Ohm's Law and V-I Characteristics

Master Ohm's law, V-I characteristics, and conductor behavior for JEE Physics with real-world circuit examples

Prerequisites

Before studying this topic, make sure you understand:

The Hook: Why Don’t All Wires Glow Like Bulbs?

Connect: Your Phone Charger → Ohm's Law

Ever noticed how your phone charger cable stays cool while charging, but a small LED bulb glows bright with the same voltage? Both have current flowing through them, so why the difference?

The answer: Different materials have different resistance to current flow. This simple relationship between voltage, current, and resistance - discovered by Georg Ohm in 1827 - powers every circuit in the world, from Iron Man’s arc reactor to your smartphone!

Challenge question: If you double the voltage across a wire, what happens to the current? Keep reading to find out!

Interactive Demo

Adjust voltage and see how current changes for different materials:

Formula Visualization: Understanding V = IR

See what each variable represents in a real circuit:


The Core Concept: Ohm’s Law

The Fundamental Relationship

$$\boxed{V = IR}$$

In simple terms: “Voltage is the push, Current is the flow, Resistance is the opposition”

Alternative forms:

  • $I = \frac{V}{R}$ (current in terms of voltage)
  • $R = \frac{V}{I}$ (resistance from V-I measurement)
The Water Pipe Analogy

Think of electricity like water flowing through pipes:

ElectricalWater System
Voltage (V)Water pressure
Current (I)Water flow rate
Resistance (R)Pipe narrowness/friction

Higher pressure (V) → More water flow (I) Narrower pipe (higher R) → Less water flow (I)

This is exactly how Ohm’s law works!

Units

  • Voltage (V): Volt (V) = Joule/Coulomb
  • Current (I): Ampere (A) = Coulomb/second
  • Resistance (R): Ohm (Ω) = Volt/Ampere

Ohmic vs Non-Ohmic Conductors

Ohmic Conductors

Definition: Materials that obey Ohm’s law - current is directly proportional to voltage.

$$I \propto V \quad \text{(at constant temperature)}$$

V-I Graph: Straight line through origin

  I ↑
    |     /
    |    /
    |   /  (slope = 1/R)
    |  /
    | /
    |/________→ V

Examples:

  • Metallic conductors (copper, aluminum, silver)
  • Carbon resistors
  • Nichrome wire (used in heaters)

Key property: Resistance $R$ remains constant (at constant temperature)

Non-Ohmic Conductors

Definition: Materials where current is NOT directly proportional to voltage.

V-I Graph: NOT a straight line

Examples:

  1. Diode (Semiconductor)

    • Current flows easily in one direction
    • High resistance in reverse direction
    • Exponential I-V relationship
  2. Filament Bulb

    • Resistance increases as it heats up
    • I-V curve bends (current increases slower than voltage)
  3. LED (Light Emitting Diode)

    • No current until threshold voltage
    • Then sudden current increase
  4. Electrolyte solutions

    • I-V characteristic depends on ion concentration
Common Misconception

Wrong thinking: “Ohm’s law applies to all materials”

Reality: Ohm’s law is NOT a universal law! It only works for ohmic conductors at constant temperature.

JEE Trap: Questions may give a non-linear V-I graph and ask if Ohm’s law applies - answer is NO!


V-I Characteristics Graphs

1. Metallic Conductor (Ohmic)

  I 
    |     /
    |    /  Straight line
    |   /   R = constant
    |  /
    | /
    |/________ V

Slope = 1/R

  • Linear relationship
  • Passes through origin
  • Constant resistance

2. Tungsten Filament Bulb

  I ↑
    |      ___---
    |   __/      Curve bends
    |  /         (R increases with I)
    | /
    |/________→ V

Why it bends:

  • Current heats the filament
  • Temperature ↑ → Resistance ↑
  • So current increases slower than voltage

3. Diode (Semiconductor)

  I ↑  Forward bias
    |        |
    |        |___
    |        |
    |________|_______→ V
    |    Reverse
    |    bias

Key features:

  • Threshold voltage needed in forward bias
  • Very little current in reverse bias
  • Used as one-way valve for current

4. Electrolyte

  I ↑
    |      /
    |    /   Initially curved
    |  /     (then becomes linear)
    | /
    |/________→ V

Behavior:

  • Non-linear at low voltages
  • Becomes approximately ohmic at higher voltages

Limitations of Ohm’s Law

When Ohm's Law Breaks Down
  1. Temperature changes:

