B5-4. Electromotive Force and Internal Resistance

1. Real Power Sources

In idealized circuits, a battery delivers a constant voltage regardless of the current drawn. In reality, the chemicals and physical structures inside a real battery naturally impede the flow of ions. This intrinsic property is called internal resistance ($r$).

Definitions:

  • Electromotive Force ($\varepsilon$): The absolute total work done per unit charge in converting chemical energy to electrical energy (V).
  • Terminal Potential Difference ($V$): The actual, usable voltage delivered to the external load $R$ (V).
  • Lost Volts ($Ir$): The unrecoverable voltage dropped completely inside the battery to overcome $r$.

The Governing Equations:

By applying Kirchhoff's Voltage Law to a circuit containing a real battery and an external load $R$:

$$\varepsilon = I(R + r)$$

Because the terminal voltage delivered to the load is defined as $V = IR$, we can substitute to find the linear relationship between voltage and current:

$$V = \varepsilon - I r$$

Example 1: Graphical Extraction of Battery Parameters

Problem: In a lab, a variable resistor is placed across a real battery. A plot of terminal potential difference ($V$) on the y-axis against current ($I$) on the x-axis yields a straight line. The y-intercept is exactly $12.0$ V, and the x-intercept (the short-circuit current) is $24.0$ A. Calculate the internal resistance of the battery.

Solution:

  • Step 1: Map the equation to a linear function ($y = mx + c$).
    The equation $V = -rI + \varepsilon$ models a linear decay. The y-intercept ($c$) is the absolute EMF ($\varepsilon$), and the gradient ($m$) is the negative internal resistance ($-r$).
  • Step 2: Identify EMF.
    $\varepsilon = \mathbf{12.0 \text{ V}}$
  • Step 3: Calculate the gradient to find $r$.
    $$\text{Gradient} = \dfrac{\Delta V}{\Delta I} = \dfrac{0 - 12.0}{24.0 - 0} = -0.5$$
    Since the gradient equals $-r$, the internal resistance $r = \mathbf{0.5 \text{ Ω}}$.

2. Maximum Power Transfer Theorem

Because energy is always lost internally as heat ($I^2 r$), there is a strict mathematical ceiling to how much power a battery can successfully deliver to an external load.

Example 2: Proving the Maximum Power Boundary

Problem: A heavy-duty battery with an EMF $\varepsilon = 20$ V and internal resistance $r = 4$ Ω is connected to a load $R$. Calculate the power delivered to the load when $R = 2$ Ω, $R = 4$ Ω, and $R = 8$ Ω. Conclude when maximum power transfer occurs.

Solution:

  • Step 1: Set up the power equation.
    $$I = \dfrac{\varepsilon}{R + r} \quad \Rightarrow \quad P_{\text{load}} = I^2 R = \left(\dfrac{\varepsilon}{R+r}\right)^2 R$$
  • Step 2: Test the three configurations.
    When $R = 2$ Ω: $P = \left(\dfrac{20}{2+4}\right)^2 \times 2 = \left(\dfrac{20}{6}\right)^2 \times 2 \approx \mathbf{22.2 \text{ W}}$
    When $R = 4$ Ω: $P = \left(\dfrac{20}{4+4}\right)^2 \times 4 = \left(\dfrac{20}{8}\right)^2 \times 4 = 6.25 \times 4 = \mathbf{25.0 \text{ W}}$
    When $R = 8$ Ω: $P = \left(\dfrac{20}{8+4}\right)^2 \times 8 = \left(\dfrac{20}{12}\right)^2 \times 8 \approx \mathbf{22.2 \text{ W}}$
  • Conclusion: The electrical power peaks in a symmetrical bell-curve distribution. Maximum power is successfully transferred exactly when the load resistance structurally matches the internal resistance ($R = r$).
  • Calculus Proof: Setting $\dfrac{dP}{dR} = \varepsilon^2 \left[ \dfrac{(R+r)^2 - 2R(R+r)}{(R+r)^4} \right] = 0$ rigorously yields $R = r$.