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Physics · Current Electricity

Ohm's Law

Ohm's law is the foundational relation of steady-current circuits: the current through a conductor is proportional to the potential difference across it, at constant temperature. Set out in NCERT Class XII Section 3.4 and NIOS Section 17.2, it gives the defining equation $V = IR$, the geometric law $R = \rho L/A$, and the microscopic statement $j = \sigma E$. Almost every Current Electricity question in NEET — from resistor networks to current density — rests on this single law, so a clean grasp of it is non-negotiable.

The statement of Ohm's law

A basic law regarding the flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for the flow of currents was understood. Imagine a conductor through which a current $I$ is flowing, and let $V$ be the potential difference between its ends. Then Ohm's law states that the current is directly proportional to the potential difference, provided physical conditions such as temperature and pressure remain unchanged:

$$V \propto I \qquad \text{or} \qquad V = RI$$

The constant of proportionality $R$ is called the resistance of the conductor. It is the property of a conductor by virtue of which it opposes the flow of current through it. For a metallic conductor obeying this law, a plot of $I$ against $V$ is a straight line through the origin, and the slope of the $V$–$I$ graph is precisely $R$.

Figure 1 · I–V graph of an ohmic conductor V I V I slope = 1/R O

For an ohmic conductor the current rises in exact proportion to the applied voltage, so the graph is a straight line through the origin. Because $I = V/R$, the slope of the $I$–$V$ line equals $1/R$; a steeper line means a smaller resistance.

Resistance and the ohm

Rearranging the law gives the defining equation for resistance, $R = V/I$. The SI unit of resistance is the ohm, denoted by the symbol $\Omega$ (read as omega) and defined as

$$1\ \Omega = 1\ \text{V A}^{-1} = \frac{1\ \text{volt}}{1\ \text{ampere}}$$

That is, a conductor has a resistance of one ohm if a potential difference of one volt across it drives a current of one ampere through it. The resistance $R$ depends not only on the material of the conductor but also on its dimensions — its length and cross-sectional area — as we derive next.

Geometry: R = ρL/A

Consider a conductor in the form of a slab of length $L$ and cross-sectional area $A$. Imagine placing two identical such slabs side by side, so the length of the combination is $2L$. The same current $I$ flows through the combination as through either slab, while the potential difference doubles to $2V$. Hence the resistance of the combination is $2V/I = 2R$: doubling the length doubles the resistance. In general, resistance is proportional to length, $R \propto L$.

Now imagine cutting the slab lengthwise into two identical slabs, each of cross-sectional area $A/2$. For the same voltage $V$, the current through each half-slab is $I/2$, so each has resistance $V/(I/2) = 2R$: halving the area doubles the resistance. Therefore resistance is inversely proportional to area, $R \propto 1/A$. Combining the two results,

$$R = \rho\,\frac{L}{A}$$

where the constant of proportionality $\rho$ is the resistivity of the material. Resistivity depends on the nature of the material (and on temperature) but not on the dimensions of the conductor, whereas the resistance of a conductor depends on its dimensions as well as on the material. Rearranged, $\rho = RA/L$, with SI unit ohm-metre ($\Omega\,\text{m}$).

Figure 2 · R = ρL/A geometry L A I R = ρL/A

Resistance scales up with the length the current must traverse and down with the cross-sectional area available to it — exactly like a longer, narrower pipe offering more resistance to water flow. The material constant $\rho$ fixes the proportionality.

Microscopic form: j = σE

Ohm's law can be recast in a point-by-point form valid inside the material. Define the current density $j$ as the current per unit area taken normal to the current, $j = I/A$, with SI unit $\text{A m}^{-2}$. Substituting $R = \rho L/A$ into $V = IR$ gives $V = I\rho L/A$. If $E$ is the magnitude of the uniform electric field in a conductor of length $L$, the potential difference across its ends is $V = EL$. Using these,

$$EL = j\,\rho\,L \quad\Rightarrow\quad E = \rho\, j$$

Since $j$ is directed along $E$, this can be written in vector form $\mathbf{E} = \rho\,\mathbf{j}$, or equivalently

$$\mathbf{j} = \sigma\,\mathbf{E}, \qquad \sigma \equiv \frac{1}{\rho}$$

where $\sigma$ is the conductivity of the material, the reciprocal of resistivity, with unit $\text{S m}^{-1}$ (siemens per metre). The relation $\mathbf{j} = \sigma\mathbf{E}$ is the microscopic form of Ohm's law: it states that the current density at every point of a conductor is proportional to the local electric field. NEET frequently tests this form directly through current-density problems.