    • Resistance varies with temperature
    • Must specify “at constant temperature”
  2. Non-ohmic materials:

    • Semiconductors (diodes, transistors)
    • Vacuum tubes
    • Electrolytes (at low voltages)
  3. Very high electric fields:

    • Breakdown voltage exceeded
    • Material properties change
  4. AC circuits with inductors/capacitors:

    • Need to use impedance, not resistance
    • Phase difference between V and I

Microscopic Form of Ohm’s Law

From drift velocity concepts:

$$J = \sigma E$$

where:

  • $J$ = current density
  • $\sigma$ = conductivity
  • $E$ = electric field

Derivation connection:

$$I = nAev_d$$ $$v_d = \mu E = \frac{e\tau}{m}E$$ $$\therefore I = nAe \cdot \frac{e\tau}{m}E$$

With $E = \frac{V}{L}$ and $J = \frac{I}{A}$:

$$J = \sigma E \quad \text{where } \sigma = \frac{ne^2\tau}{m}$$

This is Ohm’s law in vector form!


Memory Tricks & Patterns

Mnemonic for Ohm’s Law

“Very Interesting Results” → $V = I \times R$

Or think: “Voltage Is Really important” → $V = IR$

Remembering Forms

Triangle method:

    V
   ___
  | | |
  |I|R|
  |_|_|
  • Cover V → $V = I \times R$
  • Cover I → $I = V / R$
  • Cover R → $R = V / I$

Graph Recognition Patterns

Graph ShapeMaterial TypeOhm’s Law?
Straight line through originMetallic conductorYES
Curved upward (bending)Filament bulbNO
Steep rise after thresholdDiode/LEDNO
Flat then risingElectrolyteNO (initially)

JEE Exam Trick

If graph is straight: $R = \frac{V}{I} = \frac{1}{\text{slope}}$

If graph is curved: $R = \frac{V}{I}$ at that POINT (not constant!)


When to Use This

Decision Tree

Given two, find third:

  • Know V and R → Find I = V/R
  • Know I and R → Find V = IR
  • Know V and I → Find R = V/I

For V-I graph questions:

  • Straight line → Ohmic → Use $R = 1/\text{slope}$
  • Curved → Non-ohmic → Resistance varies with V or I

Word problem clues:

  • “Metallic wire” → Assume ohmic
  • “Bulb”, “Diode”, “LED” → Non-ohmic
  • “At constant temperature” → Confirms ohmic behavior

Common Mistakes to Avoid

Trap #1: Assuming All Materials are Ohmic

Wrong: “This question has V and I, so V = IR applies”

Correct: First check if material is ohmic!

  • Metals at constant T → Ohmic ✓
  • Semiconductors, bulbs, electrolytes → Non-ohmic ✗

JEE Insight: If graph is given, it’s probably testing your understanding of NON-ohmic behavior!

Trap #2: Confusing Slope and Resistance

Wrong: “Slope of V-I graph = Resistance”

Correct:

  • Slope of V-I graph = $\frac{dV}{dI} = R$
  • Slope of I-V graph = $\frac{dI}{dV} = \frac{1}{R}$ = Conductance

Memory trick: More common is I-V graph, where slope = 1/R

Which axis is which?

  • V-I graph: V on y-axis, I on x-axis → slope = R
  • I-V graph: I on y-axis, V on x-axis → slope = 1/R
Trap #3: Temperature Effect Ignored

Wrong: “Wire resistance is always constant”

Correct: Resistance changes with temperature!

  • For metals: $R$ increases with T (positive temperature coefficient)
  • For semiconductors: $R$ decreases with T (negative temperature coefficient)

JEE application: Filament bulb problems - as current increases, bulb heats up, R increases, so I-V curve bends!

Trap #4: Negative Resistance

Wrong: “Resistance can be negative if current flows backward”

Correct:

  • Resistance is ALWAYS positive (for passive elements)
  • Current direction determines sign of $I$, not $R$
  • If you get negative R, you made a sign error!

Exception: Active devices like tunnel diodes can show negative differential resistance in specific regions (but that’s beyond JEE scope).


Practice Problems

Level 1: Foundation (NCERT/Basic)

Problem 1.1

A wire carries a current of 2 A when connected to a 12 V battery. Find its resistance.

Solution:

Using Ohm’s law: $V = IR$

$$R = \frac{V}{I} = \frac{12}{2} = 6 \text{ Ω}$$

Answer: $R = 6$ Ω

Concept check: This assumes the wire is ohmic (metallic conductor).

