Topic 11.2 Simple Circuits: model a single-loop circuit with a source of EMF, internal resistance and a load, and find currents and voltages.

What this topic is asking

The College Board (Topic 11.2) wants you to model a single-loop circuit containing a source of EMF with internal resistance and an external load, and to find the current and voltages. The central idea is that a real battery's terminal voltage drops below its EMF once it delivers current.

EMF and the energy source

A battery converts chemical energy into electrical potential energy, raising the potential of charge from the negative to the positive terminal. The EMF is what that conversion would deliver per coulomb if the source were perfect. Other sources of EMF work the same way but draw on different energy: a solar cell converts light, a generator converts mechanical work, and an inductor's back-EMF converts stored magnetic energy. In every case the EMF measures the energy per coulomb the source can supply, and the symbol $\varepsilon$ stands for that energy-per-charge, not a force.

Internal resistance and terminal voltage

A real source also has an internal resistance $r$, representing the energy lost inside it. When a current $I$ flows, $Ir$ volts are dropped internally, so the voltage available outside, the terminal voltage, is

At open circuit ($I = 0$) the terminal voltage equals the EMF; as the load draws more current, $V_{term}$ falls. For a single loop with external load $R$, applying $V_{term} = IR$ gives the loop current

Energy conservation around the loop

Going once around the loop, the EMF $\varepsilon$ supplied per coulomb is shared between the internal and external resistances:

This is conservation of energy per charge (Kirchhoff's loop rule for one loop). Multiplying by $I$ gives the power balance: $\varepsilon I = I^2 R + I^2 r$, the source's total power split between the load and internal heating. The fraction delivered to the load, $\dfrac{R}{R + r}$, approaches one only when the load resistance far exceeds the internal resistance. This is why a battery with a high internal resistance (an old or cold cell) wastes much of its energy heating itself and delivers little to the circuit, and why short-circuiting a battery ($R \to 0$) dumps nearly all the power internally, heating the cell dangerously.

Try this

Q1. A $1.5$ V cell with $r = 0.20\,\Omega$ drives $0.50$ A. Find its terminal voltage. [2 points]

• Cue. $V_{term} = \varepsilon - Ir = 1.5 - (0.50)(0.20) = 1.4$ V.

Q2. State when a battery's terminal voltage equals its EMF. [1 point]

• Cue. At open circuit, when no current flows ($I = 0$).