Go deeper

The constants $\rho$ and $\sigma$ are explained microscopically — in terms of drifting electrons and relaxation time — in Drift Velocity & Resistivity.

Ohmic vs non-ohmic conductors

Most metals obey Ohm's law and the relation between voltage and current is linear; such resistors are called ohmic. Resistors which do not obey Ohm's law are called non-ohmic. Devices such as the vacuum diode, semiconductor diode, and transistors show non-ohmic character — for a semiconductor diode, the $I$–$V$ graph is not linear, and the law fails even at low voltages.

A crucial conceptual point: the equation $V = IR$ merely defines resistance and can be applied to any conducting device, ohmic or not. The assertion that "$V = IR$ is Ohm's law" is not strictly true. Ohm's law is the stronger statement that the $I$–$V$ plot is linear, i.e. that $R$ is independent of $V$. A diode has a perfectly well-defined resistance $V/I$ at any operating point, yet it disobeys Ohm's law because that resistance changes with voltage.

NEET Trap

"At constant temperature" is not optional

Ohm's law holds only when physical conditions, especially temperature, remain unchanged. A heated filament draws less current than $V/R_{\text{cold}}$ predicts because its resistance rises with temperature. A bulb's $I$–$V$ curve bends precisely for this reason — it is not a failure of $V=IR$, but a reminder that $R$ itself is temperature-dependent.

Apply $V = IR$ with the resistance at the operating temperature, not the room-temperature value.

Resistors in series

Resistors are in series when joined end-to-end so that the same current passes through all of them. Consider two resistors $R_1$ and $R_2$ in series carrying current $I$ from a battery of voltage $V$. The potential differences across them are $V_1 = IR_1$ and $V_2 = IR_2$, and their sum equals the supply voltage:

$$V = V_1 + V_2 = IR_1 + IR_2$$

If the equivalent resistance is $R_{\text{eq}}$, then $V = IR_{\text{eq}} = I(R_1 + R_2)$, so $R_{\text{eq}} = R_1 + R_2$. Extending to any number of resistors,

$$R_{\text{eq}} = R_1 + R_2 + R_3 + \cdots$$

The equivalent resistance of a series combination is the sum of the individual resistances, and is therefore larger than any single resistor in the chain.

Resistors in parallel

Resistors are in parallel when joined so that the same potential difference $V$ exists across all of them. The main current divides; for two resistors $R_1$ and $R_2$ the branch currents are $I_1 = V/R_1$ and $I_2 = V/R_2$, and the total current is their sum:

$$I = I_1 + I_2 = \frac{V}{R_1} + \frac{V}{R_2}$$

Writing $I = V/R_{\text{eq}}$ and dividing through by $V$ gives the reciprocal-sum rule, which extends to any number of resistors:

$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$

The equivalent resistance of a parallel combination is smaller than the smallest individual resistance — adding more parallel branches lowers the resistance and increases the current drawn from the source. This is why home appliances are wired in parallel: each receives the full mains voltage, and the total current rises as more are switched on.

Figure 3 · Series and parallel networks SERIES R₁ R₂ Rₑq = R₁+R₂ PARALLEL R₁ R₂ 1/Rₑq = 1/R₁+1/R₂

In series the same current threads every resistor and resistances add. In parallel the same voltage sits across every branch and reciprocals add, so the equivalent is always less than the smallest branch.

