Ohm's Law Practice Problems (Voltage, Current, Resistance, Power) with Solutions

V = I·R

I = (V)/(R)

R = (V)/(I)

P = V·I = I²·R = (V²)/(R)

Problems 10

3 multiple-choice and 7 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.

A 4.7 kΩ resistor is connected across a 12 V supply. What current flows through it, in mA?
mA
Answer: 2.553 mA
Solution
1. Solve Ohm's law for current: I = (V)/(R).
2. I = (12 V)/(4,700 Ω) = 0.002553 A
3. Multiply by 1000 for mA: 2.553 mA.
Check it with the Ohm's law and power calculator
20 mA flows through a 330 Ω resistor. What is the voltage across it, in volts?
V
Answer: 6.6 V
Solution
1. Convert units first: 20 mA = 0.020 A.
2. V = I·R = 0.020 A × 330 Ω = 6.6 V
3. Multiplying 20 × 330 without converting mA gives 6600 V, a common slip.
Check it with the Ohm's law and power calculator
A resistor on a 9 V battery draws 15 mA. What is its resistance, in ohms?
Ω
Answer: 600 Ω
Solution
1. R = (V)/(I), with the current in amps: 15 mA = 0.015 A.
2. R = (9 V)/(0.015 A) = 600 Ω
Check it with the Ohm's law and power calculator
There are 5 V across a 220 Ω resistor. How much power does it dissipate, in mW?
mW
Answer: 113.6 mW
Solution
1. With voltage and resistance known, P = (V²)/(R) is quickest.
2. P = ((5 V)²)/(220 Ω) = (25)/(220) W = 0.1136 W
3. In mW that is 113.6 mW. Finding the current first (22.7 mA) and using P = V·I gives the same.
Check it with the Ohm's law and power calculator
A resistor dissipates 0.15 W. Following the usual rule of at least twice the dissipated power, which rating should you pick?

A⅛ W (0.125 W)B¼ W (0.25 W)C½ W (0.5 W)D2 W
Answer: C. ½ W (0.5 W)
Solution
1. The rating should be at least 0.15 W × 2 = 0.3 W.
2. ¼ W (0.25 W) is above the dissipation but leaves too little margin, so the part runs hot.
3. The smallest standard rating above 0.3 W is ½ W. 2 W also works but is needlessly large.
Check it with the Ohm's law and power calculator
What is the largest current a 1 kΩ, ¼ W resistor can carry at its full power rating, in mA?
mA
Answer: 15.81 mA
Solution
1. Solving P = I²·R for current gives I = √(P)/(R).
2. I = √(0.25 W)/(1,000 Ω) = √0.00025 A = 0.01581 A
3. That is about 15.81 mA. In practice, stay around 70% of this for margin.
Check it with the Ohm's law and power calculator
If the voltage across the same resistor is doubled, what happens to the power it dissipates?

AIt doublesBIt becomes four times largerCIt halvesDIt does not change
Answer: B. It becomes four times larger
Solution
1. P = (V²)/(R) with R unchanged.
2. Doubling V makes V² grow by 2² = 4, so the power becomes four times larger. The current also doubles, so P = V·I gives 2 × 2 = 4 as well.
Check it with the Ohm's law and power calculator
You want 15 mA through a red LED (forward voltage 2.0 V) from a 5 V supply. What series resistor do you need, in ohms?
Ω
Answer: 200 Ω
Solution
1. The resistor sees the supply minus the LED voltage: 5 V − 2.0 V = 3.0 V.
2. R = (V_s − V_f)/(I) = (3.0 V)/(0.015 A) = 200 Ω
3. 200 Ω is an E24 standard value. The resistor dissipates 3.0 V × 0.015 A = 45 mW, so ¼ W is plenty.
Check it with the LED resistor calculator
A 12 V lamp draws 0.5 A while lit. What is the filament's resistance when hot, in ohms?
Ω
Answer: 24 Ω
Solution
1. R = (V)/(I) = (12 V)/(0.5 A) = 24 Ω
2. A cold filament has much lower resistance, so a multimeter on an unlit lamp reads far less. That is why lamps draw a large inrush current when switched on.
Check it with the Ohm's law and power calculator
Which of these does not have current proportional to voltage, so Ohm's law cannot be applied directly?

ACarbon-film resistorBMetal-film resistorCLEDDCopper wire at constant temperature
Answer: C. LED
Solution
1. Resistors and wire at a steady temperature are linear: double the voltage, double the current.
2. An LED is a diode: near its forward voltage the current rises exponentially. That is why an LED needs a separate current-limiting resistor.
Check it with the LED resistor calculator

Not graded yet.

Ohm's Law Practice Problems

Key formulas

Problems 10