28.5 Real Batteries

Real batteries, like ideal batteries, use chemical reactions to separate charge, create a potential difference, and provide energy to the circuit. However, real batteries also provide a slight resistance to the charges on the charge escalator. They have what is called an internal resistance, which is symbolized by r. FIGURE 28.17 shows both an ideal and a real battery.

The figure consists of two parts that illustrate the diagrammatic representation of an ideal battery and a real battery.

FIGURE 28.17 Full Alternative Text

From our vantage point outside a battery, we cannot see $ε$ and $r$ separately. To the user, the battery provides a potential difference $Δ V_{bat}$ called the terminal voltage. $Δ V_{bat} = ε$ for an ideal battery, but the presence of the internal resistance affects $Δ V_{bat}$ . Suppose the current in the battery is I. As charges travel from the negative to the positive terminal, the potential increases by $ε$ but decreases by $Δ V_{int} = - I r$ due to the internal resistance. Thus the terminal voltage of the battery is

Δ V_{bat} = ε - I r \leq ε

(28.16)

Only when $I = 0$ , meaning that the battery is not being used, is $Δ V_{bat} = ε$ .

FIGURE 28.18 shows a single resistor $R$ connected to the terminals of a battery having emf $ε$ and internal resistance r. Resistances $R$ and $r$ are in series, so we can replace them, for the purpose of circuit analysis, with a single equivalent resistor $R_{eq} = R + r$ . Hence the current in the circuit is

I = \frac{ε}{R_{eq}} = \frac{ε}{R + r}

(28.17)

If $r ≪ R$ , so that the internal resistance of the battery is negligible, then $I \approx ε / R$ , exactly the result we found before. But the current decreases significantly as $r$ increases.

The figure consists of two parts that illustrate the circuit diagram of the internal resistance in a battery and its equivalent resistance.

FIGURE 28.18 Full Alternative Text

We can use Ohm’s law to find that the potential difference across the load resistor $R$ is

Δ V_{R} = I R = \frac{R}{R + r} ε

(28.18)

Similarly, the potential difference across the terminals of the battery is

Δ V_{bat} = ε - I r = ε - \frac{r}{R + r} ε = \frac{R}{R + r} ε

(28.19)

The potential difference across the resistor is equal to the potential difference between the terminals of the battery, where the resistor is attached, not equal to the battery’s emf. Notice that $Δ V_{bat} = ε$ only if $r = 0$ (an ideal battery with no internal resistance).

EXAMPLE 28.5 Lighting up a flashlight

A $6 Ω$ flashlight bulb is powered by a 3 V battery with an internal resistance of $1 Ω$ . What are the power dissipation of the bulb and the terminal voltage of the battery?

MODEL Assume ideal connecting wires but not an ideal battery.

VISUALIZE The circuit diagram looks like Figure 28.18. $R$ is the resistance of the bulb’s filament.

SOLVE Equation 28.17 gives us the current:

I = \frac{ε}{R + r} = \frac{3 V}{6 Ω + 1 Ω} = 0.43 A

This is 15% less than the 0.5 A an ideal battery would supply. The potential difference across the resistor is $Δ V_{R} = I R = 2.6 V,$ thus the power dissipation is

P_{R} = I Δ V_{R} = 1.1 W

The battery’s terminal voltage is

Δ V_{bat} = \frac{R}{R + r} ε = \frac{6 Ω}{6 Ω + 1 Ω} 3 V = 2.6 V

REVIEW $1 Ω$ is a typical internal resistance for a flashlight battery. The internal resistance causes the battery’s terminal voltage to be 0.4 V less than its emf in this circuit.

28.5 Real Batteries

FIGURE 28.17 An ideal battery and a real battery.

FIGURE 28.18 A single resistor connected to a real battery is in series with the battery’s internal resistance, giving $R_{eq} = R + r$ .

EXAMPLE 28.5 Lighting up a flashlight