FeatureSeriesParallel
Same quantity in all resistorsCurrent $I$Voltage $V$
What adds upVoltages: $V = V_1 + V_2 + \cdots$Currents: $I = I_1 + I_2 + \cdots$
Equivalent resistanceR_eq = R₁ + R₂ + …1/R_eq = 1/R₁ + 1/R₂ + …
Compared to individualLarger than the biggestSmaller than the smallest
Two equal resistors R$2R$$R/2$
Everyday useLower voltage to one deviceHome appliances, full mains each
NEET Trap

Resistors and capacitors combine in opposite ways

For resistors, the values add in series and the reciprocals add in parallel. For capacitors the rule is exactly reversed — capacitances add in parallel, reciprocals add in series. Mixing these up is a classic NEET slip when a question puts a few resistors next to a few capacitors.

Resistors: series adds $R$, parallel adds $1/R$. Capacitors: the other way round.

Worked examples

Example 1 · resistance from V and I

In our homes, electricity is supplied at 220 V. Calculate the resistance of a bulb if the current drawn by it is 0.2 A. (NIOS Example 17.1)

By Ohm's law, $R = V/I = 220\ \text{V} / 0.2\ \text{A} = 1100\ \Omega$.

Example 2 · comparing resistances by geometry

Two copper wires A and B have the same length. The diameter of A is twice that of B. Compare their resistances. (NIOS Example 17.3)

Using $R = \rho L/(\pi r^2)$ with the same $\rho$ and $L$, $R_A/R_B = r_B^2 / r_A^2$. Since $r_A = 2r_B$, we get $R_A/R_B = 1/4$. The resistance of B is four times that of A — a thinner wire offers more resistance.

Example 3 · series–parallel reduction

Find the equivalent resistance between points a and d for a $5\ \Omega$, a parallel pair of $15\ \Omega$ and $3\ \Omega$, and a $7\ \Omega$ resistor, all in a single chain. (NIOS Example 17.6)

The $15\ \Omega$ and $3\ \Omega$ are in parallel: $\dfrac{15 \times 3}{15 + 3} = \dfrac{45}{18} = 2.5\ \Omega$. The $5\ \Omega$, $2.5\ \Omega$ and $7\ \Omega$ are then in series: $R = 5 + 2.5 + 7 = 14.5\ \Omega$.

Quick Recap

Ohm's law in one glance

  • Ohm's law: $V \propto I$, i.e. $V = IR$, valid at constant temperature and pressure; the $I$–$V$ graph is then a straight line through the origin with slope $1/R$.
  • Resistance: $R = V/I$, SI unit ohm, $1\ \Omega = 1\ \text{V A}^{-1}$.
  • Geometry: $R = \rho L/A$ — proportional to length, inversely to area; $\rho$ depends on material and temperature, not on dimensions.
  • Microscopic form: $\mathbf{j} = \sigma\mathbf{E}$ with $\sigma = 1/\rho$, the conductivity (unit $\text{S m}^{-1}$).
  • Series: same current; $R_{\text{eq}} = R_1 + R_2 + \cdots$ (larger than any one).
  • Parallel: same voltage; $1/R_{\text{eq}} = 1/R_1 + 1/R_2 + \cdots$ (smaller than the smallest).
  • Ohmic vs non-ohmic: metals are ohmic (linear $I$–$V$); diodes and transistors are non-ohmic. $V=IR$ defines $R$ for all of them; Ohm's law asserts $R$ is constant.

NEET PYQ Snapshot — Ohm's Law

Three NEET questions that turn directly on $V=IR$, the microscopic form, and series–parallel reduction.

NEET 2022

A copper wire of length 10 m and radius $(10^{-2}/\pi)$ m has electrical resistance of 10 Ω. The current density in the wire for an electric field strength of 10 V/m is

  1. $10^{6}\ \text{A/m}^2$
  2. $10^{-5}\ \text{A/m}^2$
  3. $10^{5}\ \text{A/m}^2$
  4. $10^{4}\ \text{A/m}^2$
Answer: (3) 10⁵ A/m²

Using the microscopic form $j = \sigma E = E/(\rho)$ with $\rho = RA/L$. So $j = LE/(RA)$. Here $A = \pi r^2 = \pi(10^{-2}/\pi)^2$, giving $j = (10 \times 10)/(10 \times \pi r^2) = 10^{5}\ \text{A/m}^2$. This is Ohm's law applied point-by-point via $j = \sigma E$.

