Ohm's Law and Resistance — MDCAT Physics Lesson

MDCAT Physics · curriculum topic: Current Electricity

Learning objective

By the end of this lesson you should be able to apply V = IR confidently, state the condition under which Ohm's law holds, work out how the dimensions of a wire change its resistance, combine resistors in series and parallel, and calculate power dissipation.

The law itself, and its one condition

Ohm's law states that the current through a conductor is directly proportional to the potential difference across it, provided the physical conditions — temperature above all, and also the material, dimensions and physical state of the conductor — remain constant. Written as an equation, V = IR, where V is in volts, I in amperes and R in ohms.

That condition is where questions are usually set. A filament lamp does not obey Ohm's law over its full range because its temperature rises sharply with current, so its resistance increases and the graph curves. For a metal wire at constant temperature, a graph with potential difference V on the vertical axis and current I on the horizontal axis is a straight line through the origin whose gradient equals the resistance. (Plotted the other way round, with I vertical, the gradient is 1/R.)

What resistance actually depends on

For a uniform conductor, R = ρL / A, where ρ is resistivity, L is length and A is cross-sectional area. Resistivity is a property of the material and its temperature; length and area are properties of the particular piece of wire.

  • Double the length at constant area: resistance doubles.
  • Double the diameter: area increases four times (A = πr²), so resistance falls to one quarter.
  • Stretch a wire to twice its length at constant volume, with resistivity and temperature unchanged: L doubles and A halves, so resistance becomes four times greater.
  • Heating a metallic conductor increases its resistance; for a semiconductor such as silicon, resistance falls with rising temperature.

Combining resistors

  • Series: R_total = R₁ + R₂ + … The same current passes through each resistor and the total resistance is larger than the largest single resistor.
  • Parallel: 1/R_total = 1/R₁ + 1/R₂ + … The same potential difference is across each resistor and the total resistance is smaller than the smallest single resistor.
  • Two resistors in parallel can be handled quickly as R_total = R₁R₂ / (R₁ + R₂).

Power in a resistor

Electrical power is P = VI. Substituting Ohm's law gives two more forms that are often faster in MCQs: P = I²R when you know the current, and P = V²/R when you know the potential difference. Choosing the right form saves a step and avoids an unnecessary intermediate rounding.

Worked example — a series circuit with power

A 3 Ω and a 6 Ω resistor are connected in series across an 18 V battery of negligible internal resistance. Find the current in the circuit, the potential difference across the 6 Ω resistor, and the power dissipated in it.

  1. Total resistance in series: R = 3 + 6 = 9 Ω.
  2. Current from Ohm's law: I = V / R = 18 / 9 = 2 A. The same current flows through both resistors.
  3. Potential difference across the 6 Ω resistor: V = IR = 2 × 6 = 12 V. (The 3 Ω resistor takes the remaining 6 V, and 12 + 6 = 18 V, which checks out.)
  4. Power in the 6 Ω resistor: P = I²R = 2² × 6 = 24 W. Using P = V²/R gives 12² / 6 = 24 W, the same answer.

Answer: I = 2 A, V₆ = 12 V, P₆ = 24 W.

Practice questions

Question 1. A 12 V supply is connected across a 4 Ω resistor. The current is:

  • A. 0.33 A
  • B. 3 A
  • C. 8 A
  • D. 48 A

Answer: B. 3 A — From V = IR, I = V/R = 12/4 = 3 A.

Question 2. The length of a uniform wire is doubled while its cross-sectional area stays the same. Its resistance:

  • A. Halves
  • B. Stays the same
  • C. Doubles
  • D. Becomes four times greater

Answer: C. Doubles — R = ρL/A, so at constant ρ and A the resistance is directly proportional to length: doubling L doubles R.

Question 3. A 3 Ω and a 6 Ω resistor are connected in parallel. The equivalent resistance is:

  • A. 0.5 Ω
  • B. 2 Ω
  • C. 4.5 Ω
  • D. 9 Ω

Answer: B. 2 Ω — R = R₁R₂/(R₁+R₂) = (3 × 6)/9 = 2 Ω — smaller than the smaller resistor, as parallel combinations always are.

Question 4. Ohm's law holds for a metallic conductor provided that:

  • A. The current is very large
  • B. The temperature remains constant
  • C. The wire is very long
  • D. The supply is alternating

Answer: B. The temperature remains constant — Resistance of a metal rises with temperature, so the V–I relationship is linear only while temperature (and other physical conditions) are constant. This is why a filament lamp is non-ohmic.

Question 5. A current of 2 A flows through a 5 Ω resistor. The power dissipated is:

  • A. 2.5 W
  • B. 10 W
  • C. 20 W
  • D. 50 W

Answer: C. 20 W — P = I²R = 2² × 5 = 20 W. Using V = IR = 10 V and P = VI = 10 × 2 gives the same 20 W.

Common mistakes

  • Adding parallel resistances directly. In parallel you add reciprocals, and the answer must come out smaller than the smallest resistor — use that as a sanity check.
  • Treating a filament lamp as ohmic. Its resistance changes with temperature, so its V–I graph is curved.
  • Using P = VI with mismatched values, for example the supply voltage with the current through only one branch.
  • Forgetting that area depends on the square of the radius, so doubling the diameter quarters the resistance rather than halving it.

Work through Physics with explained MCQs

Current electricity is one of 16 Physics topics in the MDCAT Complete Preparation Bundle, each with chapter-wise MCQs and worked explanations, alongside Biology, Chemistry, English and logical reasoning.

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