Practice problems

Series and parallel resistors practice problems

Most real circuits have more than one resistor, and the first job is usually to turn them into one. These ten problems start with resistors in series, move on to parallel pairs and three-way parallel groups, then reduce mixed networks and split the current between branches.

The method

In series, one current flows through every resistor, so the resistances add:

R_total = R1 + R2 + R3

In parallel, every branch gets the same voltage. For two resistors, use product over sum:

R = R1 × R2R1 + R2

n equal resistors in parallel: R over n
Three or more in parallel: add the reciprocals, 1 over R = 1 over R1 + 1 over R2 + 1 over R3
Mixed networks: replace one purely series or purely parallel group at a time, then redraw

The problems

Type your answer and tap Check. Every one has the working, step by step.

Problem 1 of 10
A 220 Ω, a 330 Ω and a 470 Ω resistor are wired in series. What's the total resistance?
Show the working
1. In series, resistances add: R_total = R1 + R2 + R3.
2. 220 Ω + 330 Ω + 470 Ω = 1020 Ω, or 1.02 kΩ.
Answer: 1020 Ω
Problem 2 of 10
A 4.7 kΩ and a 10 kΩ resistor are in series across a 9 V battery. What current flows, in milliamps?
Show the working
1. The total is 4.7 kΩ + 10 kΩ = 14.7 kΩ.
2. Volts divided by kilohms gives milliamps: I = 9 V over 14.7 kΩ ≈ 0.612 mA.
Answer: 0.612 mA
Problem 3 of 10
A 100 Ω and a 150 Ω resistor are wired in parallel. What's their combined resistance?
Show the working
1. For two resistors in parallel, use product over sum: R = R1 × R2 over R1 + R2.
2. R = 100 × 150 over 100 + 150 = 15000 over 250 = 60 Ω
3. Check: 60 Ω is below the smaller resistor, 100 Ω, as a parallel total always is.
Answer: 60 Ω
Problem 4 of 10
Four 2.2 kΩ resistors are wired in parallel. What's their combined resistance, in ohms?
Show the working
1. n equal resistors in parallel give R over n.
2. 2.2 kΩ over 4 = 0.55 kΩ = 550 Ω
Answer: 550 Ω
Problem 5 of 10
A 4.7 kΩ and a 6.8 kΩ resistor are wired in parallel. What's their combined resistance?
Show the working
1. Product over sum, working in kilohms: R = 4.7 × 6.8 over 4.7 + 6.8 = 31.96 over 11.5
2. R ≈ 2.78 kΩ, below the smaller 4.7 kΩ.
Answer: 2.78 kΩ
Problem 6 of 10
A 1 kΩ, a 2.2 kΩ and a 4.7 kΩ resistor are all wired in parallel. What's their combined resistance, in ohms?
Show the working
1. With three branches, add the reciprocals: 1 over R = 1 over R1 + 1 over R2 + 1 over R3.
2. In kilohms: 1 over 1 + 1 over 2.2 + 1 over 4.7 = 1 + 0.4545 + 0.2128 = 1.6673
3. R = 1 over 1.6673 ≈ 0.5998 kΩ, which is 600 Ω to three figures.
Answer: 600 Ω
Problem 7 of 10
A 330 Ω resistor is in series with a 1 kΩ and a 1.5 kΩ resistor that are in parallel with each other. What's the total resistance?
Show the working
1. Reduce the parallel pair first: 1 × 1.5 over 1 + 1.5 = 1.5 over 2.5 = 0.6 kΩ, which is 600 Ω.
2. Then add the series part: 330 Ω + 600 Ω = 930 Ω
Answer: 930 Ω
Problem 8 of 10
40 mA flows into a 100 Ω and a 150 Ω resistor in parallel. How much current goes through the 150 Ω resistor?
Show the working
1. Use the current divider. The other branch's resistance goes on top: I = I_total × 100 over 100 + 150.
2. I = 40 mA × 100 over 250 = 16 mA. The 100 Ω takes the other 24 mA, and 16 + 24 = 40 mA.
Answer: 16 mA
Problem 9 of 10
A 9 V battery drives a 1.2 kΩ resistor in series with two 3.3 kΩ resistors in parallel. How much current flows through each 3.3 kΩ resistor?
Show the working
1. The parallel pair is 3.3 kΩ over 2 = 1.65 kΩ, so the total is 1.2 kΩ + 1.65 kΩ = 2.85 kΩ.
2. The battery's current: I = 9 V over 2.85 kΩ ≈ 3.158 mA.
3. The two equal branches share it equally: 3.158 mA ÷ 2 ≈ 1.58 mA each.
Answer: 1.58 mA
Problem 10 of 10
A 1 kΩ and a 2.2 kΩ resistor in series form one branch, and a 4.7 kΩ resistor forms a second branch beside it. Both branches sit across 5 V. How much current does the 5 V supply deliver?
Show the working
1. Each branch gets the full 5 V. The first branch is 1 kΩ + 2.2 kΩ = 3.2 kΩ.
2. Branch currents: 5 V over 3.2 kΩ ≈ 1.563 mA and 5 V over 4.7 kΩ ≈ 1.064 mA.
3. They add up where the branches rejoin: 1.563 + 1.064 ≈ 2.63 mA. As a check, the network is 3.2 kΩ ∥ 4.7 kΩ ≈ 1.90 kΩ, and 5 V ÷ 1.90 kΩ ≈ 2.63 mA.
Answer: 2.63 mA

Build it in a free lesson

“Combining resistors” is lesson 6 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.

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Now build it yourself

Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.

Try it yourself: free lesson

“Combining resistors” is lesson 6 of 7 in Junction City, Electronics World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.

Or start from the very first lesson, no account needed

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The ideas behind it

Questions people ask

How do you work out resistors in series and parallel?

In series, add them: 1 kΩ + 2.2 kΩ = 3.2 kΩ. In parallel, use product over sum for two resistors, R1 × R2 ÷ (R1 + R2), or add the reciprocals for more: 1/R = 1/R1 + 1/R2 + 1/R3. For a mixed network, replace one series or parallel group at a time.

Why is a parallel total smaller than the smallest resistor?

Every branch you add is another path for the current, so more current flows from the same voltage. More current from the same voltage means less resistance. Two 1 kΩ resistors in parallel act like 500 Ω.

How do I tell if two resistors are in parallel?

Parallel means both ends join the same two points, however the drawing looks. Follow the wires from each end of each resistor. If you reach the same two points without passing another part, they're in parallel.