Ohm's law says that the current through a resistor is proportional to the voltage across it: double the voltage and the current doubles. The constant that links them is the resistance. It is the first formula of electricity, and it is enough to analyze most circuits in a first physics course.
This article covers the law, electric power, and the two basic ways to connect resistors, in series and in parallel. You can test every idea in the circuit lab before reading the formulas in detail.
What Ohm's law says
Three quantities describe a simple circuit. Voltage ($latex V$) is the push a battery gives to the charges, measured in volts. Current ($latex I$) is the amount of charge that flows per second, measured in amperes. Resistance ($latex R$) tells how strongly the material opposes that flow, measured in ohms ($latex \Omega$).
Ohm's law
$latex V = I \times R$
Solve for whichever quantity you need: $latex I = \dfrac{V}{R}$ gives the current and $latex R = \dfrac{V}{I}$ gives the resistance. The Ohm triangle is a quick memory aid: cover the letter you want and read the formula from the other two.
A water analogy helps. Voltage is the pressure in a pipe, current is the flow of water, and resistance is how narrow the pipe is. More pressure pushes more water through the same pipe, and a narrower pipe lets less through at the same pressure.
The law is exact for resistors at a constant temperature. A light bulb filament heats up and its resistance rises, so for a bulb the graph of voltage against current bends instead of staying straight.
| Quantity | Symbol | Unit | Abbreviation |
|---|---|---|---|
| Voltage | V | volt | V |
| Current | I | ampere | A |
| Resistance | R | ohm | Ω |
| Power | P | watt | W |
Watch the prefixes: 1 mA = 0.001 A and 1 kΩ = 1000 Ω. If prefixes feel slippery, review scientific notation.
Electric power
A resistor turns electrical energy into heat at a rate called power. Multiplying voltage and current gives it directly, and Ohm's law produces two more forms:
$$P = V \times I = I^2 \times R = \frac{V^2}{R}$$
Use the form that matches the data you have. Power in kilowatts times time in hours gives energy in kilowatt-hours: a 1.2 kW heater running for 2 hours uses 2.4 kWh.
Circuit lab: change the voltage and the resistors
Pick one resistor, two in series or two in parallel, then move the sliders. The moving dots show the current, and each resistor glows more as it dissipates more power. The readout below the drawing updates with every change.
Circuit lab: change the voltage and the resistors
Try this. In series mode, make R2 much larger than R1 and see which resistor takes most of the voltage. Then switch to parallel and notice that the total resistance ends up smaller than either resistor.
Resistors in series
In a series circuit there is a single path, so the same current flows through every resistor. The voltages add up to the battery voltage, and the resistances add:
Series connection
$latex R_{total} = R_1 + R_2 + R_3 + \dots$
Each resistor takes a share of the voltage proportional to its resistance, $latex V_1 = V \times \dfrac{R_1}{R_{total}}$. This is the voltage divider. If one resistor in a series string breaks, the current stops everywhere, which is why old holiday light strings went dark all at once.
Resistors in parallel
In a parallel circuit each resistor is connected directly across the battery, so all of them have the same voltage. The currents add, and the total resistance is smaller than the smallest resistor:
Parallel connection
$latex \dfrac{1}{R_{total}} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$
For two resistors the formula simplifies to $latex R_{total} = \dfrac{R_1 \times R_2}{R_1 + R_2}$. Every new branch gives the current one more path, so the total current grows. That is why plugging many devices into one power strip can overload it. Adding the fractions $latex \dfrac{1}{R_1} + \dfrac{1}{R_2}$ needs a common denominator, so adding fractions with unlike denominators is useful background.
| Series | Parallel | |
|---|---|---|
| Current | Same in every resistor | Splits between branches |
| Voltage | Splits between resistors | Same across every branch |
| Total resistance | Sum of the resistances | Smaller than the smallest |
| If one resistor breaks | All current stops | Other branches keep working |
Solved examples
Example 1
A 12 V battery is connected to a 40 Ω resistor. Find the current and the power dissipated.
Solve Ohm's law for the current:
$$I = \frac{V}{R} = \frac{12\ \text{V}}{40\ \Omega} = 0.3\ \text{A}$$
The power is voltage times current:
$$P = V \times I = 12\ \text{V} \times 0.3\ \text{A} = 3.6\ \text{W}$$
The resistor carries 0.3 A and dissipates 3.6 W.
Example 2
A 9 V battery drives two resistors in series, 6 Ω and 12 Ω. Find the current and the voltage across each resistor.
Add the resistances:
$$R_{total} = 6\ \Omega + 12\ \Omega = 18\ \Omega$$
The current is the same everywhere:
$$I = \frac{9\ \text{V}}{18\ \Omega} = 0.5\ \text{A}$$
Each voltage comes from Ohm's law on that resistor:
$$V_1 = 0.5 \times 6 = 3\ \text{V}$$
$$V_2 = 0.5 \times 12 = 6\ \text{V}$$
Check: 3 V + 6 V = 9 V, the battery voltage. The 12 Ω resistor takes twice the voltage of the 6 Ω one.
