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III. Long Answer Questions · Q2

Q.Obtain the macroscopic form of Ohm's law from its microscopic form and discuss its limitations.

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Step 1. Start from the microscopic form, J=σEJ=\sigma E, for a uniform conductor of length l and cross-sectional area A.

Step 2. Assuming the field is uniform along the wire, the potential difference across it is V=ElV=El, so E=V/lE=V/l; also, J=I/AJ=I/A.

Step 3. Substituting both into J=σEJ=\sigma E gives I/A=σ(V/l)I/A=\sigma(V/l), which rearranges to V=I(lσA)V=I\left(\dfrac{l}{\sigma A}\right).

Step 4. The bracketed quantity depends only on the conductor's geometry and material, and is DEFINED as its resistance, R=l/(σA)=ρl/AR=l/(\sigma A)=\rho l/A (using ρ=1/σ\rho=1/\sigma).

Step 5. This gives the macroscopic form of Ohm's law, V=IRV=IR -- the everyday relation between the terminal quantities V and I of a specific wire, as opposed to the microscopic relation between the FIELD quantities J and E at a point.

Step 6. Limitations: this macroscopic form silently assumes R is a fixed constant, independent of V, I, or temperature. It applies cleanly only to ohmic materials (straight-line I-V graph through the origin); non-ohmic devices (e.g. a diode, or a filament lamp that heats and changes resistance as more current flows) have no single constant R, so V=IRV=IR with one fixed R does not correctly describe them, even though the more fundamental microscopic form J=σEJ=\sigma E (with σ\sigma evaluated locally) remains valid.

✓Final answer

Substituting E=V/lE=V/l and J=I/AJ=I/A into the microscopic form J=σEJ=\sigma E gives V=IRV=IR with R=l/(σA)=ρl/AR=l/(\sigma A)=\rho l/A; this macroscopic form is limited to ohmic materials with a genuinely constant R and breaks down for non-ohmic devices whose resistance itself varies with V, I or temperature.

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