Basic Electrical Wiring and Installation

Lessons

Senior 3 · Physics

Basic Electrical Wiring and Installation

5 lessons

Lesson 1 of 5

Resistors in Series, in Parallel and in Combination

25 min

Why does the meter speed up?

Every socket and lamp in a house gets the same 230 V, whether one appliance is switched on or ten. Yet each extra appliance you switch on makes the prepaid meter count units faster, because more current is drawn from the supply. In S1 you met series and parallel circuits with two resistors. This lesson builds the rules properly, uses them on larger networks, and shows why a house is wired the way it is.

Resistors in series

In series, the same current $I$ passes through every resistor, one after another. The supply's voltage is shared: $V = V_{1} + V_{2} + V_{3}$ . Since $V_{1} = I R_{1}$ and so on,

IR = IR_1 + IR_2 + IR_3 \quad\Rightarrow\quad R = R_1 + R_2 + R_3

The total resistance is larger than any one resistor. The voltage divides in proportion to the resistances: the biggest resistor gets the biggest share. Two resistors in series used to supply a fraction of a voltage form a potential divider:

V_2 = V \times \frac{R_2}{R_1 + R_2}

Worked example 1: three in series

Resistors of 10 Ω, 20 Ω and 30 Ω are connected in series to a 12 V battery. Find the current and the voltage across each.

R = 10 + 20 + 30 = 60\ \Omega \qquad I = \frac{V}{R} = \frac{12}{60} = 0.20\ \text{A}

V_1 = 0.20 \times 10 = 2.0\ \text{V} \quad V_2 = 0.20 \times 20 = 4.0\ \text{V} \quad V_3 = 0.20 \times 30 = 6.0\ \text{V}

Check: $2.0 + 4.0 + 6.0 = 12\ \text{V}$ .

Resistors in parallel

In parallel, every resistor is connected across the same two points, so each has the same voltage $V$ . The current divides between the branches: $I = I_{1} + I_{2} + I_{3}$ . Since $I_1 = \frac{V}{R_1}$ and so on,

\frac{V}{R} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \quad\Rightarrow\quad \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

Two useful shortcuts:

\text{two resistors: } R = \frac{R_1 R_2}{R_1 + R_2} \qquad\qquad n \text{ equal resistors } R_0 \text{: } R = \frac{R_0}{n}

The total resistance is smaller than the smallest branch: every branch added is one more path for the current. The current divides in inverse proportion to the resistances: the smallest resistor takes the largest current.

Definition

Equivalent (total) resistance: the single resistance that would draw the same current from the same supply as a whole network of resistors.

Potential divider: two or more resistors in series across a supply, used to obtain a chosen fraction of the supply voltage across one of them. Example: 1 kΩ and 2 kΩ across 9 V give 6 V across the 2 kΩ resistor.

Worked example 2: three in parallel

Resistors of 6 Ω, 12 Ω and 12 Ω are connected in parallel to a 6.0 V supply. Find the equivalent resistance and the current in each.

\frac{1}{R} = \frac{1}{6} + \frac{1}{12} + \frac{1}{12} = \frac{2 + 1 + 1}{12} = \frac{4}{12} \quad\Rightarrow\quad R = 3.0\ \Omega

I_6 = \frac{6.0}{6} = 1.0\ \text{A} \qquad I_{12} = \frac{6.0}{12} = 0.50\ \text{A each} \qquad I = \frac{6.0}{3.0} = 2.0\ \text{A}

Check: $1.0 + 0.50 + 0.50 = 2.0\ \text{A}$ .

Combined networks

To analyse a network that mixes series and parallel parts, replace each simple group by its equivalent resistance, step by step, until one resistance remains; then work back out to find each current and voltage.

Worked example 3: the network above

Find the ammeter reading and the current in each of $R_{2}$ and $R_{3}$ .

Step 1 the parallel pair between P and Q:

R_{PQ} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4.0\ \Omega

Step 2 in series with $R_{1}$ : $R = 4.0 + 4.0 = 8.0\ \Omega$ , so the ammeter reads $I = \frac{12}{8.0} = 1.5\ \text{A}$ .

Step 3 voltages: $V_1 = 1.5 \times 4.0 = 6.0\ \text{V}$ , so $V_{PQ} = 12 - 6.0 = 6.0\ \text{V}$ .

