Series and Parallel Circuits
This topic is part of the HSC Physics course under the section Electric Circuits.
HSC Physics Syllabus
 investigate qualitatively and quantitatively series and parallel circuits to relate the flow of current through the individual components, the potential differences across those components and the rate of energy conversion by the components to the laws of conservation of charge and energy, by deriving the following relationships: (ACSPH038, ACSPH039, ACSPH044)
Series and Parallel Circuits for HSC Physics
In Year 11 HSC Physics, we delve into the intricacies of electric circuits, specifically focusing on series and parallel configurations. An electric circuit provides a path for electric current to flow. When components like resistors are connected in a circuit, they can be arranged in series, parallel, or a combination of both.
Electric Circuit Diagrams
We use simple diagrams to model components that make up an electric circuit. The diagrams below are common components you will encounter in HSC Physics.
There are two main ways to represent resistors of fixed values of electrical resistance. Both representations are equivalent and it is up to you to decide which one to use when drawing electric circuit diagrams.
Two common representations of fixed value resistors
A battery is a commonly used component to provide the voltage (potential difference) necessary for current to flow in a circuit. Conventional current flows from positive (higher) to negative (lower) potential. The direction of electron flow is the opposite to this.
A switch controls whether a circuit is open or closed. When the switch is closed, current can flow in the closed circuit. When the switch is open, current cannot open as the circuit is open.
Series Circuits
In a series circuit, components are connected endtoend, forming a single path for the current to flow. If one component fails, the circuit is broken, and the current stops flowing, much like a string of old Christmas lights.
In a series circuit, components are connected one after another, so there's only one path for the current to flow. Here's what happens to voltage and current:
 Current: Because there's only one path, the same current flows through every component. The current is constant throughout the circuit. If one light goes out in a string of oldfashioned Christmas lights wired in series, they all go out because the single pathway for current is interrupted.
 Voltage: The voltage supplied by the power source e.g. battery is divided among the components based on their resistance. The greater resistance a component has, the greater voltage drop across it. The sum of the voltage drops (V) across each component equals the total voltage supplied by the source (e.g. battery). This division of voltage is due to the energy being used to move charges through each resistor (read Kirchhoff's Voltage Law below).
 Resistance: Total resistance is the sum of individual resistances.
The resistance of a series circuit is described by:
$$R_{\text{series}} = R_1 + R_2 + \ldots + R_n$$
Kirchhoff's Voltage Law
Kirchhoff's Voltage Law states that the sum of all the electrical potential differences around any closed network (loop) is zero. This is because a charge going around a complete loop in a circuit must return to its original potential energy level by the time it completes the loop.
$$\sum V = 0$$
$$V_1 + V_2 + V_3 + V_4 =0$$
In other words, the work done on the charge by the battery (or any other source of voltage) is equal to the energy used by the charge as it passes through the rest of the circuit.
Consider a series circuit with a battery and several resistors. Here's how KVL applies:

Voltage Rise: When the charge passes through the battery, it gains electrical potential energy, which is represented as a positive voltage (let's call this `V_{\text{battery}}`).

Voltage Drops: As the charge moves through each resistor, it loses some of its potential energy, which is converted into other forms of energy, like heat due to resistance. Each of these energy losses corresponds to a voltage drop (let's call these `V_1`, `V_2`... `V_n`).

Energy Conservation: By the time the charge returns to the negative terminal of the battery, it must have used up all the potential energy it gained from the battery, so the gain in energy (voltage rise) is equal to the total loss of energy (sum of voltage drops).
This gives us the equation:
$$V_{\text{battery}}  V_1  V_2  \ldots  V_n = 0$$
$$\sum V = 0$$
Parallel Circuits
In a parallel circuit, components are connected across common points or junctions, providing multiple paths for the current to flow. If one component fails, the rest can still operate.
In a parallel circuit, components are connected across the same two points, creating multiple paths for the current to flow. Here's how voltage and current behave:

Current: The total current flowing from the source is the sum of the currents flowing through each parallel branch. Each component draws the current it needs, and the total current is the sum of these currents (read Kirchhoff's Current Law below).

Voltage: In parallel, each component experiences the same voltage. Unlike in series circuits, the voltage across each branch is equal to the full voltage of the source. This is like how the water pressure at the start of each branch of a river is the same, regardless of how many branches there are.
 Resistance: Total resistance is found by adding the reciprocals of each resistance and taking the reciprocal of the sum.
The relationships in a parallel circuit are given by:
$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2} + \ldots + \frac{1}{R_n}$$
Kirchhoff's Current Law
Kirchhoff's current law states that the total current entering a junction (or node) in an electric circuit must equal the total current leaving the junction. This law is based on the principle of charge conservation; charge can neither be created nor destroyed. Therefore, whatever charge flows into the junction must flow out.
Consider a junction in a circuit where several currents converge:

Incoming Currents: These are the currents flowing towards the junction. Let's denote them as `I_1, I_2 \ldots, I_n`.

Outgoing Currents: These are the currents flowing away from the junction. We'll denote them as `I_a, I_b \ldots, I_m`.

Charge Conservation: At the junction, the charges are not stored or lost; they simply take different paths. Thus, the total charge per unit time (current) flowing into the junction must be equal to the total charge per unit time flowing out.
This gives us the equation:
$$\sum I_{in} = \sum I_{out}$$
$$I_1 + I_2 + \ldots + I_n = I_{a} + I_{b} + \ldots + I_{m}$$
$$$\sum I = 0$$$
Ammeters and Voltmeters
To measure current and voltage in circuits, we use ammeters and voltmeters, respectively.
Ammeters
 Purpose: To measure the current flowing through a circuit.
 Connection: Connected in series with the component whose current you wish to measure.
 Ideal characteristic: Should have very low resistance to minimise any alteration to the circuit's behaviour. If an ammeter has a very high resistance, it would cause a significant voltage drop, drawing energy from the circuit.
Voltmeters
 Purpose: To measure the potential difference (voltage) across two points in a circuit or a component e.g. resistor.
 Connection: Connected in parallel with the component.
 Ideal characteristic: Should have very high resistance to draw negligible current from the circuit. If a voltmeter has a very low resistance, this would cause some amount of current to be redirected from the main circuit to flow through the voltmeter.
Circuit Example 1
(a) Calculate the current in the circuit.
(b) Calculate the potential difference across the second resistor, R_{2}.
Solution to part (a):
The total resistance of a series circuit is the sum of resistance of individual components:
$$R = 5 + 15$$
$$R = 20 \, \Omega$$
Using Ohm's law:
$$V = IR$$
$$9 = I \times 20$$
$$I = 0.45 \text{ A}$$
Solution to part (b):
The potential difference across the resistor can be calculated using Ohm's law:
$$V = IR$$
$$V = 0.45 \times 15$$
$$V = 6.75 \text{ A}$$
Circuit Example 2
(a) Calculate the total resistance of the circuit.
(b) What is the magnitude of current flowing through R_{1}?
Solution to part (a):
Resistance of a parallel circuit is given by:
$$\frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_2}$$
$$\frac{1}{R_{\text{parallel}}} = \frac{1}{2} + \frac{1}{3}$$
$$\frac{1}{R_{\text{parallel}}} = \frac{5}{6}$$
$$R_{\text{parallel}} = 1.2 \, \Omega$$
Solution to part (b):
The voltage across resistors connected in parallel is equal, so the potential difference across both resistors is 1.5 V each.
Using Ohm's law:
$$V = IR$$
$$1.5 = I \times 2$$
$$I = 0.75 \text{ A}$$