# Series and Parallel Circuits

> Physics · CIE IGCSE 0625
> Source: https://www.owlsprep.com/study/cie-0625-u4-series-and-parallel-circuits/

This guide covers core rules for current, voltage and resistance in series and parallel circuits, plus extended-only combined resistance calculations, aligned to CIE IGCSE Physics 0625 2026-2028 syllabus point 4.3.2.

**Prerequisites:** [Current, voltage and resistance fundamentals](https://www.owlsprep.com/study/cie-0625-u4-current-voltage-resistance/); [Circuit symbol and drawing conventions](https://www.owlsprep.com/study/cie-0625-u4-circuit-diagrams/)

## Learning objectives

- Describe the structural difference between series and parallel circuits
- Apply core current, voltage and resistance rules for series circuits
- Apply core current, voltage and resistance rules for parallel circuits
- Calculate combined resistance for series and parallel combinations (Extended only)
- Explain advantages of parallel circuits for household wiring

## Core: Series vs Parallel Circuit Structure

**Series Circuit** — A circuit where all components are connected end-to-end in a single continuous loop, so there is only one path for electric current to flow.

*Example:* Two bulbs connected in a single loop to a battery, with no separate branches.

**Parallel Circuit** — A circuit where components are connected across separate parallel branches, so there are multiple independent paths for electric current to flow.

*Example:* Two bulbs each connected to their own separate branch across a single battery.

> **info**
>
> Nearly all household wiring uses parallel circuits to ensure appliances work independently and receive full mains voltage.

**Worked example:** Describe the layout of a series and parallel circuit using 1 battery, 2 resistors, and 1 switch.

1. Series circuit: Connect the battery positive terminal to the switch, switch to first resistor, first resistor to second resistor, second resistor back to the battery negative terminal. This forms one single loop with no branches.
2. Parallel circuit: Connect the battery terminals to two separate branches: the first branch contains the switch and first resistor, the second branch contains the second resistor. Both branches connect back to the battery terminals.

## Core: Rules for Series Circuits

Two rules for series circuits are required at Core tier:

1. Current is identical at every point in the circuit: $I_{total} = I_1 = I_2 = ... = I_n$
2. Total resistance equals the sum of individual component resistances: $R_{total} = R_1 + R_2 + ... + R_n$

> **info**
>
> The rule that the p.d.s across the components add up to the supply p.d. ($V_{total} = V_1 + V_2 + ...$) is Extended-tier (Supplement) content, covered in the Extended section below.

**Worked example:** A 3 V battery is connected in series with 2 Ω and 4 Ω resistors. Calculate the total circuit resistance and the current flowing through the circuit.

1. First calculate total resistance using the series resistance rule:
2. $$R_{total} = R_1 + R_2 = 2 + 4 = 6 \Omega$$
3. Use Ohm's Law ($I = V/R$) to calculate current, which is the same everywhere in the series circuit:
4. $$I = \frac{V_{total}}{R_{total}} = \frac{3}{6} = 0.5 A$$

## Core: Rules for Parallel Circuits

At Core tier, parallel circuits are treated qualitatively. You need to know the following:

- A parallel circuit provides more than one path, so the current divides between the separate branches.
- The combined (total) resistance of two resistors in parallel is less than the resistance of the smallest individual resistor.
- Adding more parallel paths lowers the total resistance of the circuit.
- Each parallel branch can be switched on or off independently without breaking the others, which is why household appliances are wired in parallel.

> **info**
>
> The quantitative rules that the branch currents add to give the supply current ($I_{total} = I_1 + I_2 + ...$) and that every branch has the same p.d. as the supply ($V_1 = V_2 = V_{supply}$) are Extended-tier (Supplement) content, covered in the Extended section below.

**Worked example:** A 6 Ω resistor and a 3 Ω resistor are connected in parallel. Without calculating an exact value, state what you can say about the total resistance of the combination.

1. For any parallel combination, the total resistance is less than the smallest individual resistor.
2. The smallest resistor here is 3 Ω, so the total resistance must be less than 3 Ω.

## Extended Only: Current, P.D. and Resistance Calculations

> **Extended Only Content**
>
> This section is required only for Extended tier (Papers 2/4) candidates. Core tier candidates may skip this section.

