Ohm's law is short enough to fit on a badge and deep enough to run half the Class 10 electricity chapter. Here is what it says, when it stops being true, and the numericals it turns into.
The statement
At constant physical conditions (mainly temperature), the current I through a conductor is directly proportional to the potential difference V across it:
V ∝ I, so V = IR.
R, the constant of proportionality, is the resistance of the conductor, measured in ohms (Ω). Rearranged: I = V/R and R = V/I.
The graph of V against I for an ohmic conductor is a straight line through the origin. Its slope is R. If the graph is I against V, the slope is 1/R — students mix these up in the MCQ every year.
Resistance depends on the wire
Resistance is not a fixed property of "copper"; it depends on the piece of copper:
R = ρL/A
- L, length: double the length, double the resistance.
- A, cross-sectional area: double the area, halve the resistance.
- ρ, resistivity: a property of the material only, in Ω·m. Copper ≈ 1.7 × 10⁻⁸ Ω·m; nichrome is about 60 times higher, which is why heaters use it.
When Ohm's law fails
Ohm's law is an experimental rule for metals at steady temperature, not a law of nature. It fails for:
- A filament lamp — the filament heats up, resistance rises, the V–I curve bends.
- A diode — current flows easily one way and barely at all the other.
- Electrolytes and gases — the relation is non-linear.
Such devices are called non-ohmic. The question "does the V–I graph pass through the origin as a straight line?" settles it.
Worked example 1 — the basic one
A 12 V battery is connected across a 4 Ω resistor. Find the current.
I = V/R = 12/4 = 3 A.
Worked example 2 — find the resistance
A heater draws 5 A from a 220 V supply. Find its resistance.
R = V/I = 220/5 = 44 Ω.
Worked example 3 — series
Resistors of 2 Ω, 3 Ω and 5 Ω are in series across 20 V. Find the current and the voltage across the 5 Ω resistor.
R_total = 2 + 3 + 5 = 10 Ω. I = 20/10 = 2 A (the same through each). V across 5 Ω = IR = 2 × 5 = 10 V.
Worked example 4 — parallel
Resistors of 6 Ω and 3 Ω are in parallel across 12 V. Find the total current.
1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 2 Ω. I = 12/2 = 6 A. Check by branches: 12/6 = 2 A and 12/3 = 4 A; total 6 A. The bigger current takes the smaller resistance.
Worked example 5 — resistivity
A wire of length 2 m and area 1 mm² has resistance 0.034 Ω. Find its resistivity.
Convert area: 1 mm² = 1 × 10⁻⁶ m². ρ = RA/L = 0.034 × 10⁻⁶ / 2 = 1.7 × 10⁻⁸ Ω·m — copper.
Worked example 6 — stretching a wire
A wire of resistance R is stretched to double its length. Find the new resistance.
Volume is constant, so doubling L halves A. R' = ρ(2L)/(A/2) = 4ρL/A = 4R. (Answering "2R" is the classic slip: the area changes too.)
Where students go wrong
- Writing the slope of the I–V graph as R. It is 1/R.
- Adding parallel resistances directly. Use reciprocals, or for two resistors the product-over-sum shortcut: R = R₁R₂/(R₁ + R₂).
- Leaving area in mm² inside a formula that wants m².
- Saying Ohm's law "always holds". State the constant-temperature condition — it is usually a mark on its own.
How it is tested
Class 10 asks for the statement with the condition, the V–I graph, R = ρL/A, and series/parallel numericals like the ones above. Class 12 builds the same ideas into Kirchhoff's laws, the Wheatstone bridge and the potentiometer; the arithmetic is the same, the circuits are bigger.