TL;DR:
Ohm's law links three quantities: voltage, current and resistance, as
V = IR.Current is what flows, voltage is the push, resistance is how much the part opposes the flow.
Power in a resistor is
P = VI = I²R, and it turns into heat.
You connect a 9 V battery to a 450 Ω resistor. How much current flows? It is the most basic question in electronics, and one short equation answers it. The same equation, rearranged, also tells you which resistor to choose for a given current.
The idea in plain words
A circuit is a closed loop that lets charge flow. Voltage (V, volts) is the push each unit of charge gets from the battery. Current (I, amperes) is how much charge passes a point each second. Resistance (R, ohms, Ω) is how much a part holds the current back.
Ohm's law says that, for a resistor, the current is proportional to the voltage across it: V = IR. Double the push and the current doubles. Double the resistance and the current halves.
Why the three quantities fit together
A useful picture is water in pipes. Voltage is like the pressure difference, current is like the flow rate and resistance is like a narrow pipe. The analogy has limits, but it gets the proportions right: more pressure gives more flow, a narrower pipe gives less.
Rearranging gives three forms: I = V / R, R = V / I and V = IR. A triangle diagram helps some people remember them, but it is simpler to remember one and rearrange.
Worked example: a 9 V battery across 450 ohms gives 20 mA
Known:
V = 9 V,R = 450 Ω. Unknown: I.Rearrange:
I = V / R.Substitute:
I = 9 / 450 = 0.02 A.Convert:
0.02 A = 20 mA(since 1 A = 1000 mA).Power in the resistor:
P = VI = 9 × 0.02 = 0.18 W.Second route:
P = I²R = 0.02² × 450 = 0.0004 × 450 = 0.18 W. Same answer.
The resistor therefore turns 0.18 W into heat. A common small resistor is rated for 1/4 W, which is 0.25 W, so it copes, with a bit of margin. A resistor rated for much less would get hot.
Now add a second 450 Ω resistor in series. The total resistance is 450 + 450 = 900 Ω, so I = 9 / 900 = 0.01 A = 10 mA. The same current flows through both resistors, and each one has V = 0.01 × 450 = 4.5 V across it. The two voltages add up to the battery's 9 V.
Try a different question to practise the rearrangement. You want an LED-style load to get about 10 mA from the same 9 V supply through a plain resistor, ignoring any drop across the load. Then R = V / I = 9 / 0.010 = 900 Ω. The question flips from finding current to choosing a part, and the same single equation does both jobs.
A graph of current against voltage for a true resistor is a straight line through the origin, and the slope is 1/R. A steeper line means smaller resistance. Parts that give a curve instead, like a bulb whose filament heats up, are not following Ohm's law over the whole range.
How to check your answer
Units. Volts divided by ohms gives amperes. Watts equal volts times amperes.
Convert prefixes carefully. 20 mA is 0.020 A. Using 20 A would be a thousand times too big.
Two routes. Compute power as
VIand asI²R. If they disagree, one value is wrong.Sum of voltages. In a series loop the voltage drops add up to the supply: 4.5 + 4.5 = 9.
Sense. More resistance must mean less current, as with 10 mA versus 20 mA.
Common mistakes
Forgetting the units conversion between mA and A, or between kΩ and Ω.
Using Ohm's law for parts that are not resistors. A diode or LED does not have a constant resistance, so
V = IRwith one fixed R does not describe it.Assuming a battery always supplies a fixed current. It supplies a fixed voltage (roughly) and the current follows from the load.
Mixing up voltage across a part with voltage at a point. Voltage is always between two points.
Ignoring power ratings, which is how parts overheat.
Safety: Stay with batteries for practice, never mains outlets. Do not short a battery with a wire: a very low resistance means a very large current, and the wire and the battery get hot. Never leave a circuit connected unattended, and do not use damaged or swollen batteries.
Where you will use it next
The next step is combining parts. Series and parallel circuits shows how resistances add in a line and share in branches. Like the trade in torque and levers, where force and distance swap, voltage and current swap here: for a fixed supply, raising R lowers I.
If you want to talk through a circuit while you work on a schematic, DuckyHelper is always there: press one key in any app, ask out loud, and it can see your screen and draw on it.
Frequently asked questions
Is a high resistance good or bad?
Neither. It depends on the job. A heating element needs a modest resistance so current can make heat. A sensor input often wants high resistance so it draws very little current.
Why is power I²R useful?
It shows that heat grows with the square of the current. Doubling the current in the same resistor gives four times the heat, which is why thin wires carrying large currents get hot.
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