NEET 2021

The effective resistance of a parallel connection that consists of four wires of equal length, equal area of cross-section and same material is 0.25 Ω. What will be the effective resistance if they are connected in series?

  1. 4 Ω
  2. 0.25 Ω
  3. 0.5 Ω
  4. 1 Ω
Answer: (1) 4 Ω

All four wires are identical, each of resistance $R$. In parallel, $R_P = R/4 = 0.25\ \Omega$, so $R = 1\ \Omega$. In series, $R_S = R + R + R + R = 4R = 4\ \Omega$. This contrasts the two combination rules directly.

NEET 2023

10 resistors, each of resistance R, are connected in series to a battery of emf E and negligible internal resistance. Then those are connected in parallel to the same battery; the current is increased n times. The value of n is

  1. 1000
  2. 10
  3. 100
  4. 1
Answer: (3) 100

Series: $R_{\text{eq}} = 10R$, so $i = E/10R$. Parallel: $R_{\text{eq}} = R/10$, so $i' = 10E/R$. Therefore $i' = 100\,i$, giving $n = 100$. A clean test of how the series and parallel equivalents differ by a factor of $10^2$ for ten equal resistors.

FAQs — Ohm's Law

Six high-yield clarifications drawn straight from NCERT and NIOS.

What is Ohm's law?

Ohm's law states that the electric current I flowing through a conductor is directly proportional to the potential difference V across its ends, provided physical conditions such as temperature and pressure remain unchanged. This is written V ∝ I, or V = RI, where the constant of proportionality R is the resistance of the conductor.

Is V = IR always a statement of Ohm's law?

No. The equation V = IR merely defines resistance and can be applied to any conducting device, whether or not it obeys Ohm's law. Ohm's law is the stronger assertion that the plot of I versus V is linear, i.e. R is independent of V. A resistor obeys Ohm's law; a diode does not, even though V = IR can still be used to define its resistance at any operating point.

Why does the resistance of a wire depend on its length and area?

Resistance is proportional to length L and inversely proportional to cross-sectional area A, giving R = ρL/A. Doubling the length doubles the resistance, while halving the area doubles it. The constant ρ is the resistivity, a property of the material that does not depend on the dimensions of the conductor.

What is the microscopic form of Ohm's law?

The microscopic form is j = σE, where j is the current density (current per unit area normal to the flow), E is the electric field in the conductor, and σ = 1/ρ is the conductivity. It follows from V = El and I = jA combined with R = ρL/A, and it states Ohm's law point by point inside the material rather than over the whole conductor.

How do resistors combine in series and in parallel?

In series the same current flows through all resistors and the voltages add, so the equivalent resistance is the sum R_eq = R₁ + R₂ + R₃ + …. In parallel the same voltage is across all resistors and the currents add, so the reciprocals add: 1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + …. The series equivalent is larger than any single resistor, and the parallel equivalent is smaller than the smallest one.

What is the difference between ohmic and non-ohmic conductors?

An ohmic conductor obeys Ohm's law: its I–V graph is a straight line through the origin, so its resistance is constant. Most metals are ohmic over a useful range. A non-ohmic conductor does not give a linear I–V graph; examples are the vacuum diode, semiconductor diode, and transistors, where the resistance changes with the applied voltage.

Related Topics
Current Electricity Back to the full chapter hub. Electric Current in Conductors How a steady current arises from a steady field. Drift Velocity & Resistivity The microscopic origin of ρ and σ. Limitations of Ohm's Law Where V ∝ I breaks down — diodes and GaAs. Electrical Energy & Power P = VI = I²R = V²/R from Ohm's law. Kirchhoff's Rules When series–parallel reduction is not enough.