Year 11 PhysicsModule 4⏱ ~35 min5 MC · 3 Short AnswerLesson 7 of 17
Parallel Circuits
Sydney Harbour Bridge LED upgrade (2018): 228 LED floodlights (50 W each, 240 V each) wired in parallel, total power 228 × 50 = 11,400 W. The previous sodium floodlights consumed 228 × 400 = 91,200 W, an 87.5% reduction. Because the LEDs are in parallel, each receives the full 240 V and operates independently: one failing does not affect the other 227.
Today's hook: In 2018, Sydney Harbour Bridge replaced its old sodium floodlights with 228 LED floodlights, each 50 W, each running at 240 V, all wired in parallel. Total power dropped from 91,200 W to 11,400 W, an 87.5% energy saving. Because each LED is on its own parallel branch, one failing does not affect the others. Why does adding more parallel branches decrease total resistance rather than increase it?
0/5TASKS
1
You’re here
Warm up first
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Before you read, predict
Two identical 10 Ω resistors are connected in parallel to a 12 V battery.
Predict 1: What is the voltage across each resistor?
Predict 2: Is the total resistance 20 Ω, 10 Ω, or 5 Ω? Explain your guess.
In a parallel circuit with two branches, the voltage across each branch is:
Total resistance is always less than the smallest branch resistance
Understand
Why adding branches decreases total resistance
Why voltage is the same across all parallel branches
Current divides in inverse proportion to resistance
Can Do
Calculate total resistance in a parallel circuit
Find branch currents from the supply voltage
Predict effects of adding or removing a parallel branch
3
You’re here
Key Terms
Key Terms
Parallel circuitA circuit in which components are connected side-by-side, providing multiple paths for current to flow.
Branch currentThe current through an individual branch of a parallel circuit. $I_n = V/R_n$.
Current dividerIn parallel, current divides between branches in inverse proportion to their resistance: large $R$ → small $I$.
Junction ruleAt any junction, total current in = total current out (Kirchhoff's Current Law; conservation of charge).
Cross-lesson links: L06 covered series circuits (same current, voltages add, R_total increases). L07 is the mirror image, parallel circuits (same voltage, currents add, R_total decreases). The 2018 Sydney Harbour Bridge upgrade is a real example: 228 LEDs in parallel, each at 240 V, each independent.
Misconceptions to fix
✗ Wrong: Adding more parallel branches makes total resistance larger.
✓ Right: Each new branch provides an additional current path, reducing total resistance. $R_{\text{total}}$ is always less than the smallest branch resistance.
✗ Wrong: In a parallel circuit, the current through each branch is the same.
✓ Right: Each branch has the same voltage ($V_{\text{supply}}$) but different currents. $I_n = V/R_n$, more resistance means less current in that branch.
4
You’re here
Parallel Circuit Rules
Core Content
1
Parallel Circuit Rules
+5 XP
Plug one lamp into a power board, then plug a second lamp into the same power board, the first lamp does not dim. Plug in a third, a fourth, each lamp glows at full brightness regardless of how many are connected. Each lamp sees the full 230 V because each is connected directly across both supply rails, independently. The current from the wall goes up with each lamp added, but the voltage across every lamp stays the same. Three rules describe all parallel circuits.
Parallel circuit rules
1. Same voltage: $V_1 = V_2 = V_3 = V_{\text{supply}}$
In a parallel circuit: voltage is the same across every branch ($V_1 = V_2 = V_{\text{supply}}$); total current equals the sum of branch currents; $1/R_{\text{total}} = 1/R_1 + 1/R_2 + \ldots$, so total resistance is always less than any single branch. Shortcut for two resistors: $R = R_1 R_2/(R_1+R_2)$.
Add the highlighted rules to your notes before the check below.
In a parallel circuit, the voltage across each branch equals the supply voltage.
Adding a resistor in parallel always increases total resistance.
5
You’re here
Activity 1, Parallel Circuit Calculations
Activity 1, Parallel Circuit Calculations
ApplyBand 3
Two resistors, $R_1 = 8\ \Omega$ and $R_2 = 24\ \Omega$, are connected in parallel to a 24 V battery.
Calculate the total resistance using the product-over-sum rule.
Calculate the current through each branch.
Calculate the total current drawn from the battery and verify using $I = V/R_{\text{total}}$.
Two 20 Ω resistors are connected in parallel. The total resistance is:
6
You’re here
Activity 2, Current Divider Analysis
Activity 2, Current Divider Analysis
AnalyseBand 4
Three resistors, 5 Ω, 10 Ω, and 20 Ω, are connected in parallel to a 20 V supply.
