Ch 7 · 4.3.2 Series and Parallel Circuits · O Level Physics 5054
Topic 7 of 17 in Electricity and Magnetism — O Level Physics 5054 — 38 lessons.
Introduction
Old fairy lights went out all at once when one bulb broke: they were wired in series. The lights in your home work independently: they are in parallel. The two arrangements share current, p.d. and resistance in different ways, and four simple rules let you work out everything. (Syllabus section 4.3.2.)
Key ideas — Series circuits: one path
Components connected end to end, in a single loop.
🧑🏫 Teacher's explanation
Imagine a single-lane road where cars are traveling from one end to the other. In a series circuit, there is only one path for the 'cars' (electrons) to follow. Because there are no side roads or exits, every single car that enters the circuit must pass through every single component. This means the flow rate, or current, is identical at every single point along that loop. If 5 cars pass a tree at the start, 5 cars must pass a tree at the end.
📖 Definitions
🔍 Worked example — Easy
A circuit consists of a 12 V battery connected in series with two resistors, R_1 = 4 Omega and R_2 = 2 Omega. Calculate the total resistance and the current in the circuit.
🔍 Worked example — Standard
Three resistors of 5 Omega, 10 Omega, and 15 Omega are connected in series to a 60 V power supply. Calculate the potential difference across the 10 Omega resistor.
🔍 Worked example — Exam-style
A 24 V battery is connected to two resistors in series. The current in the circuit is 1.5 A. One resistor is 8 Omega. Calculate the resistance of the second resistor.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Parallel circuits: more than one path
Components connected side by side, each on its own branch.
🧑🏫 Teacher's explanation
Imagine a river that splits into two smaller streams around an island and then merges back together. The total amount of water flowing through the river per second (the current) must be the same before and after the split. If 5 liters per second enter the split, and 3 liters go through the left stream, exactly 2 liters must go through the right stream. This is the principle of conservation of charge at a junction.
📖 Definitions
🔍 Worked example — Easy
Two resistors, R_1 = 6 Omega and R_2 = 3 Omega, are connected in parallel. Calculate the total resistance R_total.
🔍 Worked example — Standard
A 12 V battery is connected across two parallel resistors, R_1 = 4 Omega and R_2 = 6 Omega. Calculate the total current I_total leaving the battery.
🔍 Worked example — Exam-style
In a parallel circuit, a 10 V source is connected to three branches. Branch A has a current of 0.5 A, Branch B has a current of 1.2 A, and Branch C has a current of 0.3 A. Calculate the total current I_total and the total resistance R_total.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — A method for circuit calculations
🧑🏫 Teacher's explanation
Think of electricity like water flowing through pipes. In a single loop (a series circuit), the water has only one path; therefore, the amount of water passing through any point must be the same. If you squeeze the pipe (add resistance), the flow slows down everywhere.
📖 Definitions
🔍 Worked example — Easy
A circuit consists of a 12 V battery connected in series to two resistors, R_1 = 4 Omega and R_2 = 2 Omega. Calculate the total current in the circuit.
🔍 Worked example — Standard
Two resistors, R_1 = 6 Omega and R_2 = 3 Omega, are connected in parallel across a 12 V supply. Calculate the total resistance of the arrangement.
🔍 Worked example — Exam-style
A 6 V battery is connected to a 2 Omega resistor in series with a parallel combination of two 4 Omega resistors. Calculate the total current drawn from the battery.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the total resistance of the circuit. (b) Calculate the total current flowing from the battery. [4 marks]
Worked example
Example 1: series. A 6.0 Ω and a 4.0 Ω resistor are in series with a 12 V supply. Find the current and the p.d. across each resistor.
🔍 Worked example — Easy
A 6.0 Omega and a 4.0 Omega resistor are connected in series with a 12 V supply. Calculate the total resistance, the circuit current, and the potential difference across each resistor.
🔍 Worked example — Exam-style
A 4.0 Omega resistor is in series with a parallel arrangement of a 12 Omega and a 6.0 Omega resistor. The total supply voltage is 12 V. Calculate the total current in the circuit.
⚠️ Common errors
Common mistakes
Practice
Answers: (1) 10 Ω. (2) 10 × 40/10 + 40 = 8.0\ Omega. (3) 2.5 - 0.9 = 1.6 A. (4) I = 9.0/18 = 0.50 A; V = 0.50 × 12 = 6.0 V. (5) It provides an extra path for the current, so more total current flows for the same p.d.; a larger current for the same p.d. means a smaller resistance.
Summary
All chapters in Electricity and Magnetism
- 4.1 Simple Magnetism and Magnetic Fields
- 4.2.1 Electrical Charge
- 4.2.2 Electrical Current
- 4.2.3 Electromotive Force and Potential Difference
- 4.2.4 Resistance
- 4.3.1 Circuit Diagrams and Circuit Components
- 4.3.2 Series and Parallel Circuits
- 4.3.3 Action and Use of Circuit Components
- 4.4.1 Uses of Electricity
- 4.4.2 Electrical Safety
- 4.5.1 Electromagnetic Induction
- 4.5.2 The a.c. Generator
- 4.5.3 Magnetic Effect of a Current
- 4.5.4 Forces on a Current-Carrying Conductor
- 4.5.5 The d.c. Motor
- 4.5.6 The Transformer
- 4.6 Uses of an Oscilloscope