Ohm law for a complete circuit

A real source of current has an internal resistance $r$, so the current in a closed circuit depends both on the external resistance $R$ and on the source itself. Charge carriers pass in series through the interior of the source and through the external part, so the total resistance of the circuit is $R + r$. The electromotive force $\varepsilon$ equals the work done by the non-electrostatic forces in carrying a unit charge around the whole closed circuit, hence

$$I = \frac{\varepsilon}{R + r}.$$

The voltage across the source terminals (the same as the voltage across the external part) is

$$U = IR = \varepsilon - Ir,$$

that is, it is smaller than the EMF and decreases as the current grows. The relation for the external part, $U = IR$, still holds, while the "missing" part of the EMF, $Ir$, is spent inside the source. As $R \to 0$ the current is limited only by the internal resistance: $I_{sc} = \varepsilon / r$ — this is the short-circuit current.

A cell with EMF $\varepsilon = 12$ V and internal resistance $r = 0.5\ \Omega$ is connected to an external resistor $R = 5.5\ \Omega$. Find the current in the circuit and the terminal voltage of the source.

With the terminals open, a voltmeter across a battery reads $4.5$ V. When the battery is connected to a resistor $R = 8\ \Omega$, the voltmeter across its terminals reads $4.0$ V. Find the current in the circuit and the internal resistance of the battery.

A battery with EMF $\varepsilon = 6.0$ V and internal resistance $r = 0.40\ \Omega$ feeds a variable external resistor $R$. (a) Express the power dissipated in the external resistor as a function of $R$. (b) Find the value of $R$ that makes this power maximal and compute that maximum. What fraction of the total power is useful there?

Self-check with a simulation (safe: a short circuit is modelled, not created).

1. Open a circuit simulation (for example, "Circuit Construction Kit: DC — Virtual Lab"). Build the circuit: battery — ammeter — variable resistor $R$ — back to the battery; connect a voltmeter in parallel with the battery terminals.
2. In the battery settings set the EMF $\varepsilon = 9.0$ V and the internal resistance $r = 1.5\ \Omega$.
3. Vary $R$: 0.5; 1.5; 3.0; 6.0; 15 $\Omega$. For each value record the current $I$ and the terminal voltage $U$ in a table.
4. Plot $U(I)$ and check that the points lie on the straight line $U = \varepsilon - Ir$; read $\varepsilon$ as the intercept on the $U$ axis and $r$ as the magnitude of the slope. Compare with the values you set.
5. Test a prediction: for which $R$ is the current equal to half the short-circuit current $I_{sc} = \varepsilon / r$? Set that value of $R$ and compare the measurement with the calculation.
6. On the graph, estimate the point where the useful power $P = UI$ is maximal and compare it with the condition $R = r$.

Do not short-circuit real batteries: they can heat up and leak. Check every dangerous regime in the simulation only.

To show that a real source consists of an ideal EMF source $\varepsilon$ in series with an internal resistance $r$, and that the terminal voltage falls linearly as the current grows.

Diagram and graph. On the left, a closed-circuit diagram: a dashed rectangle labelled "source" contains an ideal EMF source $\varepsilon$ in series with a resistance $r$; an external resistor $R$ with an ammeter $A$ is connected to its terminals, and a voltmeter $V$ is connected in parallel with the terminals. An arrow shows the current direction $I$ from "+" to "−" in the external part. On the right, the $U(I)$ graph: a straight line crossing the $U$ axis at $\varepsilon$ and the $I$ axis at $I_{sc} = \varepsilon / r$; the slope of the line is $-r$, and two shaded areas represent the useful power $UI$ in the external part and the loss $I^2 r$ inside the source.

While discharging, a powerbank provides an emf $\varepsilon = 5{.}0$ V and has internal resistance $r = 0{.}10\ \Omega$. A phone in fast-charge mode draws a current $I = 2{.}0$ A, so it behaves as a load of resistance $R = U/I = 2{.}4\ \Omega$. The output voltage of the powerbank is $U = \varepsilon - Ir = 5{.}0 - 0{.}10\cdot 2{.}0 = 4{.}80$ V, not 5.0 V. Substituting $R$ and $r$ into $I = \dfrac{\varepsilon}{R+r}$ gives $\dfrac{5{.}0}{2{.}4+0{.}1} = 2{.}0$ A — the same current that was measured.

The scenario exercises the complete-circuit Ohm law $I = \dfrac{\varepsilon}{R+r}$ and the terminal-voltage relation $U = \varepsilon - Ir$ that follows from it: part of the emf is lost across the internal resistance, so the real current is smaller than $\varepsilon/R$.

A car battery has $\varepsilon = 12{.}6$ V and internal resistance $r = 0{.}020\ \Omega$. When the engine is cranked, the starter draws $I = 200$ A. The starter resistance follows from the complete circuit: $R = \dfrac{\varepsilon}{I} - r = \dfrac{12{.}6}{200} - 0{.}020 = 0{.}043\ \Omega$. The terminal voltage drops to $U = 12{.}6 - 0{.}020\cdot 200 = 8{.}6$ V, which is why the headlights dim noticeably during cranking. The total power is $\varepsilon I = 2520$ W, and $I^{2}r = 200^{2}\cdot 0{.}020 = 800$ W of it is dissipated inside the battery.

Both parts of the theory are checked here: the relation $I = \dfrac{\varepsilon}{R+r}$ sets the current through the starter, while the difference $\varepsilon - Ir$ shows how a large current lowers the voltage across the load and raises the $I^{2}r$ losses inside the source.

A torch runs on two cells with an emf $\varepsilon = 3{.}0$ V; the bulb resistance is $R = 5{.}5\ \Omega$. With fresh cells $r = 0{.}5\ \Omega$, so $I = \dfrac{3{.}0}{5{.}5+0{.}5} = 0{.}50$ A, the voltage across the bulb is $U = 3{.}0 - 0{.}50\cdot 0{.}5 = 2{.}75$ V, and the power is $I^{2}R = 1{.}375$ W. As the cells discharge, the internal resistance grows, say to $r = 2{.}0\ \Omega$. Then $I = \dfrac{3{.}0}{5{.}5+2{.}0} = 0{.}40$ A, $U = 3{.}0 - 0{.}40\cdot 2{.}0 = 2{.}20$ V, and the power falls to $I^{2}R = 0{.}88$ W — the light becomes noticeably dimmer.

The scenario shows that the current in a complete circuit depends not only on the load $R$ but also on the internal resistance $r$: with $\varepsilon$ and $R$ unchanged, a larger $r$ reduces $I$ through $I = \dfrac{\varepsilon}{R+r}$ and lowers the terminal voltage to $\varepsilon - Ir$.