Parallel circuits flip the series rules on their head: now the voltage is shared and the current divides. Add a branch and, surprisingly, the total resistance drops and the current the battery delivers goes up, the opposite of series. Two short rules let you handle every branch, and they explain why every extra appliance you switch on draws more from the supply.
Two resistors are connected in parallel across one battery.
Before reading: is the total resistance of the pair larger or smaller than the smaller of the two resistors? Write your prediction and a one-sentence reason.
Why Iₜ = ΣI in a parallel circuit
Every branch connects to the same two points, so each feels the full supply voltage, and because charge is conserved at the junction the branch currents must add up to the current the battery delivers.
Each branch is connected directly across the supply, so every branch has the same voltage across it, the supply voltage. Where the branches split from the supply, charge is conserved: the current arriving must equal the total current leaving along the branches. So the total current is the sum of the branch currents:
$$I_T = I_1 + I_2 + I_3 + \dots \qquad\text{with}\qquad I_n = \frac{V}{R_n}$$
A branch with smaller resistance takes a bigger share of the current, but together the branch currents always add to the line current.
In parallel, every branch has the same voltage (the supply voltage). The total current is the sum of the branch currents, $I_T = I_1 + I_2 + \dots$, because charge is conserved at the junction. Each branch current is $I_n = V/R_n$.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: in a parallel circuit, the branch with the lower resistance carries the larger current.
Adding reciprocals, and why the total drops
We just saw that branch currents add. That raises a question: what single resistance would draw that same total current? This card answers it → combine $I_T = \Sigma I$ with $I = V/R$ to get the reciprocal rule.
Because the branches share the voltage and their currents add, dividing through by the common voltage turns the current rule into a rule for the reciprocals of resistance.
Start from $I_T = I_1 + I_2 + \dots$ and substitute $I_n = V/R_n$ (same $V$ on every branch). The total current equals $V/R_T$, so:
$$\frac{V}{R_T} = \frac{V}{R_1} + \frac{V}{R_2} + \dots \;\Rightarrow\; \frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \dots$$
Because each extra branch adds another pathway for current, the total resistance is always less than the smallest branch resistance, that is the answer to the hook.
A $12\ \text{V}$ battery is connected across two parallel resistors, $6.0\ \Omega$ and $3.0\ \Omega$. Find the total resistance, each branch current, and the total current.
For parallel resistors, $1/R_T = 1/R_1 + 1/R_2 + \dots$, so the total resistance is smaller than the smallest branch. Find each branch current with $I_n = V/R_n$; they add to the total current $I_T = V/R_T$.
Add the highlighted rule to your notes before the check below.
Quick check: Two $4.0\ \Omega$ resistors are connected in parallel. What is their total resistance?
Pattern A, Calculate then check
A $24\ \text{V}$ supply is connected across three parallel resistors: $12\ \Omega$, $8.0\ \Omega$ and $24\ \Omega$. In your book: (1) find each branch current using $I = V/R$; (2) add them to get the total current; (3) find the total resistance with $1/R_T = \Sigma 1/R$; (4) confirm $V/R_T$ gives the same total current and note that $R_T$ is less than $8.0\ \Omega$.
Match the pairs: Match each parallel-circuit quantity to its rule.
Dot point: Use the relationship between the total resistance and resistance in each branch of a parallel circuit, 1/R_total = 1/R1 + 1/R2 + …, to calculate resistances
The opposite behaviour to a series circuit
We just learned the reciprocal rule. That raises a question: what happens to the supply current as you add appliances at home? This card answers it → each new parallel branch adds another current path, so the total current rises.
Adding a parallel branch never removes the existing paths, it only adds a new one, so the supply must deliver more total current, and the combined resistance falls.
Every parallel branch keeps the same full voltage and adds its own current $V/R_n$ to the total. So switching on another appliance at home (all wired in parallel) increases $I_T$ and lowers $R_T$, the exact opposite of adding a series resistor, which raised the resistance and lowered the current. This is why circuits have a current limit and use fuses or circuit breakers: too many parallel appliances draw too much total current.
Adding a parallel branch adds another current path, so the total current $I_T$ increases and the total resistance $R_T$ decreases, the reverse of a series circuit. Household appliances in parallel each add their own current to the supply.
Pause, copy the highlighted contrast into your book before the check below.
Last check: Decide if the statement is true or false.
Connecting an extra resistor in parallel decreases the total resistance and increases the total current drawn from the supply.
Odd one out: three of these happen when an extra branch is added in parallel. Which one does not?
