Electrical circuits - voltage & current

What you will learn

what electric current and voltage mean, and their units,
how to read and draw simple circuit diagrams with standard symbols,
the difference between series and parallel circuits,
how voltmeters and ammeters are connected in a circuit,
how to use Ohm’s law, $V = IR$ , for simple calculations.

Worked example 0 Real-world example: a torch

A torch has a $3$ V battery and a small bulb. When the switch is closed, the bulb lights. Explain what is happening in terms of voltage and current.

The battery provides $3$ V of “push” (voltage) between its terminals.
Closing the switch completes a closed loop.
Charged particles (electrons) flow around the loop — this is the current.
As the current passes through the bulb, electrical energy is transformed into light and heat.

Key idea: a circuit needs a closed loop, a voltage source, and something to carry the current.

1. Current and voltage

Current ( $I$ ) is the flow of charge through a wire. Unit: ampere (A). Measured with an ammeter.

$1$ A $=$ a lot of electrons passing a point every second.

Voltage ( $V$ ) is the electrical “push” that drives current. Unit: volt (V). Measured with a voltmeter.

A $9$ V battery provides more push than a $1.5$ V battery.

Resistance ( $R$ ) is how much a component opposes current flow. Unit: ohm ( $\Omega$ ).

A thin wire or a bulb filament has high resistance; a thick copper wire has low resistance.

Ohm's law

Three forms of V = IR

V = IR, \qquad I = \dfrac{V}{R}, \qquad R = \dfrac{V}{I}.

Units

Voltage in volts (V), current in amperes (A), resistance in ohms ( $\Omega$ ).

Worked example 1 Applying Ohm's law

A $12$ V battery drives current through a resistor of $4$ $\Omega$ . Find the current.

I = \dfrac{V}{R} = \dfrac{12}{4} = 3 \text{ A}.

Worked example 2 Finding resistance

A bulb runs at $6$ V and draws $0.5$ A. Find its resistance.

R = \dfrac{V}{I} = \dfrac{6}{0.5} = 12 \text{ }\Omega.

2. Circuit diagrams and symbols

Real circuits are drawn as tidy diagrams using standard symbols.

Common circuit symbols: cell, battery, bulb, resistor, switch, ammeter (A) in series, voltmeter (V) in parallel.

A cell provides one voltage (e.g. 1.5 V). A battery is two or more cells joined together (e.g. a “9 V” battery has six 1.5 V cells in series).

3. Measuring: ammeters and voltmeters

An ammeter measures current. It is connected in series — the current flows through it. It should have very low resistance so it does not disturb the circuit.
A voltmeter measures voltage across a component. It is connected in parallel — placed across the component, not inside the loop. It should have very high resistance so almost no current goes through it.

4. Series vs parallel circuits

Series — components are connected end-to-end on the same loop.

Current is the same at every point: $I_1 = I_2 = I_3$ .
Voltage is shared among components: $V_{\text{total}} = V_1 + V_2 + V_3$ .
If one breaks, the whole circuit is broken (like old Christmas lights).

Parallel — components sit on separate branches.

Voltage is the same across every branch: $V_1 = V_2 = V_{\text{total}}$ .
Current splits between branches: $I_{\text{total}} = I_1 + I_2 + I_3$ .
If one branch breaks, the others keep working (like house wiring).

Worked example 3 Series voltage sharing

Two identical bulbs are in series with a $6$ V battery. What voltage is across each bulb?

Total voltage is shared equally between identical bulbs.
$V_1 + V_2 = 6$ V, so each gets $3$ V.

Each bulb will therefore glow less brightly than it would alone on $6$ V.

Worked example 4 Parallel current splitting

A $12$ V battery drives two parallel bulbs. Each bulb has resistance $6$ $\Omega$ . Find the current through each branch and the total current from the battery.

In parallel, each branch has the full $12$ V.
Each branch current: $I = V/R = 12/6 = 2$ A.
Total current: $2 + 2 = 4$ A.

Key idea: in parallel, adding another bulb does not reduce the others — each one still sees the full voltage.

5. What components do

Bulbs convert electrical energy into light and heat.
Resistors convert electrical energy into heat; they limit current.
Switches open or close a circuit.
Motors convert electrical energy into movement.
Buzzers / speakers convert electrical energy into sound.

Designing a useful circuit means choosing the right components and arrangement for the job.

Practice: Year 8

Fluency

Definitions and units

Give the units for (a) current, (b) voltage, (c) resistance.
Name the instrument used to measure (a) current, (b) voltage.
How is an ammeter connected? How is a voltmeter connected?
What is the difference between a “cell” and a “battery”?
Name two useful components that transform electrical energy.

Fluency

Ohm's law

Find $I$ when $V = 12$ V and $R = 3$ $\Omega$ .
Find $V$ when $I = 0.5$ A and $R = 20$ $\Omega$ .
Find $R$ when $V = 9$ V and $I = 0.3$ A.
A $240$ V kettle draws $10$ A. Find its resistance.
A $1.5$ V AA cell drives $0.1$ A through a small bulb. Find the bulb’s resistance.

