Series and Parallel Circuits

This lesson covers the rules for series and parallel circuits, including how current, potential difference and resistance behave in each arrangement, as required by AQA GCSE Combined Science Trilogy (8464, section 6.2.1).

---

Series Circuits

A series circuit has only one loop — all components are connected end-to-end, and there is only one path for the current to follow.

flowchart LR
    A["Battery"] --> B["Lamp 1"]
    B --> C["Lamp 2"]
    C --> D["Lamp 3"]
    D --> A

Rules for Series Circuits

1. Current

The current is the same at every point in a series circuit.

$I_{\text{total}} = I_1 = I_2 = I_3$Itotal​=I1​=I2​=I3​

This is because there is only one path for the charge to flow through — every electron must pass through every component.

2. Potential Difference

The potential differences across the components add up to give the total p.d. of the supply.

$V_{\text{total}} = V_1 + V_2 + V_3$Vtotal​=V1​+V2​+V3​

The voltage is shared between the components. A component with a higher resistance gets a larger share of the voltage.

3. Resistance

The total resistance is the sum of the individual resistances.

$R_{\text{total}} = R_1 + R_2 + R_3$Rtotal​=R1​+R2​+R3​

Adding more resistors in series increases the total resistance and therefore decreases the current.

---

Parallel Circuits

A parallel circuit has branches — the current can follow more than one path.

flowchart LR
    A["Battery"] --> B{"Junction"}
    B --> C["Lamp 1"]
    B --> D["Lamp 2"]
    B --> E["Lamp 3"]
    C --> F{"Junction"}
    D --> F
    E --> F
    F --> A

Rules for Parallel Circuits

1. Potential Difference

The potential difference across each branch is the same and equals the supply voltage.

$V_{\text{total}} = V_1 = V_2 = V_3$Vtotal​=V1​=V2​=V3​

2. Current

The total current is the sum of the currents through each branch.

$I_{\text{total}} = I_1 + I_2 + I_3$Itotal​=I1​+I2​+I3​

The branch with the lowest resistance carries the largest current.

3. Resistance

The total resistance of parallel resistors is less than the smallest individual resistance.

For two resistors in parallel:

$\frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2}$Rtotal​1​=R1​1​+R2​1​

For two identical resistors of value R in parallel:

$R_{\text{total}} = \frac{R}{2}$Rtotal​=2R​

Adding more resistors in parallel decreases the total resistance and therefore increases the total current drawn from the supply.

---

Comparison: Series vs Parallel

Feature	Series Circuit	Parallel Circuit
Number of paths	One	More than one
Current	Same through all components	Splits between branches; total = sum of branch currents
Potential difference	Shared between components; adds up to supply voltage	Same across every branch; equals supply voltage
Total resistance	Sum of individual resistances (increases)	Less than the smallest individual resistance (decreases)
Effect of adding more components	Total resistance increases; current decreases; each component gets less voltage	Total resistance decreases; total current increases; each component still gets full voltage
If one component breaks	Whole circuit stops	Other branches continue to work

---

Worked Examples

Worked Example 1 — Series

A 12 V battery is connected in series with a 4 Ω resistor and an 8 Ω resistor. Calculate: (a) the total resistance, (b) the current, (c) the voltage across each resistor.

(a) $R_{\text{total}} = 4 + 8 = 12 \text{ Ω}$Rtotal​=4+8=12 Ω

(b) $I = \frac{V}{R} = \frac{12}{12} = 1.0 \text{ A}$I=RV​=1212​=1.0 A

(c) $V_1 = I \times R_1 = 1.0 \times 4 = 4 \text{ V}$V1​=I×R1​=1.0×4=4 V

$V_2 = I \times R_2 = 1.0 \times 8 = 8 \text{ V}$V2​=I×R2​=1.0×8=8 V

Check: $4 + 8 = 12 \text{ V}$4+8=12 V ✓

Worked Example 2 — Parallel

A 6 V battery is connected to two resistors in parallel: 10 Ω and 15 Ω. Calculate: (a) the current in each branch, (b) the total current.

(a) $I_1 = \frac{V}{R_1} = \frac{6}{10} = 0.6 \text{ A}$I1​=R1​V​=106​=0.6 A

$I_2 = \frac{V}{R_2} = \frac{6}{15} = 0.4 \text{ A}$I2​=R2​V​=156​=0.4 A

(b) $I_{\text{total}} = 0.6 + 0.4 = 1.0 \text{ A}$Itotal​=0.6+0.4=1.0 A

Worked Example 3 — Parallel Total Resistance

Two resistors of 6 Ω and 12 Ω are connected in parallel. Calculate the total resistance.

$\frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$Rtotal​1​=61​+121​=122​+121​=123​=41​

$R_{\text{total}} = 4 \text{ Ω}$Rtotal​=4 Ω

---

Why Household Wiring Uses Parallel Circuits

• Each appliance gets the full mains voltage (230 V).
• Each appliance can be switched on and off independently.
• If one device fails, the others continue to work.

---

Common Mistakes

Mistake	Correction
Saying current is "shared" in a series circuit	Current is the same everywhere in series; it is the voltage that is shared
Adding resistances in parallel like series	In parallel, use $1/R_{\text{total}} = 1/R_1 + 1/R_2$1/Rtotal​=1/R1​+1/R2​ — the total is always less than the smallest
Thinking adding more parallel branches increases total resistance	More branches means more paths → less total resistance and more total current
Forgetting that voltage is the same across parallel branches	Each branch gets the full supply voltage in a parallel circuit

---

Summary

• In a series circuit: current is the same everywhere; voltages add up; resistances add up.
• In a parallel circuit: voltage is the same across each branch; currents add up; total resistance is less than the smallest branch resistance.
• Adding resistors in series increases total resistance; adding resistors in parallel decreases total resistance.
• Household wiring is parallel so each appliance gets the full voltage and can be switched independently.

---

Extended Worked Examples

Worked Example 1 — Series Circuit

A 12 V battery is connected to three resistors of 2 Ω, 4 Ω and 6 Ω in series.

Quantity	Calculation	Value
Total resistance	$2 + 4 + 6$2+4+6	12 Ω
Current	$I = V/R = 12/12$I=V/R=12/12	1.0 A
p.d. across 2 Ω	$V = IR = 1.0 \times 2$V=IR=1.0×2	2.0 V
p.d. across 4 Ω	$V = IR = 1.0 \times 4$V=IR=1.0×4	4.0 V
p.d. across 6 Ω	$V = IR = 1.0 \times 6$V=IR=1.0×6	6.0 V