# Series and parallel resistor combinations

> Physics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9702-u10-series-and-parallel-resistor-combinations/

This subtopic explains the rules for combining resistors in series and parallel arrangements, how to calculate equivalent resistance for mixed networks, and how to apply these rules to common CIE A-Level Physics DC circuit problems.

**Prerequisites:** [Ohm's Law and resistance definition](https://www.owlsprep.com/study/cie-9702-u10-ohms-law-resistance/); [Potential difference and electric current](https://www.owlsprep.com/study/cie-9702-u10-pd-current/)

## Learning objectives

- Derive resistance formulas for series and parallel resistor combinations
- Calculate total equivalent resistance for mixed resistor networks
- Apply combination rules to solve CIE DC circuit problems
- Recall current and voltage distribution rules for each combination type

## Series Resistor Combinations

**Series combination of resistors** — Resistors connected in a single end-to-end path, so the same current flows through every resistor, and total potential difference equals the sum of potential differences across individual resistors.

*Example:* Three 2 Ω resistors connected in a line between a battery's positive and negative terminals

**Derivation:** Derive equivalent resistance for n resistors in series

*Starting from:* Kirchhoff's Voltage Law and Ohm's Law

1. For resistors $R_1, R_2, ... R_n$ in series, current $I$ is the same through all. By KVL:
2. $$V = V_1 + V_2 + ... + V_n$$
3. Substitute Ohm's law $V = IR$ for each term:
4. $$I R_{eq} = I R_1 + I R_2 + ... + I R_n$$
5. Divide both sides by common current $I$:
6. $$R_{eq} = R_1 + R_2 + ... + R_n$$

*Conclusion:* Total equivalent resistance of series resistors is the sum of all individual resistances.

**Worked example:** Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series to a 10 V battery. Calculate total equivalent resistance and the current drawn from the battery.

1. Use the series resistance rule to find $R_{eq}$:
2. $$R_{eq} = 2 + 3 + 5 = 10 \ \Omega$$
3. Apply Ohm's law to find total current $I$:
4. $$I = \frac{V}{R_{eq}} = \frac{10}{10} = 1.0 \ A$$

> **Exam tip:** Use the voltage divider rule $V_i = V_{total} \times \frac{R_i}{R_{eq}}$ to quickly find voltage across any individual series resistor, this saves time in multiple choice questions.

*Calculator:* allowed

## Parallel Resistor Combinations

**Parallel combination of resistors** — Resistors connected across the same two nodes, so the same potential difference acts across every resistor, and total current equals the sum of currents through individual resistors.

**Derivation:** Derive equivalent resistance for n resistors in parallel

*Starting from:* Kirchhoff's Current Law and Ohm's Law

1. For resistors $R_1, R_2, ... R_n$ in parallel, potential difference $V$ is the same across all. By KCL:
2. $$I = I_1 + I_2 + ... + I_n$$
3. Substitute Ohm's law $I = V/R$:
4. $$\frac{V}{R_{eq}} = \frac{V}{R_1} + \frac{V}{R_2} + ... + \frac{V}{R_n}$$
5. Divide by common potential difference $V$:
6. $$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$$

*Conclusion:* The reciprocal of equivalent resistance equals the sum of reciprocals of individual resistances. For two resistors, this simplifies to $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$.

**Worked example:** A 4 Ω and a 6 Ω resistor are connected in parallel across a 12 V supply. Calculate equivalent resistance and total current from the supply.

1. Apply the reciprocal rule for parallel resistance:
2. $$\frac{1}{R_{eq}} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$$
3. Take the reciprocal to find $R_{eq}$:
4. $$R_{eq} = \frac{12}{5} = 2.4 \ \Omega$$
5. Calculate total current using Ohm's law:
6. $$I = \frac{V}{R_{eq}} = \frac{12}{2.4} = 5.0 \ A$$

> **warning**
>
> Equivalent parallel resistance is always smaller than the smallest individual resistor in the combination. If your result is larger than the smallest resistor, you have made an error.

*Calculator:* allowed

## Mixed Resistor Networks

Most CIE exam problems have combinations of series and parallel resistors in the same network. To solve these, you simplify step-by-step, starting from the innermost combination and working outwards.

