Series and Parallel Circuits

Components can be joined in two arrangements, and each follows its own set of rules.

• Series: components are connected in a single loop, one after another, so charge flows through each in turn along one path.
• Parallel: components are connected on separate branches between the same two points, so charge can flow through more than one path.

The diagram below shows a series loop with two lamps; a parallel circuit would instead split into branches so each lamp sits on its own branch.

The rest of this lesson works through the current, potential difference and resistance rules for each arrangement, then applies them to worked calculations. The key equation throughout is $V=IR$.

You must recall and apply $V=IR$. It is not given on the Physics equation sheet.

In a series circuit there is only one loop, so:

• Current is the same through every component (one path, so the same charge passes each point).
• Potential difference is shared between the components: the individual potential differences add up to the source potential difference.
• Total resistance is the sum of the individual resistances:

$R_{total}=R_1+R_2$

You must recall the series rule $R_{total}=R_1+R_2$. It is not on the equation sheet.

Adding a resistor in series increases the total resistance, because the charge, following its single path, has to pass through more opposition. A larger total resistance means a smaller current from the source (for the same source potential difference).

The potential difference splits in proportion to resistance: the largest resistor takes the largest share of the potential difference.

Two resistors, 4 Ω and 8 Ω, are connected in series to a 6 V battery. Find the total resistance, the current, and the potential difference across each resistor.

Step 1 — Total resistance.

$R_{total}=4+8=12\Omega$

Step 2 — Current from the source (same everywhere in series). Rearranging $V=IR$:

$I=\frac{V}{R_{total}}=\frac{6}{12}=0.5\text{ A}$

Step 3 — Potential difference across each resistor (same 0.5 A through each):

$V_{4\Omega }=IR=0.5\times 4=2\text{ V}$ $V_{8\Omega }=IR=0.5\times 8=4\text{ V}$

Check: the two potential differences add to $2+4=6$ V, the source potential difference, as they must in series. The 8 Ω resistor, being larger, takes the bigger share.

In a parallel circuit each branch connects across the same two points, so:

• Potential difference across each branch is the same, and equal to the source potential difference.
• Current splits between the branches: the branch currents add up to the total current drawn from the source.
• Total resistance is less than the smallest individual resistor.

Adding a resistor in parallel decreases the total resistance, because each new branch gives the current another path. More paths make it easier overall for charge to flow, so the equivalent resistance falls.

(Note) For AQA you are not required to calculate the total resistance of two resistors in parallel. You need only state that it is less than the smallest resistor, and explain why qualitatively.

A useful consequence: because each branch gets the full source potential difference, a branch with lower resistance carries a larger current, following $I=V/R$.

A 12 V supply is connected to two resistors in parallel: 6 Ω on one branch and 4 Ω on the other. Find the current in each branch and the total current.

Step 1 — Note the potential difference across each branch. In parallel, each branch has the full supply potential difference:

$V=12\text{ V across both branches}$

Step 2 — Current in the 6 Ω branch. Using $I=V/R$:

$I_{6\Omega }=\frac{12}{6}=2\text{ A}$

Step 3 — Current in the 4 Ω branch.

$I_{4\Omega }=\frac{12}{4}=3\text{ A}$

Step 4 — Total current (branch currents add):

$I_{total}=2+3=5\text{ A}$

The supply delivers 5 A, which splits into 2 A and 3 A. The lower-resistance 4 Ω branch carries the larger current, as expected.

Feature	Series	Parallel
Number of paths	One	More than one
Current	Same through all components	Splits; branch currents add to the total
Potential difference	Shared between components	Same across every branch
Total resistance	Sum: $R_{total}=R_1+R_2$	Less than the smallest resistor
Effect of adding a component	Total resistance increases	Total resistance decreases

A real-world reason homes are wired in parallel: each appliance gets the full mains potential difference, and switching one off does not break the path for the others.

1. Swapping the series and parallel rules

In series, current is the same and potential difference is shared. In parallel, potential difference is the same and current is shared. Mixing these two is the most common error in this topic.

2. Adding resistances in parallel

You do not add resistances in parallel. Adding in parallel gives a total less than the smallest resistor. Only series resistances add: $R_{total}=R_1+R_2$.

3. Trying to calculate a parallel total resistance

AQA does not require the total resistance of two parallel resistors. State that it is less than the smallest resistor and explain why; do not attempt a full calculation.

4. Using the wrong potential difference in a series branch

In series, each resistor has only its share of the source potential difference, not the full source value. Find the current first, then use $V=IR$ for each resistor.

5. Forgetting the branch currents must add up

In parallel, check that your branch currents add to the total supply current. If they do not, a value is wrong.