Know how to construct and use series and parallel circuits

Published by Patrick Mutisya · 14 days ago

Cambridge IGCSE Physics 0625 – 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, applying the appropriate formulas for voltage, current and resistance.

Key Concepts

  • Series circuit: components are connected end‑to‑end so the same current flows through each.
  • Parallel circuit: components are connected across the same two points, giving each its own path for current.
  • Understanding how total (equivalent) resistance, voltage distribution and current distribution differ between the two configurations.

Constructing a Simple Circuit

  1. Identify the power source (battery or supply).
  2. Choose the required components (resistors, bulbs, switches, etc.).
  3. Connect the components using conducting wires, ensuring good contacts.
  4. For safety, include a switch or a fuse where appropriate.

Series Circuits

In a series circuit the same current \$I\$ flows through every component. The total resistance \$R_{\text{total}}\$ is the sum of the individual resistances.

\$\$

R{\text{total}} = R1 + R2 + R3 + \dots + R_n

\$\$

Using Ohm’s law, the total voltage supplied \$V_{\text{s}}\$ is divided among the components:

\$\$

V{\text{s}} = I R{\text{total}} = I(R1 + R2 + \dots + R_n)

\$\$

Each component’s voltage drop \$Vk\$ can be found from \$Vk = I R_k\$.

Characteristics of Series Circuits

  • Current is the same through all components.
  • Voltage across each component may differ.
  • If one component fails (opens), the whole circuit stops working.
  • Total resistance is always greater than any individual resistance.

Suggested diagram: A simple series circuit with a battery, two resistors \$R1\$ and \$R2\$, and a switch.

Parallel Circuits

In a parallel circuit each component is connected across the same two points, so the voltage across each component is the same as the source voltage \$V_{\text{s}}\$.

The total current supplied by the source is the sum of the branch currents:

\$\$

I{\text{total}} = I1 + I2 + I3 + \dots + I_n

\$\$

Each branch current can be expressed using Ohm’s law \$Ik = \dfrac{V{\text{s}}}{R_k}\$.

The equivalent resistance \$R_{\text{eq}}\$ of \$n\$ parallel resistors is given by:

\$\$

\frac{1}{R{\text{eq}}} = \frac{1}{R1} + \frac{1}{R2} + \dots + \frac{1}{Rn}

\$\$

Characteristics of Parallel Circuits

  • Voltage across each branch equals the source voltage.
  • Current divides between branches according to their resistances.
  • If one branch fails (opens), the other branches continue to operate.
  • Total resistance is always less than the smallest individual resistance.

Suggested diagram: A simple parallel circuit with a battery and two resistors \$R1\$ and \$R2\$ connected in parallel.

Comparison of Series and Parallel Circuits

FeatureSeriesParallel
Current through each componentSame throughoutDifferent; divides according to resistance
Voltage across each componentMay differ; sum equals source voltageSame as source voltage for each branch
Total resistance\$R{\text{total}} = \sum Rk\$ (greater than any \$R_k\$)\$\displaystyle\frac{1}{R{\text{eq}}}= \sum \frac{1}{Rk}\$ (less than any \$R_k\$)
Effect of a broken componentAll components stop workingOnly the broken branch stops; others continue

Worked Example – Series Circuit

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

  1. Calculate total resistance:

    \$R_{\text{total}} = 2 + 3 + 5 = 10\ \Omega\$

  2. Find the current using Ohm’s law:

    \$I = \frac{V{\text{s}}}{R{\text{total}}} = \frac{12}{10} = 1.2\ \text{A}\$

  3. Voltage across each resistor:

    • \$V1 = I R1 = 1.2 \times 2 = 2.4\ \text{V}\$
    • \$V2 = I R2 = 1.2 \times 3 = 3.6\ \text{V}\$
    • \$V3 = I R3 = 1.2 \times 5 = 6.0\ \text{V}\$

Worked Example – Parallel Circuit

Two resistors \$R1 = 4\ \Omega\$ and \$R2 = 6\ \Omega\$ are connected in parallel across a \$12\ \text{V}\$ supply. Determine the total current supplied and the current through each resistor.

  1. Find equivalent resistance:

    \$\frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}\$

    \$R_{\text{eq}} = \frac{12}{5} = 2.4\ \Omega\$

  2. Total current:

    \$I{\text{total}} = \frac{V{\text{s}}}{R_{\text{eq}}} = \frac{12}{2.4} = 5\ \text{A}\$

  3. Branch currents:

    • \$I1 = \dfrac{V{\text{s}}}{R_1} = \dfrac{12}{4} = 3\ \text{A}\$
    • \$I2 = \dfrac{V{\text{s}}}{R_2} = \dfrac{12}{6} = 2\ \text{A}\$

Common Mistakes to Avoid

  • Assuming the current is the same in parallel branches – it is the voltage that is common.
  • Adding resistances in parallel as if they were in series.
  • Forgetting that a broken component in series stops the whole circuit.
  • Mixing up total voltage with individual voltage drops in parallel circuits.

Practice Questions

  1. A series circuit contains a \$9\ \text{V}\$ battery and three resistors of \$1\ \Omega\$, \$2\ \Omega\$, and \$3\ \Omega\$. Calculate the current and the voltage across each resistor.
  2. Two resistors, \$R1 = 10\ \Omega\$ and \$R2 = 15\ \Omega\$, are connected in parallel to a \$6\ \text{V}\$ supply. Find the total resistance, the total current, and the current through each resistor.
  3. In a mixed circuit, a \$12\ \text{V}\$ battery supplies a series combination of \$R1 = 4\ \Omega\$ and a parallel branch consisting of \$R2 = 6\ \Omega\$ and \$R_3 = 12\ \Omega\$. Determine the current through each resistor and the voltage across each branch.

Summary

Understanding series and parallel circuits is fundamental for analysing real‑world electrical systems. Remember:

  • Series: same current, voltage divides, resistances add.
  • Parallel: same voltage, current divides, reciprocal addition of resistances.
  • Use Ohm’s law and the appropriate formulas to calculate unknown quantities.