Dot point: explain the relationship between the total current of a parallel circuit and the current passing through each component using I_total = I1 + I2 + …
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 5(4 marks) 1. A $6.0\ \text{V}$ battery is connected across two parallel resistors, $2.0\ \Omega$ and $3.0\ \Omega$. Calculate the current in each branch, the total current, and the total resistance, and explain why $I_T = I_1 + I_2$.
1 mark: $I_1 = 3.0\ \text{A}$ and $I_2 = 2.0\ \text{A}$ · 1 mark: $I_T = 5.0\ \text{A}$ · 1 mark: $R_T = 1.2\ \Omega$ · 1 mark: branch currents add because charge is conserved at the junction
ExplainBand 5(3 marks) 2. Explain why the total resistance of a parallel combination is always less than the smallest individual branch resistance, and relate this to the total current.
1 mark: each added branch provides an additional path for current · 1 mark: more paths means more total current for the same voltage · 1 mark: since $R_T = V/I_T$, a larger total current for the same voltage means a smaller total resistance, necessarily below any single branch
Dot point: Use the relationship between the total resistance and resistance in each branch of a parallel circuit, 1/R_total = 1/R1 + 1/R2 + …, to calculate resistances
ExplainBand 5(3 marks) 3. A parallel circuit has two branches carrying $1.2\ \text{A}$ and $0.80\ \text{A}$. State the total current drawn from the supply and explain, using conservation of charge at a junction, why the branch currents combine in this way. Then state what happens to the total current if a third branch is added, and why.
1 mark: total current is $2.0\ \text{A}$ · 1 mark: charge is conserved at the junction, so the rate of charge arriving equals the total rate leaving along the branches · 1 mark: adding a third branch increases the total current, because the existing branches are unaffected and the new branch carries additional current
Dot point: explain the relationship between the total current of a parallel circuit and the current passing through each component using I_total = I1 + I2 + …
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Both resistors share the full supply voltage of $6.0\ \text{V}$. The branch currents are $I_1 = V/R_1 = 6.0/2.0 = 3.0\ \text{A}$ and $I_2 = V/R_2 = 6.0/3.0 = 2.0\ \text{A}$. The total current is $I_T = I_1 + I_2 = 3.0 + 2.0 = 5.0\ \text{A}$. The total resistance follows from $1/R_T = 1/2.0 + 1/3.0 = 3/6 + 2/6 = 5/6$, so $R_T = 6/5 = 1.2\ \Omega$, less than the smaller branch, as expected. The branch currents add because charge is conserved at the junction where the branches split: all the current arriving from the supply must continue along one branch or the other, so the line current equals the sum of the branch currents.
Q2 (3 marks): Every branch connected in parallel provides an additional independent path along which charge can flow between the same two points. Adding a branch does not remove or restrict any existing path, so all the current that flowed before still flows, and the new branch carries extra current on top of it. For the same applied voltage the total current therefore increases whenever a branch is added. Total resistance is defined by $R_T = V/I_T$, so with the voltage fixed and the total current larger, the total resistance must be smaller. Because the total current is always at least as large as the current in the single lowest-resistance branch, plus whatever the other branches carry, the total resistance must come out below even the smallest branch resistance. Adding branches always lowers the total resistance and always raises the total current.
Q3 (3 marks): The total current is $I_T = 1.2 + 0.80 = 2.0\ \text{A}$. The reason the branch currents add is conservation of charge. Charge cannot be created, destroyed or accumulated at a junction, so in any given second exactly as much charge must leave the junction along the two branches as arrived at it along the supply wire. Expressed as rates of flow, that is the statement $I_T = I_1 + I_2$. Adding a third branch increases the total current. Each existing branch still has the full supply voltage across it and its own unchanged resistance, so by $I = V/R$ each still carries exactly the current it did before. The new branch carries an additional current on top of those, so the supply must deliver more, and the total current rises.
You can now use the parallel rules, shared voltage, $I_T = \Sigma I$ and $1/R_T = \Sigma 1/R$, to analyse a parallel circuit. Next lesson combines series and parallel sections into one network and measures it in the lab.
↩ Back to focus-area hubAt the start you predicted whether the parallel total resistance is larger or smaller than the smaller branch. The answer: it is always smaller, because each branch is an extra path for current to flow. If you predicted larger, revisit Card 2, adding pathways makes it easier, not harder, for charge to move.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
Tell us what you would like help with. Register interest; support details will be confirmed before enrolment.
Register interest →