1. Identify the innermost purely series or parallel combination that contains no other nested combinations
2. Calculate the equivalent resistance for this innermost combination
3. Replace the original combination with its equivalent resistance on the circuit diagram
4. Repeat the process until you get a single equivalent resistance for the entire network

**Worked example:** Calculate the total equivalent resistance of this network: a 2 Ω resistor is in series with a parallel combination of 3 Ω and 6 Ω resistors.

1. First simplify the innermost parallel combination of 3 Ω and 6 Ω:
2. $$\frac{1}{R_{parallel}} = \frac{1}{3} + \frac{1}{6} = \frac{2 +1}{6} = \frac{1}{2}$$
3. The equivalent resistance of the parallel section is $R_{parallel} = 2 \ \Omega$.
4. This 2 Ω equivalent is in series with the 2 Ω resistor, so add the resistances:
5. $$R_{total} = 2 + 2 = 4 \ \Omega$$

**Check your understanding**

Check your understanding of simplification order

1. A network has a 10 Ω resistor in series with a parallel branch. The parallel branch contains a 2 Ω resistor in series with two parallel 4 Ω resistors. What do you calculate first?

   - The two 4 Ω resistors in parallel
   - The 2 Ω resistor and the parallel equivalent in series
   - The whole parallel branch with the 10 Ω resistor

   *Answer:* The two 4 Ω resistors in parallel

   *Why:* Correct! You always start from the innermost nested combination, then work outwards step by step.

> **Exam tip:** Always redraw the circuit after each simplification step to avoid mixing up series and parallel connections. CIE examiners award method marks for correct working even if your final answer is wrong.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Adding resistances directly for parallel combinations instead of summing reciprocals
  - Why it fails: Confusing series and parallel rules leads to an equivalent resistance that is far too large
  - Correct: Remember the core rule: series = sum of resistances, parallel = sum of reciprocals
- **Wrong:** Stopping after summing reciprocals for parallel resistance and forgetting to take the final reciprocal
  - Why it fails: This leaves you with a resistance value that is far too small, costing easy exam marks
  - Correct: Always double-check: after calculating $\sum 1/R_i$, take the reciprocal to get $R_{eq}$
- **Wrong:** Simplifying outer combinations before innermost nested combinations
  - Why it fails: This leads to incorrect grouping of series and parallel resistors, producing wrong results
  - Correct: Always work from the inside out: simplify the deepest nested group first, then move outwards
- **Wrong:** Claiming total voltage is the sum of voltages across parallel resistors
  - Why it fails: Confusing voltage and current rules for parallel combinations
  - Correct: Voltage is equal across all parallel resistors; total current is the sum of individual currents
- **Wrong:** Assuming any two resistors with equal current are always in series
  - Why it fails: Equal current can occur in non-series resistors in balanced networks, this is not the definition of series
  - Correct: A series combination is defined by sharing a single current path with no branches between the resistors

## Cheatsheet

| Combination Type | Current Rule | Voltage Rule | Equivalent Resistance |
| --- | --- | --- | --- |
| Series | Same current through all | $V_{total} = \sum V_i$ | $R_{eq} = \sum R_i$ |
| Parallel | $I_{total} = \sum I_i$ | Same voltage across all | $\frac{1}{R_{eq}} = \sum \frac{1}{R_i}$ |
| Two resistors parallel | $I_1/I_2 = R_2/R_1$ | Same voltage across both | $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$ |
| n equal resistors R parallel | Total current = n × I per resistor | Same voltage | $R_{eq} = R/n$ |

## What's next

Understanding series and parallel resistor combinations is the foundation for all more complex DC circuit topics, including potential dividers, internal resistance of batteries, and Kirchhoff's laws for multi-loop circuits. This subtopic appears in almost every CIE A-Level Physics paper 1 and paper 2, so mastering step-by-step simplification for mixed networks will earn you consistent easy marks. Next, you will apply these combination rules to potential dividers, a common exam topic for sensor and measurement circuit problems, before moving on to more complex multi-loop circuits. Building a solid understanding of combination rules now will make all subsequent DC circuit topics far easier to master.

- [Internal Resistance of Batteries](https://www.owlsprep.com/study/cie-9702-u10-internal-resistance/)
- [Potential divider](https://www.owlsprep.com/study/cie-9702-u10-potential-divider/)
- [Particle physics (AS)](https://www.owlsprep.com/study/cie-9702-u11-overview/)

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