EMF, internal resistance and series-parallel resistors

This week you start solving circuits. The exam has two question types on this topic: the terminal voltage and internal resistance of a real battery, and the equivalent resistance, current or power of a network of resistors. Solve every circuit that can be handled by series and parallel reduction with this method, without going into Kirchhoff; it is much faster. Kirchhoff’s rules (week 14) are needed only for circuits that cannot be reduced.

Golden rules:

1. In series the current is common, in parallel the voltage is common. Everything follows from these two sentences.
2. The series equivalent is larger than the largest, the parallel equivalent smaller than the smallest. Check your answer with this.
3. The terminal voltage of a battery is smaller than the emf: $\Delta V = \varepsilon - Ir$. With no current they are equal.
4. Reduce, find the total current, expand back. At every step write $\Delta V = IR$.
5. Energy check: the power supplied by the battery must equal the sum of the powers dissipated in the resistors.

Electromotive force and internal resistance

A battery uses chemical energy to carry charges from low potential to high potential. This work per unit charge is the emf (electromotive force), with symbol $\varepsilon$ and unit volt. Despite the name “force”, it is not a force but a potential difference.

The charges also meet some resistance inside a real battery. We model this as a small internal resistance $r$ in series with an ideal emf source. When an external (load) resistance $R$ is connected to the battery:

$I = \frac{\varepsilon}{R + r}, \qquad \Delta V = \varepsilon - Ir = IR$

$\Delta V$ is the terminal voltage measured between the terminals of the battery. Three cases are the three end points you must memorize:

Example. A battery with $\varepsilon = 9.0$ V and $r = 0.40\ \Omega$ is connected to a 4.1 Ω resistor. What are the current, the terminal voltage and the power in the load?

I  = ε/(R + r) = 9.0 / (4.1 + 0.40) = 2.0 A
ΔV = ε − I r   = 9.0 − (2.0)(0.40) = 8.2 V      (check: I R = 8.2 V)
P  = I² R      = 4.0 × 4.1 = 16 W

PITFALL · emf versus terminal voltage. If you measure the terminals of a battery that is supplying current with a voltmeter, you read $\varepsilon - Ir$, not $\varepsilon$. The information “11.4 V while delivering 3.0 A, 12.0 V when idle” directly gives $r = 0.6/3.0 = 0.20\ \Omega$. $11.4/3.0$ is the load resistance, not the internal resistance.

Power and maximum power transfer

The power balance is $\varepsilon I = I^2 R + I^2 r$: the power produced chemically by the battery turns into heat in the load and inside the battery. The power in the load:

$P_R = I^2R = \frac{\varepsilon^2 R}{(R + r)^2}$

As $R \to 0$ the current is large, but since $R$ is zero the power is zero; as $R \to \infty$ the current is zero and the power is again zero. There is a maximum in between. Solving $dP/dR = 0$:

$R = r \quad\Rightarrow\quad P_{\max} = \frac{\varepsilon^2}{4r}$

At this point the efficiency $R/(R + r)$ is $50\%$: half of the power goes to heat inside the battery. If you want high efficiency, you need $R \gg r$.

Example. Let us connect three different loads to a source with $\varepsilon = 9.0$ V and $r = 1.5\ \Omega$:

The power peaks at $R = r$ and comes out the same for the two cases where $R$ is $r/3$ and $3r$. But the efficiency keeps rising as $R$ grows. “Most power” and “most efficient” are not the same thing; read which one the question asks.

Resistors in series

Resistors connected end to end with no branching in between are in series. Because all the charge has to pass through the same path, the current is common and the voltages add:

$\Delta V = IR_1 + IR_2 + \cdots \quad\Rightarrow\quad R_{eq} = R_1 + R_2 + \cdots$

The voltage is shared in proportion to the resistances (voltage divider): the larger resistance takes the larger share of the voltage. As you add bulbs to a series chain, the power $I^2R$ of each bulb falls quadratically. If one bulb breaks, the circuit opens and all go out.

Resistors in parallel

If both ends of the resistors are connected to the same two nodes, they are in parallel. Because all are connected to the same two points, the voltage is common and the currents add:

$I = \frac{\Delta V}{R_1} + \frac{\Delta V}{R_2} + \cdots \quad\Rightarrow\quad \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$

Shortcut for two resistors: $R_{eq} = \dfrac{R_1R_2}{R_1 + R_2}$ (“product over sum”). For $N$ identical resistors, $R/N$.

The current is shared inversely with the resistance (current divider):

$I_1 = I\,\frac{R_2}{R_1 + R_2}, \qquad I_2 = I\,\frac{R_1}{R_1 + R_2}$

The small resistance takes the large current. The outlets in a house are in parallel: every appliance sees the mains voltage, and switching one appliance on or off does not affect the others.

TIP · don’t forget the reciprocal in parallel. It is very common to find $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} = 0.5$ and write “0.5 Ω”. That is $1/R_{eq}$. $R_{eq} = 2.0\ \Omega$. Check whether the result is smaller than the smallest resistance.

A worked example: reduce and expand back

The circuit is this: the wire from the positive terminal of a battery with $\varepsilon = 18$ V and $r = 1.0\ \Omega$ first passes through $R_1 = 3.0\ \Omega$. Then at node P it splits in two: $R_2 = 6.0\ \Omega$ in one branch, $R_3 = 3.0\ \Omega$ in the other. The two branches join at node Q, and from there it returns to the negative terminal of the battery through $R_4 = 3.0\ \Omega$. Find all the currents and the terminal voltage of the battery.

1. Reduce. $R_2$ and $R_3$ are in parallel between P and Q:

$R_{23} = \frac{6.0 \times 3.0}{9.0} = 2.0\ \Omega$

What remains is a series chain: $R_{\text{ext}} = 3.0 + 2.0 + 3.0 = 8.0\ \Omega$.

2. Total current.

$I = \frac{\varepsilon}{R_{\text{ext}} + r} = \frac{18}{9.0} = 2.0\ \text{A}$

3. Expand back.

Terminal voltage:   ΔV = ε − I r = 18 − 2.0 = 16 V
Across R1:          2.0 × 3.0 = 6.0 V
Between P and Q:    2.0 × 2.0 = 4.0 V   (R2 and R3 see the same voltage)
Across R4:          2.0 × 3.0 = 6.0 V   (6 + 4 + 6 = 16 V, matches the terminal voltage)
I2 = 4.0/6.0 = 0.67 A
I3 = 4.0/3.0 = 1.33 A              (0.67 + 1.33 = 2.0 A, the node checks out)

4. Energy check.

Produced by the battery:  ε I = 18 × 2.0 = 36 W
Dissipated:  r: 4.0   R1: 12   R2: 2.7   R3: 5.3   R4: 12   → total 36 W

Since both checks (the sum of the voltages and the power balance) hold, you can be sure of the answers. On the exam this check takes 20 seconds and catches a wrong answer.

Bulb questions

Think of bulbs as resistors with constant resistance; brightness is the power dissipated.

• Adding a bulb in parallel (ideal battery): the voltage across each bulb is $\varepsilon$ and does not change. The brightnesses stay the same, and the current drawn from the battery increases.
• Adding a bulb in series: the total resistance increases, the current falls, and all dim. Going from two bulbs to three, the power of each falls to $(2/3)^2 = 4/9$ of its value.
• If the battery has an internal resistance: adding a bulb in parallel increases the battery current, $Ir$ grows, and the terminal voltage falls. The other bulbs dim slightly.

Common mistakes