Electric Circuits Revision Notes for HSC SSCE Physics

Circuits and Circuit Diagrams

What are circuit diagrams?

Circuit diagrams are visual representations used to show how various components in an electrical circuit are connected together. Unlike a photograph or realistic drawing, a circuit diagram uses standardised symbols to represent each component, similar to how a train network map shows connections rather than exact physical locations.

infoNote

Think of a circuit diagram like a subway map - it doesn't show the exact physical layout, but rather the relationships and connections between different points. This makes it much easier to understand how the system works without getting distracted by unnecessary details.

The key purpose of a circuit diagram is to show the relationships between components and how they are connected, rather than their actual physical sizes or positions. This makes circuit diagrams an essential tool for understanding, designing, and analysing electrical circuits.

Circuit symbols

Each type of electrical component has its own unique symbol. Learning these symbols is essential for reading and drawing circuit diagrams. The most commonly used circuit symbols include:

chatImportant

You should familiarise yourself with these symbols as you'll need to recognise them in diagrams and use them when drawing your own circuits. Just like learning the alphabet before reading, mastering these symbols is fundamental to working with electrical circuits.

Drawing circuit diagrams

When drawing a circuit diagram, follow these guidelines:

Use the correct standardised symbol for each component
Connect components with straight lines representing wires
Arrange components in a clear, logical layout (usually rectangular)
Show the direction of conventional current flow (from positive to negative terminal)
Ensure all connections are clearly shown
Wire lengths shown in the diagram don't represent actual wire lengths

Example: Drawing a series circuit

lightbulbExample

Worked Example: Drawing a Series Circuit

Question: A battery is connected in series with a light globe, a resistor and an ammeter. Draw a circuit diagram for this circuit.

Solution:

The circuit consists of four components connected one after another (in series):

Battery
Light globe (filament lamp)
Resistor
Ammeter

To draw the diagram, we arrange these components in a single loop, using the correct symbol for each component.

The completed circuit diagram shows all components connected in a continuous path, forming a complete loop.

Practice: Identifying circuit components

Looking at a circuit diagram, you should be able to identify each component. Consider the following circuit:

This circuit contains:

A voltmeter (V) - measures potential difference
An ammeter (A) - measures current
An LED (light-emitting diode) - shown with arrows indicating light emission
A resistor - rectangular symbol
A battery - provides electrical energy

Kirchhoff's laws and conservation principles

Two fundamental conservation principles help us analyse circuits: conservation of energy and conservation of charge. When applied to circuits, these are known as Kirchhoff's laws, named after German physicist Gustav Kirchhoff.

Kirchhoff's voltage law - conservation of energy

Kirchhoff's voltage law states that the sum of all potential differences (voltages) around any closed loop in a circuit must equal zero:

$\sum V = 0$

infoNote

This law is based on conservation of energy. When a charged particle travels around a complete loop in a circuit, it must return to its starting point with the same energy it started with. Any energy gained (from a battery) must equal the energy lost (across resistors).

Understanding voltage law with a series circuit

Consider a simple series circuit:

In this circuit, current $I$ flows from the battery through resistors $R_1$ and $R_2$ . As charged particles move through each resistor, they lose potential energy, creating a potential difference (voltage drop) across each resistor.

When applying Kirchhoff's voltage law around the loop (from point A → B → C → A):

$V_{battery} - V_{R_1} - V_{R_2} = 0$

Or rearranged:

$V_{battery} = V_{R_1} + V_{R_2}$

This tells us that the voltage supplied by the battery equals the sum of voltage drops across all resistors in the circuit.

lightbulbExample

Worked Example: Applying Kirchhoff's Voltage Law

Question: Consider the circuit shown above. The battery supplies a potential difference of 12 V. A voltmeter measures the potential difference across the first resistor ( $R_1$ ) as 3 V. What is the potential difference across $R_2$ ?

Solution:

Given information:

$V_{battery} = 12$ V
$V_{R_1} = 3$ V

Using Kirchhoff's voltage law around the loop:

$\sum V = 0$

$V_{battery} - V_{R_1} - V_{R_2} = 0$

Rearranging for $V_{R_2}$ :

$V_{R_2} = V_{battery} - V_{R_1}$

$V_{R_2} = 12 \text{ V} - 3 \text{ V}$

$V_{R_2} = 9 \text{ V}$

The potential difference across $R_2$ is 9 V.

Kirchhoff's current law - conservation of charge

Kirchhoff's current law states that the total current entering a junction must equal the total current leaving that junction:

$\sum I = 0$

Or equivalently:

$\sum I_{in} = \sum I_{out}$

infoNote

This law is based on conservation of charge. Charge cannot accumulate at junctions, nor can it be created or destroyed. Therefore, whatever charge flows into a junction per unit time must flow out in the same time period.

