Know how to construct and use series and parallel circuits

4.3.2 Series and Parallel Circuits

Learning Objective

Students will be able to construct, analyse and use series and parallel circuits, apply the correct formulas for voltage, current and resistance, and interpret circuit diagrams using the standard symbols prescribed by the Cambridge IGCSE Physics (0625) syllabus.

Key Concepts (Cambridge IGCSE Physics 0625)

  • Series circuit: components are joined end‑to‑end. The same current flows through every component.

  • Parallel circuit: components are joined across the same two points. Each branch experiences the full source voltage.

  • Junction (Kirchhoff’s) rule: the sum of currents entering a junction equals the sum leaving it

    \(\displaystyle \sum I{\text{in}} = \sum I{\text{out}}\).

  • Voltage‑division rule (series):

    \(\displaystyle Vk = Vs\frac{R_k}{\sum R}\).

  • Current‑division insight (parallel): the source current equals the sum of the branch currents, therefore it is larger than the current in any single branch.

    \(\displaystyle I{\text{source}} = I1 + I2 + \dots + In\).

  • In a parallel network the equivalent resistance is always less than the smallest individual resistance.

    \(\displaystyle \frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\)

  • Advantages of wiring lamps in parallel: each lamp receives the full supply voltage, brightness is independent of the number of lamps, and a failed lamp does not affect the others.

Standard Circuit Symbols (quick reference)

SymbolComponent
⚡Battery / cell (EMF E, internal resistance r)
—⏚—Switch (open / closed)
⏚Fuse (rated current)
ΩFixed resistor
↔︎Variable resistor / potentiometer
💡Lamp (bulb)
—‖—Ideal conducting wire

Constructing a Simple Circuit (practical checklist)

  1. Identify the power source (battery or DC supply). Record its EMF \(E\) and, if required, its internal resistance \(r\).

  2. Select the components needed (resistors, lamps, switches, fuses, potentiometers, etc.).

  3. Draw a correct schematic using the symbols above.

  4. Connect the components with clean, secure wires.

  5. Incorporate safety devices:

    • Switch to open the circuit before any modification.

    • Fuse rated for the expected maximum current.

    • Insulated wires and firm mounting.

  6. Check continuity and correct polarity with a multimeter before applying full supply.

Series Circuits

Fundamental Relations

  • Current is the same through every component:

    \(\displaystyle I{\text{total}} = I1 = I2 = \dots = In\).

  • Total resistance (ideal wires):

    \(\displaystyle R{\text{total}} = R1 + R2 + \dots + Rn\).

  • Sum of voltage drops equals the source voltage:

    \(\displaystyle Vs = V1 + V2 + \dots + Vn\).

  • Voltage‑division rule:

    \(\displaystyle Vk = Vs\frac{R_k}{\sum R}\).

  • Ohm’s law for the whole circuit:

    \(\displaystyle Vs = I{\text{total}}R_{\text{total}}\).

Characteristics

  • Same current through all components.

  • Voltage across each component may differ; the algebraic sum equals the source voltage.

  • If any component opens, the entire circuit stops.

  • Total resistance is always greater than the largest individual resistance.

Typical Use

Series connections are chosen when a single current must pass through several devices (e.g., a string of Christmas lights designed to operate on the same current).

Worked Example – Voltage Division

Three resistors \(R1 = 2\;\Omega\), \(R2 = 3\;\Omega\) and \(R_3 = 5\;\Omega\) are connected in series across a 12 V battery. Find the voltage across each resistor using the voltage‑division rule.

  1. Sum of resistances: \(\displaystyle \sum R = 2+3+5 = 10\;\Omega\).

  2. Apply the rule:

    • \(V_1 = 12\;\text{V}\times\frac{2}{10}=2.4\;\text{V}\)

    • \(V_2 = 12\;\text{V}\times\frac{3}{10}=3.6\;\text{V}\)

    • \(V_3 = 12\;\text{V}\times\frac{5}{10}=6.0\;\text{V}\)

  3. Check: \(2.4+3.6+6.0 = 12\;\text{V}\) – the rule works.

Diagram (example)

Simple series circuit: ⚡ → —⏚— → Ω (R₁) → Ω (R₂) → 💡 → back to ⚡.

Parallel Circuits

Fundamental Relations

  • Voltage across every branch equals the source voltage:

    \(\displaystyle V1 = V2 = \dots = Vn = Vs\).

  • Branch currents follow Ohm’s law:

    \(\displaystyle Ik = \frac{Vs}{R_k}\).

  • Total current supplied by the source (junction rule):

    \(\displaystyle I{\text{total}} = I1 + I2 + \dots + In\).

  • Equivalent resistance of \(n\) parallel resistors:

    \(\displaystyle \frac{1}{R{\text{eq}}}= \frac{1}{R1}+ \frac{1}{R_2}+ \dots\)

    (always less than the smallest individual resistor).

Characteristics

  • Same voltage across each branch; currents divide according to resistance.

  • If one branch opens, the remaining branches continue to operate.

  • Total resistance is always less than the smallest individual resistance.

  • Ideal for lighting: each lamp receives the full supply voltage, so brightness is independent of the number of lamps.

Diagram (example)

Simple parallel circuit: ⚡ → split into two branches (Ω R₁ and Ω R₂) → recombine → back to ⚡.

Combining EMF of Cells in Series

When ideal cells are connected end‑to‑end, both their EMFs and internal resistances add:

\[

E{\text{total}} = E1 + E2 + \dots + En,\qquad

r{\text{total}} = r1 + r2 + \dots + rn

\]

Example: two 1.5 V AA cells each with \(r = 0.2\;\Omega\) give \(E{\text{total}} = 3.0\;\text{V}\) and \(r{\text{total}} = 0.4\;\Omega\).