Problem 1.2

A resistor of 100 Ω is connected across a 220 V supply. Calculate the current through it.

Solution:

$$I = \frac{V}{R} = \frac{220}{100} = 2.2 \text{ A}$$

Answer: $I = 2.2$ A

Safety note: This is a significant current - the resistor will heat up! (We’ll study this in electrical power).

Problem 1.3

Which of the following obey Ohm’s law? (a) Copper wire (b) LED (c) Diode (d) Nichrome wire

Solution:

Ohmic conductors: (a) Copper wire and (d) Nichrome wire

  • Both are metallic conductors
  • Show linear V-I characteristics at constant temperature

Non-ohmic: (b) LED and (c) Diode

  • Semiconductor devices
  • Non-linear V-I characteristics

Answer: (a) and (d) obey Ohm’s law

Level 2: JEE Main

Problem 2.1

The V-I graph for a conductor makes an angle of 30° with the V-axis. Find the resistance of the conductor.

Solution:

In V-I graph (V on y-axis, I on x-axis):

$$\text{slope} = \tan(30°) = \frac{V}{I} = R$$ $$R = \tan(30°) = \frac{1}{\sqrt{3}} = 0.577 \text{ Ω}$$

Answer: $R = \frac{1}{\sqrt{3}}$ Ω

Key insight: Angle with V-axis gives you the resistance directly!

Problem 2.2

A 220 V, 100 W bulb is connected to a 110 V supply. What is the power consumed? (Assume resistance remains constant)

Solution:

First find resistance of bulb:

$$P = \frac{V^2}{R}$$ $$R = \frac{V^2}{P} = \frac{(220)^2}{100} = 484 \text{ Ω}$$

At 110 V:

$$P' = \frac{V'^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25 \text{ W}$$

Answer: $P' = 25$ W

Key insight: Power ∝ V² for constant R, so halving voltage reduces power to 1/4!

Reality check: This assumes resistance stays constant, but for a real bulb, resistance decreases when cooler, so actual power would be slightly higher than 25 W.

Problem 2.3

A battery of 12 V is connected to a resistor. The ammeter shows 3 A. If the voltmeter is connected across the resistor, what will it read?

Solution:

This tests understanding of circuit measurement!

Ammeter reading: $I = 3$ A (measures current through resistor)

By Ohm’s law:

$$R = \frac{V}{I}$$

But we need V! Since battery is 12 V and no internal resistance mentioned:

Voltmeter reading = Voltage across resistor = 12 V

(Assuming ideal battery with no internal resistance)

Using this:

$$R = \frac{12}{3} = 4 \text{ Ω}$$

Answer: Voltmeter reads 12 V

Circuit concept: Voltmeter measures potential difference across the resistor, which equals battery voltage (for ideal battery).

Level 3: JEE Advanced

Problem 3.1

The I-V characteristic of a non-ohmic conductor is given by $I = 2V + 3V^2$ (where I is in A and V is in volts). Find the resistance when V = 1 V and V = 2 V.

Solution:

For non-ohmic conductor, resistance varies with voltage.

At V = 1 V:

$$I = 2(1) + 3(1)^2 = 2 + 3 = 5 \text{ A}$$ $$R = \frac{V}{I} = \frac{1}{5} = 0.2 \text{ Ω}$$

At V = 2 V:

$$I = 2(2) + 3(2)^2 = 4 + 12 = 16 \text{ A}$$ $$R = \frac{V}{I} = \frac{2}{16} = 0.125 \text{ Ω}$$

Answer:

  • At V = 1 V: R = 0.2 Ω
  • At V = 2 V: R = 0.125 Ω

Advanced insight: Resistance DECREASES as voltage increases for this material. This is characteristic of certain semiconductors and electrolytes!

Differential resistance:

$$r = \frac{dV}{dI} = \frac{1}{\frac{dI}{dV}} = \frac{1}{2 + 6V}$$

At V = 1 V: $r = 1/8 = 0.125$ Ω (dynamic resistance)

Problem 3.2

Two wires of the same material have lengths in the ratio 1:2 and radii in the ratio 2:1. They are connected (a) in series, (b) in parallel across a battery. Find the ratio of currents through them in both cases.

