Series & parallel resistors: practice problems with step-by-step solutions

Also called: equivalent resistance

Free Series & parallel resistors practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: 3 resistors, 4 resistors, ladder of 6 resistors.

Free practice problemEasy
R1​ = 10 Ω+−V1​8 VR2​4 ΩIR2​AR3​6 Ω

Find the current IR2 through R2.

V1 = 8 VR1 = 10 ΩR2 = 4 ΩR3 = 6 Ω
A
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What you practise

  1. Easy3 resistors
  2. Medium4 resistors
  3. Hardladder of 6 resistors

How to solve Series & parallel resistors problems

  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.

Common mistakes

  • Calling two resistors in series when a branch connects between them.
  • Adding parallel resistors directly.
  • Forgetting to take the reciprocal at the end of the parallel formula.

Key formulas

Resistors in series
Req = R1 + R2 + …
Resistors in parallel
1/Req = 1/R1 + 1/R2 + …
Two in parallel
Req = R1R2 / (R1 + R2)
Ohm's law
V = I·R
Power
P = V·I = I²R = V²/R

Learn Series & parallel resistors from scratch

1Voltage, current and resistance

A circuit is a closed loop that charge can flow around. Three quantities describe what happens in it.

  • Voltage V, in volts (V): the push that drives the charge, like water pressure in a pipe.
  • Current I, in amperes (A): how much charge flows past a point per second, like litres per second.
  • Resistance R, in ohms (Ω): how hard it is for current to flow, like a narrow part of the pipe.
+−VRI

They are linked by Ohm’s law:

V = I · R

If you know two of the three, you can find the third: I = V/R and R = V/I.

Self-check A 12 V source drives current through a 4 Ω resistor. What is the current?
  1. 0.33 A
  2. 16 A
  3. 3 A
  4. 48 A

Answer: 3 A. I = V/R = 12/4 = 3 A. Multiplying (48 A) or dividing the wrong way round (0.33 A) are the usual slips.

2Resistors in series

Resistors are in series when they sit one after another on the same path, with nothing branching off between them.

R1​R2​IIab

The same current has to pass through both, so their resistances simply add up:

Req = R1 + R2 + R3 + …

The equivalent resistance is always larger than each resistor on its own. A longer narrow pipe is harder to push water through.

Self-check What is the equivalent resistance of 3 Ω, 5 Ω and 7 Ω in series?
  1. 1.4 Ω
  2. 105 Ω
  3. 15 Ω
  4. 7 Ω

Answer: 15 Ω. In series you add: 3 + 5 + 7 = 15 Ω.

3Resistors in parallel

Resistors are in parallel when both of their ends are connected to the same two nodes. The current splits between them.

R1​R2​ab

Each resistor gets the same voltage. Because the current has more than one path, the total resistance goes down:

1/Req = 1/R1 + 1/R2 + …

For exactly two resistors there is a shortcut, “product over sum”:

Req = R1·R2 / (R1 + R2)

A quick sanity check: the result is always smaller than the smallest resistor. Two equal resistors in parallel give half of one.

Self-check What is 6 Ω in parallel with 3 Ω?
  1. 0.5 Ω
  2. 2 Ω
  3. 4.5 Ω
  4. 9 Ω

Answer: 2 Ω. Product over sum: 6·3/(6 + 3) = 18/9 = 2 Ω. It is smaller than 3 Ω, as it must be. 0.5 is 1/R: do not forget to take the reciprocal.

4The method step by step

Real networks mix series and parallel parts. You simplify them one small step at a time.

  1. Mark the two terminals you are looking from (often a and b, or the source).
  2. Start at the end of the network furthest from those terminals.
  3. Find two resistors that are clearly in series or clearly in parallel, and replace them by one.
  4. Redraw the circuit with the new resistor.
  5. Repeat until one resistor is left between the terminals. That is Req.

If a question asks for the source current, finish with Ohm’s law: I = V/Req.

5A worked example

Follow every step. Try to predict the next one before you open it.

Worked example

Find the equivalent resistance between a and b. Then find the current a 22 V source would deliver when connected to a–b.

1

Look at the network

abR1​ = 5 ΩR2​12 ΩR3​ = 4 ΩR4​8 Ω

Start at the far right. R3 and R4 are on one path with nothing branching off between them.

2

Combine R₃ and R₄ (series)

R34 = 4 + 8 = 12 Ω
abR1​ = 5 ΩR2​12 ΩR34​12 Ω

Now R34 connects between the same two nodes as R2.

3

Combine R₂ and R₃₄ (parallel)

Rp = 12·12 / (12 + 12) = 6 Ω

Two equal resistors in parallel give half of one, so 6 Ω is right.

abR1​ = 5 ΩRp​6 Ω
4

Combine R₁ and Rₚ (series)

Req = 5 + 6 = 11 Ω
5

Use Ohm’s law

I = V / Req = 22 / 11 = 2 A
Answer

Req = 11 Ω and the source delivers I = 2 A.

6Common mistakes

  • Calling two resistors “in series” when a branch leaves the node between them. Then they are not in series.
  • Adding parallel resistors directly, as if they were in series.
  • Forgetting the last step of 1/Req = …: you still have to take the reciprocal.
  • Not redrawing. After each step, draw the smaller circuit before you look for the next pair.
Self-check Two 10 Ω resistors are connected in parallel. What is the equivalent resistance?
  1. 0.2 Ω
  2. 10 Ω
  3. 20 Ω
  4. 5 Ω

Answer: 5 Ω. Equal resistors in parallel give half of one: 10·10/(10 + 10) = 5 Ω. 20 Ω would be series, and 0.2 is 1/R before taking the reciprocal.