Example 3
The same 6 Ω and 12 Ω resistors are now connected in parallel to a 12 V battery. Find the total resistance, each branch current and the total current.
Total resistance of two parallel resistors:
$$R_{total} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\ \Omega$$
Each branch has the full 12 V:
$$I_1 = \frac{12}{6} = 2\ \text{A}$$
$$I_2 = \frac{12}{12} = 1\ \text{A}$$
The currents add: $latex I = 2 + 1 = 3$ A. Check with the total resistance: $latex 12 / 4 = 3$ A. The battery now delivers 36 W. With the same resistors in series it would deliver only 8 W ($latex 12^2 / 18$), because the parallel circuit has less total resistance.
Example 4
A heater is rated 1200 W at 120 V. What current does it draw and what is its resistance?
From $latex P = V \times I$:
$$I = \frac{P}{V} = \frac{1200\ \text{W}}{120\ \text{V}} = 10\ \text{A}$$
Then Ohm's law gives the resistance:
$$R = \frac{V}{I} = \frac{120\ \text{V}}{10\ \text{A}} = 12\ \Omega$$
The same result comes from $latex R = V^2 / P$: $latex 120^2 / 1200 = 12$ Ω.
Common mistakes
- Mixed units. Convert before you calculate: 4 mA is 0.004 A and 2.5 kΩ is 2500 Ω.
- Adding parallel resistances directly. In parallel you add the reciprocals, or use the product over sum shortcut for two resistors.
- Forgetting the last step. After adding $latex 1/R$ terms you have $latex 1/R_{total}$. Invert it to get $latex R_{total}$.
- Thinking current is used up. A resistor converts energy, not charge. The current entering a resistor equals the current leaving it.
- Using the wrong voltage. For the power of one resistor use the voltage across that resistor, not the battery voltage, unless it is directly connected to the battery.
Practice problems
Problem 1
A 220 Ω resistor is connected to a 9 V battery. Find the current in mA.
Answer
$$I = \frac{9}{220} = 0.0409\ \text{A}$$
That is 40.9 mA.
Problem 2
A bulb draws 0.25 A when connected to 6 V. Find its resistance and its power.
Answer
$$R = \frac{6}{0.25} = 24\ \Omega$$
$$P = 6 \times 0.25 = 1.5\ \text{W}$$
Resistance 24 Ω, power 1.5 W.
Problem 3
Three resistors of 100 Ω, 150 Ω and 50 Ω are in series with a 12 V battery. Find the current and the voltage across the 150 Ω resistor.
Answer
$$R_{total} = 100 + 150 + 50 = 300\ \Omega$$
$$I = \frac{12}{300} = 0.04\ \text{A}$$
$$V_{150} = 0.04 \times 150 = 6\ \text{V}$$
Current 0.04 A, voltage 6 V.
Problem 4
A 30 Ω and a 60 Ω resistor are in parallel across a 9 V battery. Find the total resistance and the current in each branch.
Answer
$$R_{total} = \frac{30 \times 60}{90} = 20\ \Omega$$
$$I_1 = \frac{9}{30} = 0.3\ \text{A}$$
$$I_2 = \frac{9}{60} = 0.15\ \text{A}$$
Total resistance 20 Ω, branch currents 0.3 A and 0.15 A, total 0.45 A.
Problem 5
A 10 Ω resistor dissipates 40 W. Find the current through it and the voltage across it.
Answer
$$I = \sqrt{\frac{P}{R}} = \sqrt{\frac{40}{10}} = 2\ \text{A}$$
$$V = I \times R = 2 \times 10 = 20\ \text{V}$$
Current 2 A, voltage 20 V.
Problem 6
A 2.5 kΩ resistor carries 4 mA. What is the voltage across it?
Answer
$$V = I \times R = 0.004 \times 2500 = 10\ \text{V}$$
Convert first: 4 mA = 0.004 A and 2.5 kΩ = 2500 Ω. The voltage is 10 V.
Frequently asked questions
What is the difference between voltage and current?
Voltage is the push that drives the charges, and current is the flow that results. A battery can have voltage with no current if the circuit is open, like a pressurized pipe with a closed tap.
Does Ohm's law work for every device?
No. It describes ohmic materials, whose resistance stays constant. Diodes, LEDs and light bulbs do not follow a straight line, although Ohm's law still gives the ratio V/I at each operating point.
Why is the total resistance smaller in parallel?
Each new branch is one more path for the current. More paths mean more total current for the same voltage, and by Ohm's law more current at the same voltage means lower resistance.
How do I solve a circuit with series and parallel parts?
Reduce it step by step. Replace each parallel group with its equivalent resistance, add the series resistances, find the total current, and then work back to each part with Ohm's law.
Related calculators
- Ohm's law calculator: find voltage, current, resistance or power from any two values.
See also
- Examples of direct proportion: current is directly proportional to voltage.
- Examples of inverse proportion: for a fixed voltage, current is inversely proportional to resistance.
- Adding fractions with unlike denominators: the skill behind the parallel formula.
- Scientific notation: handle prefixes such as milli and kilo.