ScienceAges 13-15

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 13-15

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 20 Ω and 20 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 20 Ω ∥ 20 Ω · 1/R_eq = 1/20 + 1/20
2. Two resistors of 9 Ω and 11 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 9 Ω, R₂ = 11 Ω · R_eq = R₁ + R₂ = ?
3. 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
4. Two resistors of 6 Ω and 23 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 6 Ω, R₂ = 23 Ω · R_eq = R₁ + R₂ = ?
5. Two resistors of 14 Ω and 9 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 14 Ω, R₂ = 9 Ω · R_eq = R₁ + R₂ = ?
6. 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
7. 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
8. Two resistors of 60 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 60 Ω ∥ 60 Ω · 1/R_eq = 1/60 + 1/60
9. Two resistors of 23 Ω and 6 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 23 Ω, R₂ = 6 Ω · R_eq = R₁ + R₂ = ?
10. Two resistors of 8 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 8 Ω ∥ 24 Ω · 1/R_eq = 1/8 + 1/24
11. Two resistors of 25 Ω and 100 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 25 Ω ∥ 100 Ω · 1/R_eq = 1/25 + 1/100
12. Two resistors of 32 Ω and 96 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 32 Ω ∥ 96 Ω · 1/R_eq = 1/32 + 1/96

12 questions · Seed 54 · Age guidance, not curriculum certification

View the answer key

Worksheet StudioAnswer key · Ages 13-15

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 20 Ω and 20 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
10 Ω
For two parallel resistors, R_eq = (20 × 20) ÷ (20 + 20) = 400 ÷ 40 = 10 Ω.
2. Two resistors of 9 Ω and 11 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
20 Ω
In a series circuit, resistances add directly: R_eq = 9 Ω + 11 Ω = 20 Ω.
3. 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.
4. Two resistors of 6 Ω and 23 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
29 Ω
In a series circuit, resistances add directly: R_eq = 6 Ω + 23 Ω = 29 Ω.
5. Two resistors of 14 Ω and 9 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
23 Ω
In a series circuit, resistances add directly: R_eq = 14 Ω + 9 Ω = 23 Ω.
6. 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.
7. 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.
8. Two resistors of 60 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
30 Ω
For two parallel resistors, R_eq = (60 × 60) ÷ (60 + 60) = 3600 ÷ 120 = 30 Ω.
9. Two resistors of 23 Ω and 6 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
29 Ω
In a series circuit, resistances add directly: R_eq = 23 Ω + 6 Ω = 29 Ω.
10. Two resistors of 8 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
6 Ω
For two parallel resistors, R_eq = (8 × 24) ÷ (8 + 24) = 192 ÷ 32 = 6 Ω.
11. Two resistors of 25 Ω and 100 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (25 × 100) ÷ (25 + 100) = 2500 ÷ 125 = 20 Ω.
12. Two resistors of 32 Ω and 96 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
24 Ω
For two parallel resistors, R_eq = (32 × 96) ÷ (32 + 96) = 3072 ÷ 128 = 24 Ω.

12 questions · Seed 54 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 13-15

Series & parallel resistor networks

Take your time. Show your thinking.

1. Two resistors of 20 Ω and 20 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 20 Ω ∥ 20 Ω · 1/R_eq = 1/20 + 1/20
2. Two resistors of 9 Ω and 11 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 9 Ω, R₂ = 11 Ω · R_eq = R₁ + R₂ = ?
3. 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
4. Two resistors of 6 Ω and 23 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 6 Ω, R₂ = 23 Ω · R_eq = R₁ + R₂ = ?
5. Two resistors of 14 Ω and 9 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 14 Ω, R₂ = 9 Ω · R_eq = R₁ + R₂ = ?
6. 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
7. 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
8. Two resistors of 60 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 60 Ω ∥ 60 Ω · 1/R_eq = 1/60 + 1/60
9. Two resistors of 23 Ω and 6 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
Series: R₁ = 23 Ω, R₂ = 6 Ω · R_eq = R₁ + R₂ = ?
10. Two resistors of 8 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 8 Ω ∥ 24 Ω · 1/R_eq = 1/8 + 1/24
11. Two resistors of 25 Ω and 100 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 25 Ω ∥ 100 Ω · 1/R_eq = 1/25 + 1/100
12. Two resistors of 32 Ω and 96 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
Parallel: 32 Ω ∥ 96 Ω · 1/R_eq = 1/32 + 1/96

12 questions · Seed 54 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 13-15

Series & parallel resistor networks

Answers and reasoning. Review before teaching.

1. Two resistors of 20 Ω and 20 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
10 Ω
For two parallel resistors, R_eq = (20 × 20) ÷ (20 + 20) = 400 ÷ 40 = 10 Ω.
2. Two resistors of 9 Ω and 11 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
20 Ω
In a series circuit, resistances add directly: R_eq = 9 Ω + 11 Ω = 20 Ω.
3. 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.
4. Two resistors of 6 Ω and 23 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
29 Ω
In a series circuit, resistances add directly: R_eq = 6 Ω + 23 Ω = 29 Ω.
5. Two resistors of 14 Ω and 9 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
23 Ω
In a series circuit, resistances add directly: R_eq = 14 Ω + 9 Ω = 23 Ω.
6. 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.
7. 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.
8. Two resistors of 60 Ω and 60 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
30 Ω
For two parallel resistors, R_eq = (60 × 60) ÷ (60 + 60) = 3600 ÷ 120 = 30 Ω.
9. Two resistors of 23 Ω and 6 Ω are connected in series across a circuit. Calculate their combined equivalent resistance.
29 Ω
In a series circuit, resistances add directly: R_eq = 23 Ω + 6 Ω = 29 Ω.
10. Two resistors of 8 Ω and 24 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
6 Ω
For two parallel resistors, R_eq = (8 × 24) ÷ (8 + 24) = 192 ÷ 32 = 6 Ω.
11. Two resistors of 25 Ω and 100 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
20 Ω
For two parallel resistors, R_eq = (25 × 100) ÷ (25 + 100) = 2500 ÷ 125 = 20 Ω.
12. Two resistors of 32 Ω and 96 Ω are connected in parallel. Using R_eq = (R₁ × R₂) ÷ (R₁ + R₂), calculate their equivalent resistance.
24 Ω
For two parallel resistors, R_eq = (32 × 96) ÷ (32 + 96) = 3072 ÷ 128 = 24 Ω.

12 questions · Seed 54 · Age guidance, not curriculum certification