Understanding current law with a parallel circuit

In this parallel circuit, current $I_A$ approaches a junction where it splits into two paths: $I_B$ and $I_C$ . According to Kirchhoff's current law:

$I_A = I_B + I_C$

At the bottom of the circuit, the currents recombine, so:

$I_B + I_C = I_D$

lightbulbExample

Worked Example: Applying Kirchhoff's Current Law

Question: The current at point A in the circuit above is 2.0 A. Three times as much current passes through B as through C. Calculate the currents at B, C and D.

Solution:

Given information:

$I_A = 2.0$ A
$I_B = 3I_C$

Applying Kirchhoff's current law at the junction:

$I_A = I_B + I_C$

Substituting $I_B = 3I_C$ :

$I_A = 3I_C + I_C = 4I_C$

Solving for $I_C$ :

$I_C = \frac{I_A}{4} = \frac{2.0 \text{ A}}{4} = 0.5 \text{ A}$

Now finding $I_B$ :

$I_B = 3I_C = 3(0.5 \text{ A}) = 1.5 \text{ A}$

Check: $I_A = I_B + I_C = 1.5 + 0.5 = 2.0$ A ✓

At the second junction where B and C meet:

$I_D = I_B + I_C = I_A = 2.0 \text{ A}$

Answer: $I_C = 0.5$ A, $I_B = 1.5$ A, $I_D = 2.0$ A

Important principle: Current in series circuits

chatImportant

A crucial consequence of Kirchhoff's current law is that current is constant throughout a series circuit. Since there are no junctions in a single-loop circuit, the same current flows through every component. Current is not "used up" as it passes through resistors or other components - it remains constant around the entire loop.

Series and parallel circuits

There are two main types of circuit configurations: series circuits and parallel circuits. Understanding the difference is essential for circuit analysis.

Series circuits

A series circuit has only one path for current to flow through. All components are connected one after another, end to end, forming a single continuous loop.

Key characteristics of series circuits:

Same current flows through all components
Total voltage is divided among components
If one component fails, the entire circuit breaks

Resistors in series

When resistors are connected in series, we can calculate the total equivalent resistance. Starting with a circuit containing resistors $R_1$ and $R_2$ in series:

The total voltage across both resistors is:

$V_{total} = V_{R_1} + V_{R_2}$

Since the same current $I$ flows through both resistors, we can divide by $I$ :

$\frac{V_{total}}{I} = \frac{V_{R_1}}{I} + \frac{V_{R_2}}{I}$

Using Ohm's law ( $R = \frac{V}{I}$ ):

$R_{total} = R_1 + R_2$

For any number of resistors in series:

$R_{total} = R_1 + R_2 + R_3 + ... + R_n$

chatImportant

The total resistance of resistors in series is simply the sum of all individual resistances. Adding more resistors in series increases the total resistance and decreases the current.

lightbulbExample

Worked Example: Series Resistors

Question: A 6 V battery is connected to two resistors in series: $R_1 = 8$ Ω and $R_2 = 4$ Ω.

Calculate:

The total resistance
The current through the circuit
The potential difference across each resistor

Solution:

Part 1: Total resistance

$R_{total} = R_1 + R_2$

$R_{total} = 8 \text{ Ω} + 4 \text{ Ω} = 12 \text{ Ω}$

Part 2: Current through circuit

Using Ohm's law for the total circuit:

$R_{total} = \frac{V_{total}}{I}$

Rearranging:

$I = \frac{V_{total}}{R_{total}} = \frac{6 \text{ V}}{12 \text{ Ω}} = 0.5 \text{ A}$

Part 3: Potential difference across each resistor

For $R_1$ :

$V_1 = IR_1 = (0.5 \text{ A})(8 \text{ Ω}) = 4 \text{ V}$

For $R_2$ , we can use either Ohm's law or Kirchhoff's voltage law:

Using Kirchhoff's voltage law:

$V_{battery} = V_{R_1} + V_{R_2}$

$V_{R_2} = V_{battery} - V_{R_1} = 6 \text{ V} - 4 \text{ V} = 2 \text{ V}$

Answers: Total resistance = 12 Ω, Current = 0.5 A, $V_1 = 4$ V, $V_2 = 2$ V

Parallel circuits

A parallel circuit has multiple paths for current to flow through. Components are connected side by side, with multiple branches.

