ScienceAges 10-12

Series & parallel resistor networks

Calculate equivalent resistance for resistors in series (R₁ + R₂) or in parallel ((R₁ × R₂) ÷ (R₁ + R₂)). Review suitability for your learners before teaching.

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Age band:Ages 10-12 Ages 13-15

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Worksheet StudioPractice pack · Ages 10-12

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 18 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 18 Ω ∥ 36 Ω · 1/R_eq = 1/18 + 1/36
2. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 72 Ω · 1/R_eq = 1/24 + 1/72
3. Two resistors of 3 Ω and 6 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 3 Ω ∥ 6 Ω · 1/R_eq = 1/3 + 1/6
4. Two resistors of 18 Ω and 90 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 18 Ω ∥ 90 Ω · 1/R_eq = 1/18 + 1/90
5. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 120 Ω · 1/R_eq = 1/24 + 1/120
6. Two resistors of 6 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 6 Ω, R₂ = 18 Ω · R_eq = R₁ + R₂ = ?
7. Two resistors of 9 Ω and 18 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 9 Ω ∥ 18 Ω · 1/R_eq = 1/9 + 1/18
8. Two resistors of 40 Ω and 40 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 40 Ω ∥ 40 Ω · 1/R_eq = 1/40 + 1/40
9. Two resistors of 30 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 30 Ω ∥ 60 Ω · 1/R_eq = 1/30 + 1/60
10. Two resistors of 10 Ω and 10 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 10 Ω ∥ 10 Ω · 1/R_eq = 1/10 + 1/10
11. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 24 Ω · 1/R_eq = 1/12 + 1/24
12. Two resistors of 17 Ω and 14 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 17 Ω, R₂ = 14 Ω · R_eq = R₁ + R₂ = ?

12 questions · Seed 59 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 10-12

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 18 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
12 Ω
For two parallel resistors, R_eq = (18 × 36) ÷ (18 + 36) = 648 ÷ 54 = 12 Ω.
2. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
18 Ω
For two parallel resistors, R_eq = (24 × 72) ÷ (24 + 72) = 1728 ÷ 96 = 18 Ω.
3. Two resistors of 3 Ω and 6 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
2 Ω
For two parallel resistors, R_eq = (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω.
4. Two resistors of 18 Ω and 90 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
15 Ω
For two parallel resistors, R_eq = (18 × 90) ÷ (18 + 90) = 1620 ÷ 108 = 15 Ω.
5. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (24 × 120) ÷ (24 + 120) = 2880 ÷ 144 = 20 Ω.
6. Two resistors of 6 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
24 Ω
In a series circuit, resistances add directly: R_eq = 6 Ω + 18 Ω = 24 Ω.
7. Two resistors of 9 Ω and 18 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
6 Ω
For two parallel resistors, R_eq = (9 × 18) ÷ (9 + 18) = 162 ÷ 27 = 6 Ω.
8. Two resistors of 40 Ω and 40 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (40 × 40) ÷ (40 + 40) = 1600 ÷ 80 = 20 Ω.
9. Two resistors of 30 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (30 × 60) ÷ (30 + 60) = 1800 ÷ 90 = 20 Ω.
10. Two resistors of 10 Ω and 10 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
5 Ω
For two parallel resistors, R_eq = (10 × 10) ÷ (10 + 10) = 100 ÷ 20 = 5 Ω.
11. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
8 Ω
For two parallel resistors, R_eq = (12 × 24) ÷ (12 + 24) = 288 ÷ 36 = 8 Ω.
12. Two resistors of 17 Ω and 14 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
31 Ω
In a series circuit, resistances add directly: R_eq = 17 Ω + 14 Ω = 31 Ω.

12 questions · Seed 59 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 10-12

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 18 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 18 Ω ∥ 36 Ω · 1/R_eq = 1/18 + 1/36
2. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 72 Ω · 1/R_eq = 1/24 + 1/72
3. Two resistors of 3 Ω and 6 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 3 Ω ∥ 6 Ω · 1/R_eq = 1/3 + 1/6
4. Two resistors of 18 Ω and 90 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 18 Ω ∥ 90 Ω · 1/R_eq = 1/18 + 1/90
5. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 24 Ω ∥ 120 Ω · 1/R_eq = 1/24 + 1/120
6. Two resistors of 6 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 6 Ω, R₂ = 18 Ω · R_eq = R₁ + R₂ = ?
7. Two resistors of 9 Ω and 18 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 9 Ω ∥ 18 Ω · 1/R_eq = 1/9 + 1/18
8. Two resistors of 40 Ω and 40 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 40 Ω ∥ 40 Ω · 1/R_eq = 1/40 + 1/40
9. Two resistors of 30 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 30 Ω ∥ 60 Ω · 1/R_eq = 1/30 + 1/60
10. Two resistors of 10 Ω and 10 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 10 Ω ∥ 10 Ω · 1/R_eq = 1/10 + 1/10
11. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 12 Ω ∥ 24 Ω · 1/R_eq = 1/12 + 1/24
12. Two resistors of 17 Ω and 14 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 17 Ω, R₂ = 14 Ω · R_eq = R₁ + R₂ = ?

12 questions · Seed 59 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 10-12

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 18 Ω and 36 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
12 Ω
For two parallel resistors, R_eq = (18 × 36) ÷ (18 + 36) = 648 ÷ 54 = 12 Ω.
2. Two resistors of 24 Ω and 72 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
18 Ω
For two parallel resistors, R_eq = (24 × 72) ÷ (24 + 72) = 1728 ÷ 96 = 18 Ω.
3. Two resistors of 3 Ω and 6 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
2 Ω
For two parallel resistors, R_eq = (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω.
4. Two resistors of 18 Ω and 90 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
15 Ω
For two parallel resistors, R_eq = (18 × 90) ÷ (18 + 90) = 1620 ÷ 108 = 15 Ω.
5. Two resistors of 24 Ω and 120 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (24 × 120) ÷ (24 + 120) = 2880 ÷ 144 = 20 Ω.
6. Two resistors of 6 Ω and 18 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
24 Ω
In a series circuit, resistances add directly: R_eq = 6 Ω + 18 Ω = 24 Ω.
7. Two resistors of 9 Ω and 18 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
6 Ω
For two parallel resistors, R_eq = (9 × 18) ÷ (9 + 18) = 162 ÷ 27 = 6 Ω.
8. Two resistors of 40 Ω and 40 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (40 × 40) ÷ (40 + 40) = 1600 ÷ 80 = 20 Ω.
9. Two resistors of 30 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (30 × 60) ÷ (30 + 60) = 1800 ÷ 90 = 20 Ω.
10. Two resistors of 10 Ω and 10 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
5 Ω
For two parallel resistors, R_eq = (10 × 10) ÷ (10 + 10) = 100 ÷ 20 = 5 Ω.
11. Two resistors of 12 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
8 Ω
For two parallel resistors, R_eq = (12 × 24) ÷ (12 + 24) = 288 ÷ 36 = 8 Ω.
12. Two resistors of 17 Ω and 14 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
31 Ω
In a series circuit, resistances add directly: R_eq = 17 Ω + 14 Ω = 31 Ω.

12 questions · Seed 59 · Age guidance, not curriculum certification