Series and Parallel Resistor Practice Problems with Solutions

Series values add; parallel values add as reciprocals. Work mixed networks from the inside out.

Key formulas

R_series = R₁ + R₂ + …

(1)/(R_par) = (1)/(R₁) + (1)/(R₂) + …

R₁ ∥ R₂ = (R₁ × R₂)/(R₁ + R₂)

A parallel combination is always smaller than its smallest resistor. If yours is larger, recheck.
n equal resistors in parallel give the value divided by n.
Capacitors are the other way round: parallel values add, series values combine as reciprocals.

Problems 10

2 multiple-choice and 8 numeric problems. Type numbers in the unit shown next to the box (for example 2.55); SI prefixes such as 4.7k also work. Small differences from rounding along the way count as correct (±2%). When you are done, press Submit and grade at the bottom to see the answers and step-by-step solutions.

1 kΩ, 2.2 kΩ and 4.7 kΩ resistors are connected in series. What is the total resistance, in kΩ?
kΩ
Answer: 7.9 kΩ
Solution
1. Series resistances simply add: R = R₁ + R₂ + R₃.
2. R = 1 + 2.2 + 4.7 = 7.9 kΩ
Check it with the Series and parallel calculator
Two 10 kΩ resistors are connected in parallel. What is the total resistance, in kΩ?
kΩ
Answer: 5 kΩ
Solution
1. For two resistors in parallel, R = (R₁ × R₂)/(R₁ + R₂).
2. R = (10 × 10)/(10 + 10) kΩ = 5 kΩ
3. Two equal resistors in parallel always give half the value.
Check it with the Series and parallel calculator
1 kΩ and 2.2 kΩ are connected in parallel. What is the total resistance, in ohms?
Ω
Answer: 687.5 Ω
Solution
1. R = (1,000 × 2,200)/(1,000 + 2,200) = (2,200,000)/(3,200) = 687.5 Ω
2. A parallel combination is always smaller than its smallest resistor (1 kΩ). If you get more than 1 kΩ, recheck.
Check it with the Series and parallel calculator
1 kΩ, 2 kΩ and 4 kΩ are all connected in parallel. What is the total resistance, in ohms?
Ω
Answer: 571.4 Ω
Solution
1. For three or more, add the reciprocals: (1)/(R) = (1)/(R₁) + (1)/(R₂) + (1)/(R₃).
2. In kΩ: (1)/(1) + (1)/(2) + (1)/(4) = 1 + 0.5 + 0.25 = 1.75 (in kΩ⁻¹).
3. R = (1)/(1.75) kΩ = 0.5714 kΩ, or about 571.4 Ω.
Check it with the Series and parallel calculator
You add one more resistor in parallel with an existing network. What happens to the total resistance?

AIt increasesBIt stays the sameCIt decreasesDIt may rise or fall, depending on the new value
Answer: C. It decreases
Solution
1. A parallel branch adds another path for current, so more total current flows at the same voltage.
2. A positive term is added to the sum of reciprocals, so the total always decreases, however large the new resistor is.
Check it with the Series and parallel calculator
A 470 Ω resistor is in series with two 1 kΩ resistors that are in parallel with each other. What is the total resistance, in ohms?
Ω
Answer: 970 Ω
Solution
1. Start with the parallel pair: 1 kΩ ∥ 1 kΩ = 500 Ω.
2. Then add in series: 470 + 500 = 970 Ω.
Check it with the Series and parallel calculator
1 kΩ and 2 kΩ are in series across a 9 V supply. What current flows, in mA?
mA
Answer: 3 mA
Solution
1. The current is the same everywhere in a series loop, and the total resistance is 1 + 2 = 3 kΩ.
2. I = (9 V)/(3 kΩ) = 3 mA
3. Volts divided by kΩ gives mA directly.
Check it with the Series and parallel calculator
100 Ω and 300 Ω are in parallel across a 12 V supply. What total current does the supply deliver, in mA?
mA
Answer: 160 mA
Solution
1. Total resistance: (100 × 300)/(100 + 300) = 75 Ω
2. I = (12 V)/(75 Ω) = 0.16 A = 160 mA
3. Check: each branch carries (12)/(100) = 120 mA and (12)/(300) = 40 mA, which add to 160 mA.
Check it with the Series and parallel calculator
Two 100 nF capacitors are connected in series. What is the total capacitance?

A200 nFB100 nFC50 nFD25 nF
Answer: C. 50 nF
Solution
1. Capacitors work the opposite way to resistors: parallel values add, series values combine as reciprocals.
2. Two equal capacitors in series give half, so 50 nF.
Check it with the Series and parallel calculator
You want to bring a 10 kΩ resistor down to 6 kΩ by adding a resistor Rx in parallel. What is Rx, in kΩ?
kΩ
Answer: 15 kΩ
Solution
1. Solve the reciprocal formula for Rx: (1)/(R_x) = (1)/(6) − (1)/(10) (in kΩ).
2. (1)/(6) − (1)/(10) = 0.06667, so R_x = (1)/(0.06667) = 15 kΩ.
3. Check: (10 × 15)/(10 + 15) = 6 kΩ. 15 kΩ is an E24 value.
Check it with the Series and parallel calculator

Not graded yet.

Scores are kept only in this browser and are never sent anywhere.