Solution:

Resistance: $R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$

For wire 1: $R_1 = \rho \frac{L_1}{\pi r_1^2}$

For wire 2: $R_2 = \rho \frac{L_2}{\pi r_2^2}$

$$\frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2} = \frac{2}{1} \times \frac{(2)^2}{(1)^2} = 2 \times 4 = 8$$

So $R_2 = 8R_1$

(a) In Series: Same current through both wires!

$$\frac{I_1}{I_2} = \frac{1}{1} = 1:1$$

(b) In Parallel: Same voltage across both wires.

$$I = \frac{V}{R}$$ $$\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{8}{1} = 8:1$$

Answer:

  • Series: $I_1 : I_2 = 1:1$
  • Parallel: $I_1 : I_2 = 8:1$

Key concepts:

  • Series → Same current
  • Parallel → Current inversely proportional to resistance
Problem 3.3 - JEE Advanced Level

The I-V characteristic of a material is shown below. Identify the region where the material obeys Ohm’s law, and calculate the resistance in that region.

(Assume graph shows: Linear from 0-2V, then curved from 2-5V)

Solution:

Ohmic region: 0 to 2 V (linear portion)

Let’s say at V = 2 V, I = 0.5 A (from graph)

In linear region:

$$R = \frac{V}{I} = \frac{2}{0.5} = 4 \text{ Ω}$$

Beyond 2 V: Material becomes non-ohmic

  • Resistance is not constant
  • Must calculate $R = V/I$ at each point

Answer:

  • Ohmic region: 0-2 V
  • Resistance in ohmic region: 4 Ω
  • Beyond 2 V: Non-ohmic behavior

JEE Strategy: Always identify the LINEAR portion of V-I graphs - that’s where Ohm’s law applies!


Quick Revision Box

SituationFormula/Approach
Basic Ohm’s law$V = IR$
Finding resistance from V-I graph (linear)$R = \frac{V}{I} = \frac{1}{\text{slope of I-V graph}}$
Metallic conductor at constant TOhmic (linear V-I)
Filament bulb, diode, LEDNon-ohmic (curved V-I)
Resistance from I-V curve$R = V/I$ at that point
Microscopic form$J = \sigma E$
Power from Ohm’s law$P = VI = I^2R = V^2/R$

JEE Strategy: High-Yield Points

What JEE Loves to Test
  1. Graph interpretation

    • Identifying ohmic vs non-ohmic from V-I curves
    • Calculating resistance from slope
    • Appears in 2-3 questions every year!
  2. Conceptual understanding

    • Why Ohm’s law fails for certain materials
    • Temperature effects on resistance
    • Often asked as assertion-reason questions
  3. Numerical problems

    • Quick calculations with $V = IR$
    • Combined with power formulas
    • Multi-step problems with series/parallel circuits
  4. Common traps

    • Confusing slope of V-I vs I-V graphs
    • Assuming all materials are ohmic
    • Forgetting temperature dependence

Time-saving tricks:

  • Recognize graph shapes instantly (practice 10-15 graphs)
  • Use triangle method for quick formula recall
  • For non-ohmic: Calculate R at specific points, don’t assume constant

Real-World Applications

Where You See Ohm's Law
  1. Phone chargers

    • Converts 220V AC to 5V DC
    • Resistance of charging cable determines current
  2. Filament bulbs (Jawan, Oppenheimer scenes)

    • Non-ohmic behavior due to heating
    • That’s why bulbs often fail when switched ON (cold → low R → high surge current)
  3. LED lights

    • Don’t follow Ohm’s law
    • Need current-limiting resistors
  4. Electric heaters (Nichrome wire)

    • Ohmic at steady state
    • Constant resistance = predictable heating
  5. Iron Man’s arc reactor

    • Superconducting components (zero resistance!)
    • Normal conductors would waste too much energy as heat

Within Current Electricity

Connected Chapters

Math Connections


Teacher’s Summary

Key Takeaways
  1. Ohm’s law $V = IR$ relates voltage, current, and resistance - but only for ohmic conductors at constant temperature!

  2. Ohmic conductors show linear V-I graphs - metallic wires, resistors, nichrome. Non-ohmic show curved graphs - diodes, LEDs, bulbs, electrolytes.

  3. Resistance from graphs: For I-V graph (most common), slope = 1/R. For V-I graph, slope = R.

  4. Temperature matters! Metals increase resistance when heated (filament bulb curve bends). Semiconductors decrease resistance when heated.

  5. The microscopic form $J = \sigma E$ is Ohm’s law in vector form - connects to drift velocity and material properties.

“Ohm’s law is like a straight road - only some materials follow it. Others take curved paths. Know who’s who, and you’ll ace JEE questions!”