Ohms Law Practice Problems and Worked Solutions

Level 1 — single resistor problems

Problem 1

A 12 V supply is connected across a 60 Ω resistor. Find the current.

Solution: I = V ÷ R = 12 ÷ 60 = 0.2 A (200 mA)

Problem 2

A resistor carries 25 mA when 5 V is across it. Find its resistance.

Solution: Convert first — 25 mA = 0.025 A. R = V ÷ I = 5 ÷ 0.025 = 200 Ω

Problem 3

What voltage appears across a 4.7 kΩ resistor carrying 3 mA?

Solution: V = I × R = 0.003 × 4700 = 14.1 V

Level 2 — bringing power in

Problem 4

A 230 V kettle is rated 2200 W. What current does it draw?

Solution: I = P ÷ V = 2200 ÷ 230 = 9.57 A

Problem 5

Find the power dissipated by a 330 Ω resistor carrying 40 mA.

Solution: P = I² × R = 0.04² × 330 = 0.0016 × 330 = 0.528 W

A quarter-watt resistor would overheat here — see resistor power rating for the margin to allow.

Problem 6

A 100 W bulb runs from 120 V. What is its hot resistance?

Solution: R = V² ÷ P = 14400 ÷ 100 = 144 Ω

Level 3 — applied problems

Problem 7

An LED needs 20 mA and drops 2.1 V. It runs from 5 V. What series resistor is required?

Solution: The resistor must drop 5 − 2.1 = 2.9 V.
R = V ÷ I = 2.9 ÷ 0.02 = 145 Ω → use the nearest standard 150 Ω.

Full method in our LED series resistor guide.

Problem 8

Two resistors, 220 Ω and 470 Ω, sit in series across 9 V. Find the current and the voltage across each.

Solution: Total R = 690 Ω. I = 9 ÷ 690 = 13.04 mA.
Across 220 Ω: 0.01304 × 220 = 2.87 V. Across 470 Ω: 6.13 V. Sum = 9 V ✓

Problem 9

A 24 V motor draws 3 A through a cable with 0.35 Ω total resistance. What voltage reaches the motor?

Solution: Cable drop = 3 × 0.35 = 1.05 V. Motor sees 24 − 1.05 = 22.95 V, and the cable wastes 3.15 W as heat.

Problem 10

A 60 W appliance runs 5 hours a day for 30 days at 0.28 per kWh. What does it cost?

Solution: Energy = 60 × 5 × 30 ÷ 1000 = 9 kWh. Cost = 9 × 0.28 = 2.52

Answer summary

Building speed

Once the method is reliable, the goal is to make it quick. Three exercises help more than simply doing more problems.

• Predict before you solve. Guess the order of magnitude first, then check. Being consistently within a factor of two means the intuition is forming.
• Solve the same circuit two ways. Find power with P = VI and again with P = I²R. Matching answers confirm both the arithmetic and the understanding.
• Work backwards. Start from an answer and invent the question. It forces you to see which pairs of values are actually sufficient.

Keep a note of every problem you got wrong and what the cause was. Almost everyone finds the same two or three mistakes repeating, and naming them is usually enough to stop them.

Where people lose marks

• Forgetting the milli- prefix. 25 mA is 0.025 A, not 25.
• Squaring the resistance in P = I²R. Only current is squared.
• Using supply voltage across one resistor in a series chain. Find the current first.
• Rounding too early. Carry full precision until the final line.