7How exam questions look

  • Easy: three resistors, one series step and one parallel step. You find Req or the source current.
  • Medium: four resistors. The same idea, with one more step.
  • Hard: a ladder of six resistors. Always start at the far end and work back towards the source.

When the source current is asked, first find Req and then use I = V/Req.

Give the answer with its unit, and check that it is sensible: a parallel pair is smaller than either resistor.

Learn it step by step in the appLearn it step by step in the app

Worked examples

Example 1Easy
R1​ = 5 ΩIR1​+−V1​18 VR2​6 ΩAR3​12 Ω

Find the current IR1 delivered by the source, which flows through R1.

V1 = 18 VR1 = 5 ΩR2 = 6 ΩR3 = 12 Ω

Find IR1 (A)

Show hints
  1. Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
  2. R₂ and R₃ both join node A to the bottom wire, so they are in parallel. That pair is in series with R₁.
  3. Req = 9 Ω.
Show the answer and the step-by-step solution

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 5 Ω+−V1​18 VR2​6 ΩAR3​12 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R2 ∥ R3 = 6·12/(6+12) = 4 Ω
Req = R1 + R2 ∥ R3 = 5 + 4 = 9 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 18/9 = 2 A

Answer

IR1 = 2 A
Example 2Medium
R1​ = 6 ΩIR1​R3​ = 3 Ω+−V1​18 VR2​3 ΩAR4​6 ΩB

Find the current IR1 delivered by the source, which flows through R1.

V1 = 18 VR1 = 6 ΩR2 = 3 ΩR3 = 3 ΩR4 = 6 Ω

Find IR1 (A)

Show hints
  1. Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
  2. R₃ and R₄ carry the same current, so they are in series. That branch is in parallel with R₂, and the result is in series with R₁.
  3. Req = 8.25 Ω.
Show the answer and the step-by-step solution

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 6 ΩR3​ = 3 Ω+−V1​18 VR2​3 ΩAR4​6 ΩB

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R3 + R4 = 3 + 6 = 9 Ω
R2 ∥ 9 = 3·9/(3+9) = 2.25 Ω
Req = R1 + 2.25 = 6 + 2.25 = 8.25 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 18/8.25 = 2.182 A

Answer

IR1 = 2.182 A
Example 3Hard
R1​ = 10 ΩIR1​R3​ = 15 ΩR5​ = 20 Ω+−V1​12 VR2​20 ΩAR4​5 ΩBR6​20 Ω

Find the current IR1 delivered by the source, which flows through R1.

V1 = 12 VR1 = 10 ΩR2 = 20 ΩR3 = 15 ΩR4 = 5 ΩR5 = 20 ΩR6 = 20 Ω

Find IR1 (A)

Show hints
  1. Reduce the network step by step. In series, resistances add. Two resistors in parallel combine as R₁R₂/(R₁ + R₂).
  2. Start with R₅ + R₆ (series), combine with R₄ (parallel), add R₃, combine with R₂ and finally add R₁.
  3. Req = 19.86 Ω.
Show the answer and the step-by-step solution

Series and parallel resistors: in series the same current flows through each resistor and R = R₁ + R₂ + …. In parallel each resistor has the same voltage and 1/R = 1/R₁ + 1/R₂ + …. A ladder network is reduced one pair at a time until a single equivalent resistance remains.

R1​ = 10 ΩR3​ = 15 ΩR5​ = 20 Ω+−V1​12 VR2​20 ΩAR4​5 ΩBR6​20 Ω

Combine resistors from the far end

Start at the end farthest from the source and work back towards it. Resistors carrying the same current are in series; resistors across the same two nodes are in parallel.

R5 + R6 = 20 + 20 = 40 Ω
R4 ∥ 40 = 4.444 Ω
R3 + 4.444 = 19.44 Ω
R2 ∥ 19.44 = 9.859 Ω
Req = R1 + 9.859 = 10 + 9.859 = 19.86 Ω

Current from the source

The source sees only Req, so Ohm’s law gives the current it delivers. That current flows through R₁.

IR1 = V1/Req = 12/19.86 = 0.6043 A

Answer

IR1 = 0.6043 A

Questions and answers

What is Series & parallel resistors?

A circuit is a closed loop that charge can flow around. Three quantities describe what happens in it.

They are linked by Ohm’s law:

How do I solve Series & parallel resistors problems step by step?
  1. Start at the end of the network furthest from the terminals.
  2. Combine resistors in series by adding them, and resistors in parallel with 1/R = 1/R₁ + 1/R₂ + …
  3. Redraw the circuit after each step.
  4. Repeat until one resistor is left between the terminals.
What are common mistakes in Series & parallel resistors?
  • Calling two resistors in series when a branch connects between them.
  • Adding parallel resistors directly.
  • Forgetting to take the reciprocal at the end of the parallel formula.
What do the Series & parallel resistors exercises cover?
  • Easy: 3 resistors
  • Medium: 4 resistors
  • Hard: ladder of 6 resistors

Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.

How practising works

  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Your progress, mistakes and saved problems are kept, so you can review before the exam.

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Printable worksheet

Series & parallel resistors worksheet with answers
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