Direct Current Circuits

Lessons

Senior 4 · Physics

Direct Current Circuits

5 lessons

Lesson 3 of 5

Resistors in Series, in Parallel and in Mixed Networks

25 min

Why houses are wired in parallel

In a house every lamp, socket and appliance is connected in parallel across the supply, so each works at the full mains voltage and can be switched on or off without affecting the others. A string of decorative lights is often wired in series instead: one broken lamp and the whole string goes dark. Real circuits mix the two. In S3 you combined resistors in simple cases; this lesson derives the rules from the conservation of charge and energy, applies them to networks, and uses them to build the potential divider at the heart of most sensor circuits.

Series and parallel: the rules and why they hold

Figure 3.1: Resistors in series (a) and in parallel (b).

Series. There is only one path, and charge cannot pile up, so the same current $I$ flows through each resistor. Each coulomb loses energy in each resistor in turn, so the p.d.s add: $V = V_{1} + V_{2} + V_{3} = I R_{1} + I R_{2} + I R_{3}$ . The single equivalent resistance has $V = I R$ , so

R = R_{1} + R_{2} + R_{3}

Parallel. Each resistor is connected between the same two junctions, so each has the same p.d. $V$ . Charge is conserved at the junction, so the currents add: $I = I_1 + I_2 + I_3 = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$ . Dividing by $V$ :

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}, \qquad \text{and for two resistors}\quad R = \frac{R_1R_2}{R_1 + R_2}

The equivalent resistance of a parallel group is always smaller than the smallest resistance in it (adding a path makes it easier for charge to flow); $n$ equal resistors $R_{0}$ in parallel give $\frac{R_0}{n}$ . In a parallel pair the current divides in inverse proportion to the resistances: $I_1 = I\frac{R_2}{R_1 + R_2}$ .

Definition

The equivalent resistance of a combination of resistors is the single resistance that would draw the same current from the same supply. In series $R = R_1 + R_2 + \cdots$ ; in parallel $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$ . A potential divider is a pair of resistors in series across a supply, used to give an output p.d. that is a fraction of the input. Example: $4.0\,\Omega$ , $6.0\,\Omega$ and $12\,\Omega$ give $22\,\Omega$ in series and $2.0\,\Omega$ in parallel.

Worked example 1: series and parallel

Find the equivalent resistance of $4.0\,\Omega$ , $6.0\,\Omega$ and $12\,\Omega$ (a) in series, (b) in parallel.

\text{(a)}\;\; R = 4.0 + 6.0 + 12 = 22\,\Omega \qquad \text{(b)}\;\; \frac{1}{R} = \frac{1}{4.0} + \frac{1}{6.0} + \frac{1}{12} = \frac{3 + 2 + 1}{12} = \frac{1}{2.0} \;\Rightarrow\; R = 2.0\,\Omega

The parallel value, $2.0\,\Omega$ , is smaller than the smallest resistor, $4.0\,\Omega$ , as it must be.

Mixed networks

To analyse a network, look for groups that are clearly in series (one path, no junction between them) or in parallel (between the same two junctions), replace each group by its equivalent, and repeat until one resistance remains. Then work back outwards to find each current and p.d.

Figure 3.2: A mixed network, with the currents found in Worked example 2.

Worked example 2: a mixed network

For Figure 3.2 (supply 18 V, negligible internal resistance), find the total resistance, the current from the supply, and the current in each branch.

R_{\parallel} = \frac{12 \times 4.0}{12 + 4.0} = 3.0\,\Omega, \qquad R_{\text{total}} = 6.0 + 3.0 = 9.0\,\Omega, \qquad I = \frac{18}{9.0} = 2.0\,\mathrm{A}

p.d. across the parallel pair: $2.0 \times 3.0 = 6.0\,\mathrm{V}$ ; across the $6.0\,\Omega$ resistor: $12\,\mathrm{V}$ (check: $12 + 6.0 = 18$ ). Branch currents: $\frac{6.0}{12} = 0.50\,\mathrm{A}$ and $\frac{6.0}{4.0} = 1.5\,\mathrm{A}$ , adding to 2.0 A.

