$) is equal to the sum of the terminal voltage ($V_{\text{terminal}}$)—the voltage actually available to the external circuit—and the voltage "lost" across the internal resistance ($Ir$).

Let's break down each term:

• $\epsilon$: The constant EMF of the battery (e.g., 9 V for a PP3 battery).
• $I$: The current flowing through the entire circuit (and through the battery itself).
• $r$: The constant internal resistance (typically in ohms, $\Omega$).
• $V_{\text{terminal}}$ = $IR$: The voltage across the external load resistance ($R$), given by Ohm's Law. This is what powers your device.

You can visualize it using a simple circuit: an ideal EMF source ($\epsilon$) in series with an internal resistor ($r$) and an external load resistor ($R$). The current $I$ is the same through all three. The voltage drop across $r$ is $Ir$, and the voltage drop across $R$ is $IR=V_{\text{terminal}}$. Kirchhoff's loop rule gives: $\epsilon -Ir-IR=0$, which rearranges to $\epsilon =IR+Ir$.

Example Calculation: A 12.0-V car battery has an internal resistance of 0.040 $\Omega$. What is the terminal voltage when the starter motor, acting as a 0.150 $\Omega$ load, is engaged?

1. Find the total resistance in the single loop: $R_{\text{total}}=r+R=0.040\Omega +0.150\Omega =0.190\Omega$.
2. Use Ohm's Law with the EMF to find the current: $I=\epsilon /R_{\text{total}}=12.0\text{V}/0.190\Omega \approx 63.2\text{A}$.
3. The voltage drop across the internal resistance is $Ir=(63.2\text{A})(0.040\Omega )\approx 2.53\text{V}$.
4. The terminal voltage is $V_{\text{terminal}}=\epsilon -Ir=12.0\text{V}-2.53\text{V}=9.47\text{V}$.

Notice the massive current causes a significant voltage drop internally, leaving less than 10 V for the starter motor. This explains the dimming of headlights.

Terminal Voltage Under Load: Why It Drops

The terminal voltage is not a fixed property of the battery; it depends on the current draw. From the equation $V_{\text{terminal}}=\epsilon -Ir$, you can see the relationship:

• Open Circuit ($I=0$): $V_{\text{terminal}}=\epsilon$. A voltmeter measures the EMF directly.
• Closed Circuit ($I>0$): $V_{\text{terminal}}<\epsilon$. The voltage available to the external circuit is less than the EMF.
• Increased Current (Higher Load): As the external load $R$ decreases, current $I$ increases. This causes a larger $Ir$ drop, making the terminal voltage decrease further.

This is the precise answer to "why voltage drops when current increases." The internal resistance acts like a narrow section in a water pipe; the greater the flow (current), the larger the pressure drop (voltage loss) across that narrow section, leaving less pressure (voltage) for the external parts of the system.

Power and Efficiency in Non-Ideal Batteries

The concept of internal resistance critically affects power distribution. The total power output by the chemical reactions in the battery is $P_{\text{total}}=I\epsilon$. This power is divided into two parts:

1. Useful Power delivered to the external load: $P_{\text{load}}=I{V}_{\text{terminal}}=I^{2}R$.
2. Dissipated Power wasted as heat inside the battery: $P_{\text{internal}}=I^{2}r$.

The efficiency of the battery in delivering energy to the external circuit is: $$\text{Efficiency}=\frac{P_{\text{load}}}{P_{\text{total}}}=\frac{IR}{\epsilon }=\frac{V_{\text{terminal}}}{\epsilon }$$ For high efficiency, you want $V_{\text{terminal}}\approx \epsilon$, which occurs when the $Ir$ drop is minimal. This happens when the external load resistance $R$ is much larger than the internal resistance $r$ ($R>>r$), resulting in a relatively small circuit current.

Common Pitfalls

1. Treating EMF and Terminal Voltage as Identical: The most frequent error is using a battery's labeled voltage (its nominal EMF, like 1.5 V) as the voltage across a load in a closed circuit calculation. Always check if internal resistance is provided or implied. If it is, you must calculate $V_{\text{terminal}}$ using $\epsilon -Ir$.
2. Misapplying Ohm's Law to the Battery: You cannot say $V_{\text{terminal}}=Ir$. Ohm's Law applies to the internal resistance itself: the voltage across the internal resistor is $Ir$. The terminal voltage is the voltage across the battery's external terminals, which is $\epsilon -Ir$.
3. Ignoring Internal Resistance in Power Calculations: When asked for the power dissipated in the battery, many students mistakenly calculate $I\epsilon$ or $I{V}_{\text{terminal}}$. The correct calculation for energy wasted as heat inside the battery is always $P=I^{2}r$.
4. Confusing Open-Circuit and Loaded Measurements: A multiple-choice trap might show a voltmeter reading 9.0 V across a battery not connected to a circuit, then ask for the current when a resistor is connected. The 9.0 V is the EMF ($\epsilon$). You cannot find the current without knowing $r$, as $I=\epsilon /(R+r)$, not $\epsilon /R$.

Summary

• A real battery is modeled as an ideal EMF source ($\epsilon$) in series with a small internal resistance ($r$).
• The core governing equation is $\epsilon =V_{\text{terminal}}+Ir$, or equivalently, $V_{\text{terminal}}=\epsilon -Ir$.
• Terminal voltage decreases as current increases because a larger portion of the EMF is "used up" driving current through the internal resistance ($Ir$ drop).
• Power is partitioned: total power from chemistry is $I\epsilon$, but only $I{V}_{\text{terminal}}$ is delivered to the external circuit; $I^{2}r$ is wasted as heat within the battery.
• For efficient operation (minimal voltage drop and heat loss), the design goal is for the external load resistance $R$ to be much greater than the internal resistance $r$ of the power source.