At Extended tier you must use these quantitative rules for current and p.d. in series and parallel circuits:

1. Series p.d.: the supply p.d. equals the sum of the p.d.s across the components: $V_{total} = V_1 + V_2 + ... + V_n$
2. Parallel p.d.: the p.d. across each branch is the same and equals the supply p.d.: $V_1 = V_2 = ... = V_{total}$
3. Parallel current (junction rule): the supply current equals the sum of the currents in the separate branches: $I_{total} = I_1 + I_2 + ... + I_n$

**Worked example:** A 6 V supply is connected to two bulbs in parallel. The current in the first bulb is 1 A and the current in the second bulb is 0.5 A. State the p.d. across each bulb and calculate the total current drawn from the supply.

1. Each parallel branch has the same p.d. as the supply, so the p.d. across each bulb is 6 V.
2. The supply current is the sum of the branch currents:
3. $$I_{total} = I_1 + I_2 = 1 + 0.5 = 1.5 \text{ A}$$

For Extended tier, you must also calculate the exact total resistance of parallel resistor combinations using the reciprocal formula:

$$\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$$

**Parallel Combined Resistance** — The equivalent single resistance that replaces a set of parallel resistors, calculated as the reciprocal of the sum of reciprocals of individual resistances.

*Notation:* $R_{total}$

*Example:* Two 4 Ω resistors in parallel have a total resistance of 2 Ω.

**Worked example:** Calculate the total resistance of 3 Ω and 6 Ω resistors connected in parallel.

1. Substitute resistance values into the parallel resistance formula:
2. $$\frac{1}{R_{total}} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$$
3. Take the reciprocal of both sides to find the total resistance:
4. $$R_{total} = 2 \Omega$$

**Check your understanding**

Test your Extended understanding

1. What is the total resistance of three 3 Ω resistors connected in parallel?

   *Why:* 1/R_total = 1/3 + 1/3 + 1/3 = 1, so R_total = 1 Ω.

## Common pitfalls

- **Wrong:** Assuming current is equal across all branches in a parallel circuit
  - Why it fails: Current splits across parallel branches, so it varies depending on the resistance of each branch.
  - Correct: Use $I_{total} = I_1 + I_2$ for parallel circuits, and remember equal current only applies to series circuits.
- **Wrong:** Adding parallel resistors directly like series resistors (e.g., 2 Ω + 2 Ω = 4 Ω for parallel)
  - Why it fails: Adding parallel branches creates more current paths, so total resistance decreases instead of increasing.
  - Correct: For Core, remember total parallel resistance is lower than the smallest individual resistor; for Extended, use the reciprocal resistance formula.
- **Wrong:** Summing voltages across parallel branches to get total supply voltage
  - Why it fails: All parallel branches receive the full supply voltage, so voltage across every branch is identical.
  - Correct: Only sum voltages for components connected in series; for parallel, $V_1 = V_2 = V_{supply}$.
- **Wrong:** Assuming a switch on one parallel branch controls the entire circuit
  - Why it fails: A switch on a parallel branch only breaks the current path for that specific branch, not the whole circuit.
  - Correct: A switch only controls the entire circuit if it is connected in series with the power supply.
- **Wrong:** Forgetting that a broken component in a series circuit turns off all other components
  - Why it fails: A broken component breaks the single current loop in a series circuit, stopping all current flow.
  - Correct: Identify if components are in series or parallel first to predict the effect of a broken component.

## Cheatsheet

| Property | Series Circuit | Parallel Circuit |
| --- | --- | --- |
| Current | $I_{total} = I_1 = I_2 = ...$ (Core) | $I_{total} = I_1 + I_2 + ...$ (Extended) |
| Voltage / p.d. | $V_{total} = V_1 + V_2 + ...$ (Extended) | $V_{total} = V_1 = V_2 = ...$ (Extended) |
| Resistance (Core) | $R_{total} = R_1 + R_2 + ...$ | Total R < smallest individual R |
| Resistance (Extended) | $R_{total} = R_1 + R_2 + ...$ | $1/R_{total} = 1/R_1 + 1/R_2 + ...$ |
| Household Use | Used for fuses and main power switches | Used for all appliance wiring |

## What's next

Now that you have mastered series and parallel circuit rules, you are ready to apply these to more complex mixed series-parallel circuit calculations, and analyse circuits with ammeters, voltmeters, and variable components like diodes and thermistors. These concepts are foundational for all electricity topics in your CIE IGCSE Physics 0625 exam, and will be tested in both written and practical paper questions where you may be asked to set up circuits and take measurements. Extended tier candidates will also use these rules to solve electrical power and energy calculations in later topics.

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