JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
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This MCQ module is based on: Electric Current Ohms Law
TOPIC 9 OF 38
Electric Current Ohms Law
🎓 Class 12
Physics
CBSE
Theory
Ch 3 – Current Electricity
⏱ ~14 min
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Electric Current Ohms Law
3.1 Introduction
In Chapter 1, all the charges (positive or negative) we considered to be at rest. Charges in motion constitute an electric current. Lightning is one such phenomenon in which charges flow from the clouds to the earth through the atmosphere, sometimes with disastrous results. The flow of charges in lightning is not steady, but in our everyday life we use many devices where charges flow in a steady manner — such as cells powering torches, mains power lighting our homes, and currents that drive trains, fans and computers. Current Electricity, the topic of this chapter, deals with the steady flow of charges through electric circuits.
3.2 Electric Current
Imagine a small area held normal to the direction of flow of charges. Both the positive and the negative charges may flow forward and backward across the area. In a given time interval \(t\), let \(q_+\) be the net amount (i.e., forward minus backward) of positive charge flowing in the forward direction across the area. Similarly, let \(q_-\) be the net amount of negative charge flowing across the area in the forward direction. The net amount of charge flowing across the area in the forward direction in the time interval \(t\) is \(q = q_+ - q_-\). This is proportional to \(t\) for steady current flow.
The current is defined as the rate of flow of charge:
\[ I = \dfrac{q}{t}\quad\text{(steady)},\qquad I = \lim_{\Delta t\to 0}\dfrac{\Delta Q}{\Delta t}\quad\text{(instantaneous)} \]
SI Unit of Current: The ampere (A) = 1 C/s. Currents in everyday life range from microamperes (in nerve impulses) to thousands of amperes (in lightning).
3.3 Electric Currents in Conductors
An electric charge experiences a force when an electric field is applied. If it is free to move, it will thus move contributing to a current. In nature, free charged particles do exist: like in upper layers of atmosphere called the ionosphere. However, in atoms and molecules, the negatively charged electrons and the positively charged nuclei are bound to each other and are thus not free to move. Bulk matter is made up of many molecules — a gram of water has about \(10^{22}\) molecules. These molecules are so closely packed that the electrons are no longer attached to individual nuclei. In some materials the electrons are still bound (insulators), but in conductors a few electrons are free to move within the bulk material. These materials, generally metals, are called conductors and develop a current when an electric field is applied.
If we consider a conductor with two ends maintained at different electric potentials, an electric field exists inside it which sets up the steady drift of electrons across the conductor — and so a steady current.
3.4 Ohm's Law
A basic law regarding flow of currents was discovered by G.S. Ohm in 1828, long before the physical mechanism responsible for flow of currents was discovered. Imagine a conductor through which a current \(I\) is flowing and let \(V\) be the potential difference between the ends of the conductor. Then Ohm's law states that
\[ V \propto I \quad\Longrightarrow\quad V = IR \]
Ohm's Law: The current through a conductor is directly proportional to the potential difference applied across its ends, provided physical conditions (temperature, mechanical strain) remain unchanged. The constant of proportionality \(R\) is called the resistance of the conductor; SI unit is the ohm (Ω).
3.4.1 Resistance and Geometry
The resistance of a conductor depends on its length \(\ell\) and the cross-sectional area \(A\). Two identical conductors joined end-to-end (length \(2\ell\), same area) double the resistance: \(R \propto \ell\). Two identical conductors joined side-by-side (same length, area \(2A\)) halve the resistance: \(R \propto 1/A\). Combining,
\[ R = \dfrac{\rho\,\ell}{A} \]
where \(\rho\) is the resistivity of the material. Its reciprocal \(\sigma = 1/\rho\) is called the conductivity (SI unit: S/m or Ω⁻¹m⁻¹).
3.5 Drift of Electrons & Origin of Resistivity
As remarked before, an electron will suffer collisions with the heavy fixed ions, but after collision, it will emerge with the same speed as before. However, the direction of its velocity after the collision is completely random. So at a given time there is no preferential direction for the velocities of the electrons — the average velocity is zero. There is no net current.
When an external electric field \(\vec E\) is applied across the conductor, the electrons experience a force \(-e\vec E\). Between two successive collisions (average time \(\tau\), called relaxation time), each electron acquires an additional velocity \(\vec v = -\dfrac{e\vec E}{m}\tau\). The average over many electrons gives the drift velocity:
\[ \vec v_d = -\dfrac{e\,\vec E\,\tau}{m} \]
The negative sign tells us drift velocity is opposite to the applied field. Its magnitude is small (~10⁻⁴ m/s), yet it is enough to give measurable currents because the number density of electrons is huge.
