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.
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:
Know
- Parallel circuit rules: same voltage, currents add, $1/R_{\text{total}} = \sum 1/R_n$
- $V_1 = V_2 = V_3 = V_{\text{supply}}$
- $I_{\text{total}} = I_1 + I_2 + I_3$
- 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
Core Content
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.
1. Same voltage: $V_1 = V_2 = V_3 = V_{\text{supply}}$
2. Currents add: $I_{\text{total}} = I_1 + I_2 + I_3 + \ldots$
3. Reciprocal resistance: $\dfrac{1}{R_{\text{total}}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \ldots$
Two-resistor shortcut
$R_{\text{total}} = \dfrac{R_1 \times R_2}{R_1 + R_2}$ (product over sum)
$R_1 = 6\ \Omega$, $R_2 = 12\ \Omega$, $R_3 = 4\ \Omega$ in parallel, $V = 12\ \text{V}$. Find $R_{\text{total}}$, $I_1$, $I_2$, $I_3$, $I_{\text{total}}$.
- $\dfrac{1}{R_{\text{total}}} = \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{4} = \dfrac{2}{12} + \dfrac{1}{12} + \dfrac{3}{12} = \dfrac{6}{12} = \dfrac{1}{2}$
- $R_{\text{total}} = 2\ \Omega$
- $I_1 = 12/6 = 2.0\ \text{A}$; $I_2 = 12/12 = 1.0\ \text{A}$; $I_3 = 12/4 = 3.0\ \text{A}$
- $I_{\text{total}} = 2.0 + 1.0 + 3.0 = 6.0\ \text{A}$
- Check: $I_{\text{total}} = V/R_{\text{total}} = 12/2 = 6.0\ \text{A}$ ✓
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.
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:
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:
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.
Show all answers
Activity 1 — Model Answers
- $R_{\text{total}} = (8 \times 24)/(8 + 24) = 192/32 = 6.0\ \Omega$
- $I_1 = 24/8 = 3.0\ \text{A}$; $I_2 = 24/24 = 1.0\ \text{A}$
- $I_{\text{total}} = 3.0 + 1.0 = 4.0\ \text{A}$. Check: $24/6 = 4.0\ \text{A}$ ✓
Activity 2 — Model Answers
- $1/R = 1/5 + 1/10 + 1/20 = 4/20 + 2/20 + 1/20 = 7/20$. $R_{\text{total}} = 20/7 \approx 2.86\ \Omega$. $I_{\text{total}} = 20/(20/7) = 7.0\ \text{A}$.
- $I_5 = 20/5 = 4.0\ \text{A}$; $I_{10} = 20/10 = 2.0\ \text{A}$; $I_{20} = 20/20 = 1.0\ \text{A}$.
- Sum = 4.0 + 2.0 + 1.0 = 7.0 A ✓. Largest current through 5 Ω because lowest resistance.
Short Answer — Model Answers
Q3 (3 marks): (a) $R_{\text{total}} = (12 \times 6)/(12 + 6) = 72/18 = 4.0\ \Omega$. (b) $I_1 = 18/12 = 1.5\ \text{A}$; $I_2 = 18/6 = 3.0\ \text{A}$. (c) $I_{\text{total}} = 1.5 + 3.0 = 4.5\ \text{A}$.
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).
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.