Key characteristics of parallel circuits:

Current divides among the different paths
Same voltage across all parallel components
If one component fails, current can still flow through other paths

Resistors in parallel

When resistors are connected in parallel, they all have the same potential difference across them. Using the circuit shown:

For a parallel circuit with two resistors:

Voltage is the same: $V_{in} = V_1 = V_2$
Total current divides: $I_T = I_1 + I_2$

Using Ohm's law for each resistor:

$I_1 = \frac{V_1}{R_1} \text{ and } I_2 = \frac{V_2}{R_2}$

For the total circuit:

$I_T = \frac{V_{total}}{R_{total}}$

Substituting into Kirchhoff's current law:

$\frac{V_{total}}{R_{total}} = \frac{V_1}{R_1} + \frac{V_2}{R_2}$

Since $V_{total} = V_1 = V_2$ in a parallel circuit:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$

For any number of resistors in parallel:

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}$

chatImportant

For parallel resistors, we add the reciprocals (1/R values). The total resistance is always less than the smallest individual resistor. Adding more resistors in parallel decreases total resistance and increases total current.

lightbulbExample

Worked Example: Parallel Resistors

Question: For the circuit shown below with two parallel resistors ( $R_1 = 9$ Ω and $R_2 = 18$ Ω) connected to a 9 V battery, calculate:

The total equivalent resistance
The current through the battery
The current through each resistor

Solution:

Part 1: Total resistance

$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_{total}} = \frac{1}{9 \text{ Ω}} + \frac{1}{18 \text{ Ω}}$

Finding a common denominator:

$\frac{1}{R_{total}} = \frac{2}{18 \text{ Ω}} + \frac{1}{18 \text{ Ω}} = \frac{3}{18 \text{ Ω}}$

$R_{total} = \frac{18 \text{ Ω}}{3} = 6 \text{ Ω}$

Part 2: Current through battery

$I = \frac{V_{battery}}{R_{total}} = \frac{9 \text{ V}}{6 \text{ Ω}} = 1.5 \text{ A}$

Part 3: Current through each resistor

For $R_1$ :

$I_1 = \frac{V_{battery}}{R_1} = \frac{9 \text{ V}}{9 \text{ Ω}} = 1 \text{ A}$

For $R_2$ , using Kirchhoff's current law:

$I_T = I_1 + I_2$

$I_2 = I_T - I_1 = 1.5 \text{ A} - 1 \text{ A} = 0.5 \text{ A}$

Answers: Total resistance = 6 Ω, Total current = 1.5 A, $I_1 = 1$ A, $I_2 = 0.5$ A

infoNote

Important observation: Notice that the total resistance (6 Ω) is less than either individual resistor (9 Ω or 18 Ω). This is always true for parallel resistors. Also, more current flows through the path with lower resistance ( $R_1$ ), demonstrating that current divides in inverse proportion to resistance.

Practical investigation: Series and parallel circuits

Understanding the behaviour of series and parallel circuits is enhanced through practical investigation. A typical investigation involves:

Aim: To investigate light globes and resistors connected in series and parallel

Part 1: Observing light globes

Compare the brightness of globes connected in:

Series configuration
Parallel configuration

Part 2: Measuring resistor properties

For resistors in both series and parallel configurations, measure:

Current through the circuit
Voltage across individual resistors
Calculate total resistance using Ohm's law

infoNote

Expected observations:

Configuration	Brightness/Current	Voltage	Total Resistance
Series	Dimmer/Lower	Divided among resistors	Sum of resistances (increases)
Parallel	Brighter/Higher	Same across all resistors	Less than smallest resistor (decreases)

Analysis: Plot graphs of current versus number of resistors to observe:

Series: Current decreases as more resistors are added
Parallel: Current increases as more resistors are added

bookmarkSummary

Key Points to Remember

Circuit diagrams use standardised symbols to show how components are connected, focusing on relationships rather than physical layout
Kirchhoff's voltage law: The sum of potential differences around any closed loop equals zero ( $\sum V = 0$ ) - based on conservation of energy
Kirchhoff's current law: Current entering a junction equals current leaving ( $\sum I_{in} = \sum I_{out}$ ) - based on conservation of charge
Series circuits have one path; current is the same throughout but voltage divides: $R_{total} = R_1 + R_2 + R_3 + ...$
Parallel circuits have multiple paths; voltage is the same across all branches but current divides: $\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...$
Current is never "used up" in a circuit - it flows continuously and is conserved

Electric Circuits (HSC SSCE Physics): Revision Notes

Circuits and Circuit Diagrams

What are circuit diagrams?

Circuit symbols

Drawing circuit diagrams

Example: Drawing a series circuit

Practice: Identifying circuit components

Kirchhoff's laws and conservation principles

Kirchhoff's voltage law - conservation of energy

Understanding voltage law with a series circuit

Kirchhoff's current law - conservation of charge

Understanding current law with a parallel circuit

Important principle: Current in series circuits

Series and parallel circuits

Series circuits

Resistors in series

Parallel circuits

Resistors in parallel

Practical investigation: Series and parallel circuits

Key Points to Remember

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