Potential Divider (Series‑Voltage Division)

A pair of series resistors can provide a stable lower voltage from a higher supply:

\[

V{\text{out}} = Vs\frac{R2}{R1+R_2}

\]

This principle underlies many sensor circuits and adjustable supplies that use a potentiometer.

Comparison of Series and Parallel Circuits

FeatureSeriesParallel
Quantity common to all componentsCurrent (same everywhere)Voltage (same across each branch)
Voltage across each componentMay differ; sum = source voltageEqual to source voltage for every branch
Total resistance\(R{\text{total}} = \sum Rk\) (greater than any individual R)\(\displaystyle \frac{1}{R{\text{eq}}}= \sum \frac{1}{Rk}\) (less than the smallest R)
Effect of a broken componentAll components stop workingOnly the broken branch stops; others continue
Source current vs. branch current (parallel)Not applicableSource current = sum of branch currents → always larger than any single branch current
Typical advantageEnsures the same current through all devicesEach lamp/device receives full voltage; independent operation

Worked Examples

Example 1 – Series Circuit

Three resistors \(R1 = 2\;\Omega\), \(R2 = 3\;\Omega\), \(R_3 = 5\;\Omega\) are connected in series across a 12 V battery. Find the current and the voltage across each resistor.

  1. Total resistance: \(R_{\text{total}} = 2+3+5 = 10\;\Omega\).

  3. k = I Rk\)):

    • \(V_1 = 1.2 \times 2 = 2.4\;\text{V}\)

    • \(V_2 = 1.2 \times 3 = 3.6\;\text{V}\)

    • \(V_3 = 1.2 \times 5 = 6.0\;\text{V}\)

    (Check: \(2.4+3.6+6.0 = 12\;\text{V}\).)

Example 2 – Parallel Circuit

Two resistors \(R1 = 4\;\Omega\) and \(R2 = 6\;\Omega\) are connected in parallel across a 12 V supply.

  1. Equivalent resistance:

    \[

    \frac{1}{R{\text{eq}}}= \frac{1}{4}+\frac{1}{6}= \frac{5}{12}\;\Rightarrow\;R{\text{eq}}= \frac{12}{5}=2.4\;\Omega

    \]

  2. Total current: \(I_{\text{total}} = \dfrac{12}{2.4}=5\;\text{A}\).

    • \(I_1 = \dfrac{12}{4}=3\;\text{A}\)

    • \(I_2 = \dfrac{12}{6}=2\;\text{A}\)

    (Check: \(3+2 = 5\;\text{A}\).)

Example 3 – Mixed (Series‑Parallel) Circuit

A 12 V battery supplies a series combination of \(R1 = 4\;\Omega\) and a parallel branch containing \(R2 = 6\;\Omega\) and \(R_3 = 12\;\Omega\).

  1. Parallel‑branch equivalent:

    \[

    \frac{1}{R{23}} = \frac{1}{6}+\frac{1}{12}= \frac{3}{12}= \frac{1}{4}\;\Rightarrow\;R{23}=4\;\Omega

    \]

  2. Total resistance: \(R{\text{total}} = R1 + R_{23}=4+4=8\;\Omega\).

  4. 1\) (same as total): \(I1 = 1.5\;\text{A}\).

    Voltage across \(R1\): \(V1 = I1R1 = 1.5 \times 4 = 6\;\text{V}\).

  5. {23}= Vs - V_1 = 12 - 6 = 6\;\text{V}\).

    • \(I_2 = \dfrac{6}{6}=1.0\;\text{A}\)

    • \(I_3 = \dfrac{6}{12}=0.5\;\text{A}\)

    (Check: \(I2+I3 = 1.5\;\text{A}=I_{\text{total}}\).)

Example 4 – Cells in Series

Two 1.5 V cells each have an internal resistance of 0.2 Ω. They are connected in series and power a 6 Ω lamp.

  • Total EMF: \(E_{\text{total}} = 1.5 + 1.5 = 3.0\;\text{V}\).

  • \[

    I = \frac{E{\text{total}}}{R{\text{lamp}} + r_{\text{total}}}= \frac{3.0}{6+0.4}=0.46\;\text{A}

    \]

Common Mistakes to Avoid

  • Mixing up what is common: current is common in series; voltage is common in parallel.

  • Adding parallel resistances as if they were in series.

  • Forgetting that the sum of voltage drops in a series circuit must equal the source voltage.

  • Neglecting the junction rule when analysing parallel networks.

  • Assuming a broken component in a parallel circuit stops the whole circuit.

Practice Questions

  1. A series circuit contains a 9 V battery and three resistors of 1 Ω, 2 Ω and 3 Ω. Calculate:

    • the current through the circuit,

    • the voltage across each resistor, and

    • verify that the sum of the drops equals 9 V.

  2. Two resistors, \(R1 = 10\;\Omega\) and \(R2 = 15\;\Omega\), are connected in parallel to a 6 V supply. Determine:

    • the equivalent resistance,

    • the total current drawn from the supply, and

    • the current through each resistor.

  3. In a mixed circuit a 12 V battery supplies a series combination of \(R1 = 4\;\Omega\) and a parallel branch consisting of \(R2 = 6\;\Omega\) and \(R_3 = 12\;\Omega\). Find:

    • the total current from the battery,

    • the voltage across each resistor, and

    • the current in each branch.

  4. Two 1.5 V cells each with internal resistance 0.2 Ω are connected in series and used to light a 6 Ω lamp. Calculate the current supplied and the terminal voltage across the lamp.