Practice problems

Ohm's law practice problems

Ohm's law, V = I × R, is the sum behind almost every circuit you'll build. These ten problems start with a single step and work up to choosing an LED resistor and checking a resistor won't overheat. Type your answer to check it, and open the working whenever you want it.

The method

V = I × R

I = V over R when you want the current
R = V over I when you want the resistance
P = V × I for the power, the heat a resistor gives off

Keep the units in step: volts, amps and ohms, or volts, milliamps and kilohms. So 9 V across 1 kΩ is 9 mA, with no zeros to count.

The problems

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

Problem 1 of 10
A 12 V supply sits across a 600 Ω resistor. What current flows?
Show the working
1. You want the current, so use I = V over R.
2. I = 12 V over 600 Ω = 0.02 A
3. 0.02 A is 20 mA.
Answer: 20 mA
Problem 2 of 10
30 mA flows through a 150 Ω resistor. What's the voltage across it?
Show the working
1. You want the voltage, so use V = I × R, with the current in amps: 30 mA = 0.03 A.
2. V = 0.03 A × 150 Ω = 4.5 V
Answer: 4.5 V
Problem 3 of 10
A part draws 25 mA from a 5 V supply. What resistance does it look like?
Show the working
1. You want the resistance, so use R = V over I, with 25 mA = 0.025 A.
2. R = 5 V over 0.025 A = 200 Ω
Answer: 200 Ω
Problem 4 of 10
0.5 A flows through a 22 Ω resistor. What's the voltage across it?
Show the working
1. V = I × R = 0.5 A × 22 Ω = 11 V
Answer: 11 V
Problem 5 of 10
A 9 V battery sits across a 4.7 kΩ resistor. What current flows, in milliamps?
Show the working
1. Volts divided by kilohms gives milliamps, so there are no zeros to count.
2. I = 9 V over 4.7 kΩ = 1.91 mA
Answer: 1.91 mA
Problem 6 of 10
A sensor draws 2 mA from 3.3 V. What resistance does it look like, in kilohms?
Show the working
1. Volts divided by milliamps gives kilohms.
2. R = 3.3 V over 2 mA = 1.65 kΩ
Answer: 1.65 kΩ
Problem 7 of 10
A resistor carries 15 mA with 6 V across it. The voltage is raised to 12 V. What current flows now?
Show the working
1. The resistance doesn't change: R = 6 V over 15 mA = 0.4 kΩ = 400 Ω.
2. I = 12 V over 0.4 kΩ = 30 mA. Double the voltage, double the current: 30 mA.
Answer: 30 mA
Problem 8 of 10
A red LED drops 2 V and should run at 20 mA from a 9 V battery. What resistor goes in series with it?
Show the working
1. The resistor gets what the LED leaves: 9 V − 2 V = 7 V.
2. R = 7 V over 0.02 A = 350 Ω. The next standard value up is 390 Ω.
Answer: 350 Ω
Problem 9 of 10
A 100 Ω resistor sits across 12 V. How much power does it turn into heat?
Show the working
1. First the current: I = 12 V over 100 Ω = 0.12 A.
2. P = V × I = 12 V × 0.12 A = 1.44 W. A 2 W resistor at least; a ¼ W one would burn.
Answer: 1.44 W
Problem 10 of 10
A 1 kΩ resistor has 5 V across it. How much power does it use, in milliwatts?
Show the working
1. Use P = V² over R, which skips finding the current.
2. P = (5 V)² over 1000 Ω = 25 over 1000 W = 0.025 W
3. 0.025 W is 25 mW, well inside a ¼ W (250 mW) resistor's rating.
Answer: 25 mW

Build it in a free lesson

“Ohm's law” is lesson 5 of 6 in Copper Flats, Electronics World 1. Free with an account. Lessons unlock in order, so you’ll do the 4 before it first.

Free lesson

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

“Ohm's law” is lesson 5 of 6 in Copper Flats, Electronics World 1. Free with an account. Lessons unlock in order, so you’ll do the 4 before it first.

Or start from the very first lesson, no account needed

Check it with a calculator

The ideas behind it

Questions people ask

How do I know which form of Ohm's law to use?

Write down the two values you know and the one you want. Want the current? I = V ÷ R. The voltage? V = I × R. The resistance? R = V ÷ I. Cover the one you want on the V-I-R triangle and the other two show the sum.

Do I have to convert milliamps to amps?

Either convert (20 mA = 0.02 A) or keep the units that fit together: volts ÷ kilohms gives milliamps, and volts ÷ milliamps gives kilohms. Mixing amps with kilohms is where most wrong answers come from.

Why doesn't Ohm's law work for the LED itself?

An LED's current isn't proportional to its voltage: it barely conducts below its forward voltage, then the current shoots up. Take its forward voltage off the supply first and apply Ohm's law to the resistor, as problem 8 does.