Fluency

Series and parallel

Three identical bulbs share a $9$ V battery in series. What voltage is across each?
In a parallel circuit with a $12$ V battery, what voltage is across each branch?
If one bulb in a series string fails (breaks the circuit), what happens to the rest?
If one bulb in a parallel circuit fails, what happens to the others?
Why is house wiring in parallel rather than series?

Reasoning

Explain

Explain why a voltmeter has very high resistance.
Explain why you should not use an ammeter in parallel with a battery.
Explain why two bulbs in series are dimmer than the same bulbs in parallel (on the same battery).
Draw (describe) a simple circuit diagram with a cell, switch, bulb, and ammeter.

Problem solving

Applied contexts

A phone charger outputs $5$ V. The phone takes $2$ A while charging. What resistance does this correspond to?
A lamp rated $240$ V draws $0.25$ A. Find its resistance. At this voltage, how much power is it using (use $P = VI$ )?
Two bulbs, each $4$ $\Omega$ , are in parallel with a $12$ V battery. Find (a) current through each, (b) total current.
Explain why a $9$ V smoke alarm still works when one bulb on a series of Christmas lights elsewhere in the house fails.

Challenge

Reasoning

Harder reasoning

Three resistors ( $2$ $\Omega$ , $3$ $\Omega$ , $4$ $\Omega$ ) are connected in series across a $9$ V battery. Find (a) total resistance, (b) current through the circuit, (c) voltage across the $4$ $\Omega$ resistor.
A car headlight runs at $12$ V and $5$ A. Find its resistance. If both headlights are on simultaneously (in parallel), find the total current supplied by the battery.
Explain why a short circuit (a wire accidentally connecting the two terminals of a battery) can cause a fire, using Ohm’s law.
Design a circuit that lets one switch turn on two bulbs independently of a third bulb. Describe your arrangement in words (or sketch mentally) and explain why it behaves as required.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

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Year 8 answers

Fluency

Definitions and units

(a) Ampere (A), (b) volt (V), (c) ohm ( $\Omega$ ).
(a) Ammeter, (b) voltmeter.
Ammeter: in series (in line with the component, so current flows through it). Voltmeter: in parallel (across the component).
A cell is one unit; a battery is two or more cells joined together.
Any two of: bulb (light + heat), resistor (heat), motor (movement), buzzer/speaker (sound), heater element (heat).

Fluency

Ohm's law

$I = 12/3 = 4$ A.
$V = 0.5 \times 20 = 10$ V.
$R = 9/0.3 = 30$ $\Omega$ .
$R = 240/10 = 24$ $\Omega$ .
$R = 1.5/0.1 = 15$ $\Omega$ .

Fluency

Series and parallel

$9/3 = 3$ V across each.
$12$ V across each branch.
The circuit is broken; all other bulbs go out.
The others keep working (each branch is still complete).
Appliances need the same full voltage, and if one device turns off or breaks, others must keep working.

Reasoning

Explain

High resistance means almost no current flows through the voltmeter itself, so it measures the voltage across the component without changing the circuit.
Ammeters have near-zero resistance. Connecting directly across a battery gives a huge current (short circuit) that can overheat wires and damage the ammeter.
In series, each bulb only gets a share of the voltage, so less current flows through both bulbs and each is dimmer. In parallel each bulb still sees the full voltage, so it glows at full brightness.
A simple loop: cell (positive terminal) $\to$ switch $\to$ ammeter $\to$ bulb $\to$ back to negative terminal. All components in a single series loop.

Problem solving

Applied contexts

$R = V/I = 5/2 = 2.5$ $\Omega$ .
$R = 240/0.25 = 960$ $\Omega$ . Power $P = VI = 240 \times 0.25 = 60$ W.
(a) Each bulb: $I = 12/4 = 3$ A. (b) Total: $3 + 3 = 6$ A.
The Christmas lights are on their own circuit; the smoke alarm is on a different parallel branch (usually battery-powered). A fault in one parallel branch does not affect another.

Reasoning

Challenge

(a) $R_{\text{total}} = 2 + 3 + 4 = 9$ $\Omega$ . (b) $I = 9/9 = 1$ A. (c) $V_4 = IR = 1 \times 4 = 4$ V.
$R = 12/5 = 2.4$ $\Omega$ per headlight. In parallel, each still draws $5$ A, so total current $= 10$ A.
A short circuit has near-zero resistance. By $I = V/R$ , a tiny resistance gives a huge current. The wire then heats rapidly ( $P = VI$ is large), can melt insulation, and start a fire.
Put bulbs 1 and 2 in parallel with each other, controlled together by a first switch. Put bulb 3 on its own branch (also parallel to the rest) controlled by a second switch. Because every branch is in parallel with the battery, each branch operates independently at full voltage.

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