The potential divider

Two resistors in series across an input p.d. $V_{in}$ share it in proportion to their resistances, because they carry the same current. The output across $R_{2}$ is

V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}

If a load is connected across the output (Figure 3.3a), it is in parallel with $R_{2}$ , lowering the resistance of the lower part and so lowering $V_{out}$ . The output stays close to the unloaded value only if the load's resistance is much larger than $R_{2}$ .

Figure 3.3: Potential dividers. Replacing a fixed resistor with a sensor makes the output respond to the surroundings.

Replace one resistor with a sensor and the output responds to the world: a thermistor (resistance falls as temperature rises) gives a temperature-dependent output, and a light-dependent resistor (resistance falls in bright light) a light-dependent one. These outputs switch on fans, alarms and street lights (S3, Unit 10).

Worked example 3: a loaded divider and a sensor

(a) A 9.0 V supply feeds $R_1 = 2.0\,\mathrm{k\Omega}$ and $R_2 = 1.0\,\mathrm{k\Omega}$ in series. Find $V_{out}$ across $R_{2}$ , then with a $1.0\,\mathrm{k\Omega}$ load across the output.

V_{out} = 9.0 \times \frac{1.0}{3.0} = 3.0\,\mathrm{V}; \qquad R_2 \parallel \text{load} = 0.50\,\mathrm{k\Omega}, \quad V_{out} = 9.0 \times \frac{0.50}{2.5} = 1.8\,\mathrm{V}

(b) In a fire alarm, a 5.0 V supply feeds a thermistor in series with a fixed $4.7\,\mathrm{k\Omega}$ resistor, and the output is taken across the fixed resistor. The thermistor's resistance is $10\,\mathrm{k\Omega}$ when cool and $1.0\,\mathrm{k\Omega}$ when hot (illustrative values). Find the output in each case.

\text{cool: } 5.0 \times \frac{4.7}{14.7} = 1.6\,\mathrm{V}, \qquad \text{hot: } 5.0 \times \frac{4.7}{5.7} = 4.1\,\mathrm{V}

The large rise in output when the thermistor warms can switch on an alarm circuit.

Experiment 3.1: checking the combination rules

Apparatus: three resistors (for example $10\,\Omega$ , $22\,\Omega$ , $47\,\Omega$ ), a digital multimeter, a low-voltage d.c. supply, connecting leads.

Measure each resistor alone with the multimeter set to ohms (with the supply disconnected).
Connect them in series and measure the total resistance; then in parallel; then as one resistor in series with the other two in parallel. Compare each reading with the value calculated from the rules.
Connect the series arrangement to the supply and measure the p.d. across each resistor and across all three. Check that the p.d.s add and that each is proportional to its resistance.

Agreement within the resistors' tolerance (often ±5%) confirms the rules.

Warning

Safety. Never measure resistance with the ohmmeter while the circuit is connected to a supply: it can damage the meter and gives wrong readings. Use only low-voltage supplies in the laboratory, and switch off before rearranging the circuit.

Warning

Common mistakes. (1) Forgetting to take the reciprocal at the end of a parallel calculation: $\frac{1}{R} = 0.5\,\Omega^{-1}$ means $R = 2\,\Omega$ . (2) Treating two resistors as "in series" just because they are next to each other on the page: there must be no junction between them. (3) Ignoring the load when finding a divider's output. (4) Saying the current is "used up" by the first resistor in a series chain.

Tip

Quick checks: a series total is bigger than the biggest resistor; a parallel total is smaller than the smallest; for two in parallel use "product over sum".

Resistors in series and parallel (Ben Finio). Watch how the equivalent resistance is found for series and parallel groups, and how a mixed circuit is simplified step by step.