Calculate the total resistance and total current.
Calculate the current through each resistor.
Show that the currents add to the total, and explain which branch carries the most current and why.
Three of these are correct parallel circuit rules. Pick the odd one out.
In a parallel circuit, as more branches are added, the total resistance _____.
In a parallel circuit, the branch with the smallest resistance carries:
Drive It: Add a Lamp and Watch the Total Resistance Fall
Two beliefs survive being told they are wrong, because in the simplest cases they still predict the right answer: that current is used up as it goes round a loop, and that adding any component must raise the total resistance. The bench below lets you set both of them up and watch them break.
Interactive · Series and parallel circuit bench
Try this: leave it on Series with two lamps and move the ammeter to each of the three probe points in turn — the reading does not change. Now press Parallel and watch the total resistance fall from 12.00 Ω to 3.000 Ω. Finally set Lamp 1 to 2 Ω and Lamp 2 to 10 Ω in series, where both carry exactly the same current, and look at which one is brighter.
On the bench set Series, two lamps, 12 V, with Lamp 1 at 2 Ω and Lamp 2 at 10 Ω. The ammeter reads 1.000 A wherever you put it. Which lamp is drawn brighter, and why?
Run the bench: one 6 Ω lamp on 12 V, then add a second identical lamp, once in series and once in parallel. Judge each statement against what the meters actually read.
Adding the second lamp in parallel drops the total resistance from 6.000 Ω to 3.000 Ω, so the supply current doubles to 4.000 A.
Adding that second lamp in parallel halves the current through the first lamp, just as adding it in series does.
✓
Show what you have learned
Quick recall, parallel circuits
+5 XP
Short Answer, 10 marks
+5 XP
ApplyBand 3(3 marks) 3. Two resistors, $R_1 = 12\ \Omega$ and $R_2 = 6\ \Omega$, are connected in parallel to an 18 V supply. Calculate (a) the total resistance, (b) the current through each branch, and (c) the total current from the battery.
AnalyseBand 4(3 marks) 4. A 30 Ω resistor and a 60 Ω resistor are connected in parallel to a 30 V supply. (a) Calculate the total resistance. (b) Show that the total current equals $I_1 + I_2$. (c) If the 30 Ω branch is disconnected, what happens to the voltage across the 60 Ω resistor? Explain.
EvaluateBand 6(4 marks) 5. Your home's mains circuits are wired in parallel, not series. Using principles of conservation of charge and conservation of energy, evaluate why parallel wiring is essential for household circuits. Include reference to voltage, current, and fault tolerance in your response.
Q4 (3 marks): (a) $R_{\text{total}} = (30 \times 60)/(30 + 60) = 1800/90 = 20\ \Omega$. (b) $I_1 = 30/30 = 1.0\ \text{A}$; $I_2 = 30/60 = 0.50\ \text{A}$; sum = 1.5 A. Check: $30/20 = 1.5\ \text{A}$ ✓. (c) Voltage stays at 30 V, the battery still provides the same EMF; the 60 Ω branch is directly connected to the battery terminals and remains at 30 V regardless of the other branch.
Q5 (4 marks): Parallel wiring is essential for three reasons. (1) Voltage, every appliance is connected directly between the two supply rails, receiving the full 230 V needed for correct operation; in series, voltage would be divided, reducing appliance brightness/power. (2) Current, each appliance draws only the current it needs ($I = V/R$); the mains fuse carries total current but appliances are independent; in series, all appliances would share the same current, causing incompatibility. (3) Fault tolerance, if one appliance or circuit fails open-circuit, other circuits remain unaffected (conservation of charge still satisfied on each branch independently); in series, one fault would cut all devices. Parallel circuits therefore satisfy both conservation of charge (current divides at junctions) and conservation of energy (same voltage = same energy per coulomb for each branch).
✓
Retrieve, reflect and finish
Check what actually stuck
Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
The 2018 Sydney Harbour Bridge LED upgrade wired 228 floodlights (50 W, 240 V each) in parallel, total power 11,400 W vs the old 91,200 W, an 87.5% saving. Every floodlight received the full 240 V because each was connected directly across the supply rails. One failing left the other 227 working, the defining advantage of a parallel connection over a series one.
Now look back at your Think First answers: both 10 Ω resistors receive the full 12 V (same voltage across every branch). Total resistance = (10×10)÷(10+10) = 5 Ω, less than either branch, not 20 Ω or 10 Ω.