Step 4 branch currents: $I_2 = \frac{6.0}{6} = 1.0\ \text{A}$ , $I_3 = \frac{6.0}{12} = 0.50\ \text{A}$ . Check: $1.0 + 0.50 = 1.5\ \text{A}$ .

Why houses are wired in parallel

Appliances in parallel (as in a house)	Appliances in series (never used for mains)
Each gets the full 230 V it was designed for	The 230 V is shared, so each gets less
Each can be switched on and off on its own	One switch turns everything on or off
If one fails, the others keep working	If one fails, the circuit is broken and all stop
Each appliance switched on adds a branch: total resistance falls, total current rises	Each appliance added raises the resistance and lowers the current

That last line is why the meter speeds up: more appliances in parallel draw a larger total current from the supply. It is also why a circuit must be protected against too many appliances being switched on at once (lesson 5).

Warning

Two common slips in parallel sums. (1) Stopping at $\frac{1}{R}$ : after adding the fractions, turn the answer upside down. (2) Using "product over sum" for three resistors at once: it works only for two. Check: a parallel total must be smaller than the smallest branch.

Tip

Series: same current, voltages add. Parallel: same voltage, currents add. Equal resistors: $n$ in series give $n R_{0}$ , $n$ in parallel give $\frac{R_0}{n}$ .

Resistors in series and parallel (Manocha Academy). Watch how the series and parallel formulas are derived from the way current and voltage are shared, and how they are used in examples.

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Practice

Resistors of 5 Ω and 15 Ω are connected in series to a 10 V supply. Find the current and the voltage across each.
Find the equivalent resistance of 4 Ω and 12 Ω in parallel.
Five identical 10 Ω resistors are connected in parallel. Find their equivalent resistance.
A 6 Ω resistor is in series with a parallel pair of 3 Ω and 6 Ω, across a 16 V supply. Find the total current and the current in each of the parallel resistors.
A potential divider is made from 1.0 kΩ and 2.0 kΩ resistors in series across a 9.0 V battery. What is the voltage across the 2.0 kΩ resistor?
Explain why the equivalent resistance of resistors in parallel is always less than the smallest one.
Give three reasons why the appliances in a house are connected in parallel.
You have three 6 Ω resistors. Using all three each time, find four different equivalent resistances you can make, and draw each arrangement.

Show answersHide answers

$R = 20\ \Omega$ , $I = 0.50\ \text{A}$ ; 2.5 V across 5 Ω and 7.5 V across 15 Ω.
$R = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3.0\ \Omega$ .
$R = \frac{10}{5} = 2.0\ \Omega$ .
Parallel pair: $\frac{3 \times 6}{3 + 6} = 2.0\ \Omega$ ; total $8.0\ \Omega$ ; $I = \frac{16}{8.0} = 2.0\ \text{A}$ . Voltage across the pair $= 2.0 \times 2.0 = 4.0\ \text{V}$ : $I_3 = \frac{4.0}{3} \approx 1.33\ \text{A}$ , $I_6 = \frac{4.0}{6} \approx 0.67\ \text{A}$ .
$V = 9.0 \times \frac{2.0}{1.0 + 2.0} = 6.0\ \text{V}$ .
Each extra branch gives the current another path, so for the same voltage the total current is larger than through any one branch alone; a larger current for the same voltage means a smaller resistance.
Each appliance gets the full 230 V; each can be switched independently; if one fails or is switched off, the others still work.
All in series: 18 Ω. All in parallel: 2 Ω. Two in parallel (3 Ω) in series with the third: 9 Ω. Two in series (12 Ω) in parallel with the third: $\frac{12 \times 6}{18} = 4\ \Omega$ .

Key points

Series: $R = R_1 + R_2 + \dots$ ; same current; voltages add and divide in proportion to $R$ .
Parallel: $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$ ; same voltage; currents add and divide in inverse proportion to $R$ .
Reduce a combined network step by step, then work back to each current and voltage.
House appliances are in parallel: full voltage, independent switching, and a total current that grows with each appliance.

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. Consulted for the physics; no text was copied.
Video: Resistors in Series and Parallel, Manocha Academy, embedded from YouTube.

Written by Ganzaa: the lesson text, the circuit diagram, the table, worked examples and practice exercises.

1Why does the meter speed up?

2Resistors in series

3Resistors in parallel

4Combined networks

5Why houses are wired in parallel

Why does the meter speed up?

Resistors in series

Resistors in parallel

Combined networks

Why houses are wired in parallel