✦

Practice

Find the equivalent resistance of $3.0\,\Omega$ and $6.0\,\Omega$ (a) in series, (b) in parallel.
Three $6.0\,\Omega$ resistors are available. List four different resistances you can make using all three.
A $10\,\Omega$ resistor is in parallel with a $15\,\Omega$ resistor, and the pair is in series with a $4.0\,\Omega$ resistor across a 12 V supply. Find the current from the supply and the current in the $15\,\Omega$ resistor.
A 12 V supply feeds a potential divider of $3.0\,\mathrm{k\Omega}$ and $1.0\,\mathrm{k\Omega}$ . Find the output across the $1.0\,\mathrm{k\Omega}$ resistor, without a load and with a $3.0\,\mathrm{k\Omega}$ load.
A 60 W lamp and a 100 W lamp are designed for 230 V. Find the resistance of each and the total current when both are switched on in parallel across 230 V.
In a street light, an LDR (resistance $400\,\Omega$ in daylight, $40\,\mathrm{k\Omega}$ in darkness) is in series with a $10\,\mathrm{k\Omega}$ resistor across 6.0 V. Across which component should the output be taken so that the output is high in the dark? Find the output in darkness.
A student connects a voltmeter of resistance $10\,\mathrm{k\Omega}$ across one of two $100\,\mathrm{k\Omega}$ resistors in series across 10 V and reads 0.83 V instead of 5.0 V. Explain the reading and evaluate the voltmeter's suitability.

Show answersHide answers

(a) $9.0\,\Omega$ ; (b) $\frac{3.0 \times 6.0}{9.0} = 2.0\,\Omega$ .
All in series $18\,\Omega$ ; all in parallel $2.0\,\Omega$ ; two in series with one in parallel across them: $\frac{12 \times 6.0}{18} = 4.0\,\Omega$ ; two in parallel in series with one: $3.0 + 6.0 = 9.0\,\Omega$ .
$R_\parallel = \frac{10 \times 15}{25} = 6.0\,\Omega$ ; total $10\,\Omega$ ; $I = 1.2\,\mathrm{A}$ . p.d. across the pair $= 1.2 \times 6.0 = 7.2\,\mathrm{V}$ ; current in $15\,\Omega$ $= \frac{7.2}{15} = 0.48\,\mathrm{A}$ .
Unloaded: $12 \times \frac{1.0}{4.0} = 3.0\,\mathrm{V}$ . Loaded: $1.0 \parallel 3.0 = 0.75\,\mathrm{k\Omega}$ , $V_{out} = 12 \times \frac{0.75}{3.75} = 2.4\,\mathrm{V}$ .
$R = \frac{V^2}{P}$ : $\frac{230^2}{60} = 880\,\Omega$ and $\frac{230^2}{100} = 530\,\Omega$ . Total current $= \frac{60 + 100}{230} = 0.70\,\mathrm{A}$ .
Across the LDR, whose share of the p.d. rises when its resistance rises in the dark. Darkness: $V_{out} = 6.0 \times \frac{40}{50} = 4.8\,\mathrm{V}$ (in daylight $6.0 \times \frac{0.4}{10.4} = 0.23\,\mathrm{V}$ ).
The voltmeter is in parallel with the $100\,\mathrm{k\Omega}$ resistor, giving $\frac{100 \times 10}{110} = 9.1\,\mathrm{k\Omega}$ ; the divider then gives $10 \times \frac{9.1}{109.1} = 0.83\,\mathrm{V}$ , exactly the reading. The voltmeter's resistance is far too low compared with the resistors it measures, so it badly disturbs the circuit. A voltmeter should have a resistance much larger than the resistance it is connected across (lesson 5).

Quick check

Key points

Series: same current, p.d.s add, $R = R_1 + R_2 + \cdots$ . Parallel: same p.d., currents add, $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots$ .
Simplify networks group by group, then work back to find each current and p.d.
Potential divider: $V_{out} = V_{in}\frac{R_2}{R_1 + R_2}$ ; a load lowers the output unless it is much larger than $R_{2}$ ; sensors make the output respond to light or temperature.

Sources and credits

21.1: Resistors in Series and Parallel, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Electronic-Axial-Lead-Resistors-Array.jpg by Evan-Amos, Wikimedia Commons, public domain.
Video: Resistors in Series and Parallel, Ben Finio, embedded from YouTube.

The derivations, the three diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1Why houses are wired in parallel

2Series and parallel: the rules and why they hold

3Mixed networks

4The potential divider

5Experiment 3.1: checking the combination rules

Why houses are wired in parallel

Series and parallel: the rules and why they hold

Mixed networks

The potential divider

Experiment 3.1: checking the combination rules