3.5.1 Current Density and Microscopic Ohm's Law
Consider a conductor of cross-section \(A\) carrying current \(I\). If \(n\) is the free-electron density, then in time \(\Delta t\) the electrons travel a distance \(v_d\Delta t\), and the number of electrons crossing a section is \(n A v_d \Delta t\). Hence,
\[ I = n\,A\,e\,v_d \quad\Longrightarrow\quad J = \dfrac{I}{A} = n\,e\,v_d \]
Substituting \(v_d = eE\tau/m\):
\[ J = \dfrac{n e^2 \tau}{m} E = \sigma E,\qquad \sigma=\dfrac{n e^2 \tau}{m},\quad \rho=\dfrac{m}{n e^2\tau} \]
Thus resistivity arises from collisions of electrons with lattice ions. The smaller the relaxation time, the higher the resistivity.
3.5.2 Mobility
The mobility of a charge carrier is defined as the magnitude of drift velocity per unit electric field:
\[ \mu = \dfrac{|v_d|}{E} = \dfrac{e\,\tau}{m}\]
Resistivity of Common Materials (at 0 °C)
| Material | ρ (Ω·m) | Type |
|---|---|---|
| Silver | 1.6 × 10⁻⁸ | Conductor |
| Copper | 1.7 × 10⁻⁸ | Conductor |
| Aluminium | 2.7 × 10⁻⁸ | Conductor |
| Tungsten | 5.6 × 10⁻⁸ | Conductor |
| Nichrome (alloy) | ~1.0 × 10⁻⁶ | Alloy (heater) |
| Silicon (pure) | ~2.3 × 10³ | Semiconductor |
| Glass | 10¹⁰ – 10¹⁴ | Insulator |
| Wood (dry) | 10⁸ – 10¹¹ | Insulator |
3.6 V–I Characteristics — Ohmic & Non-Ohmic
If we plot V (x-axis) against I (y-axis) for a metallic conductor at constant temperature, we get a straight line through the origin. Such conductors are called ohmic. Many devices, however, do not follow Ohm's law:
- Junction diode: conducts strongly only in one direction. V–I is non-linear and not symmetric about the origin.
- Gallium arsenide (GaAs): shows a negative resistance region (current decreases as V increases beyond a threshold).
- Thermistor / filament lamp: resistance changes strongly with temperature, giving curved V–I characteristics.
Interactive Simulation: Ohm's Law & Resistance Calculator L3 Apply
Vary the voltage and resistance to see how current, power and drift velocity respond. Useful for L1–L6 explorations of Ohm's law.
Current I = V/R = 2.00 A
Power P = VI = 12.00 W
Drift v_d (Cu wire 1mm² area) ≈ 0.15 mm/s
Worked Example 1: Drift Speed in a Copper Wire
An electric bulb is connected by a 0.75 m long copper wire of cross-section \(5.0 \times 10^{-7}\) m². Find the drift speed of conduction electrons when the bulb draws a current of 2.7 A. Take the number density of free electrons in copper as \(n = 8.5 \times 10^{28}\) m⁻³.
Given: I = 2.7 A; A = 5.0 × 10⁻⁷ m²; n = 8.5 × 10²⁸ m⁻³; e = 1.6 × 10⁻¹⁹ C.
\[ v_d = \dfrac{I}{n A e} = \dfrac{2.7}{(8.5\times10^{28})(5.0\times10^{-7})(1.6\times10^{-19})}\] \[ v_d \approx \boxed{4.0\times 10^{-4}\ \text{m/s} = 0.4\ \text{mm/s}}\] Despite enormous thermal speed (~10⁵ m/s), the drift is a few tenths of a millimetre per second.
\[ v_d = \dfrac{I}{n A e} = \dfrac{2.7}{(8.5\times10^{28})(5.0\times10^{-7})(1.6\times10^{-19})}\] \[ v_d \approx \boxed{4.0\times 10^{-4}\ \text{m/s} = 0.4\ \text{mm/s}}\] Despite enormous thermal speed (~10⁵ m/s), the drift is a few tenths of a millimetre per second.
Worked Example 2: Resistivity from Geometry
A wire of length 2.0 m and uniform cross-section 1.0 mm² has a resistance of 0.034 Ω. Find the resistivity of the material.
\[\rho = \dfrac{R\,A}{\ell} = \dfrac{0.034 \times 1.0\times 10^{-6}}{2.0}=\boxed{1.7\times 10^{-8}\ \Omega\cdot\text{m}}\]
This matches copper. Hence the wire is most likely made of copper.
Worked Example 3: Mobility & Conductivity
If the relaxation time for free electrons in a metal is \(\tau = 2.5\times 10^{-14}\) s, find: (a) electron mobility μ, (b) conductivity σ if n = 8.5 × 10²⁸ m⁻³.
(a) \(\mu = e\tau/m = (1.6\times10^{-19})(2.5\times10^{-14})/(9.1\times10^{-31}) = \boxed{4.4\times 10^{-3}\ \text{m}^2/\text{V·s}}\)
(b) \(\sigma = n e \mu = (8.5\times10^{28})(1.6\times10^{-19})(4.4\times10^{-3}) = \boxed{6.0\times 10^{7}\ \text{S/m}}\)
(b) \(\sigma = n e \mu = (8.5\times10^{28})(1.6\times10^{-19})(4.4\times10^{-3}) = \boxed{6.0\times 10^{7}\ \text{S/m}}\)
Activity 3.1 — Verifying Ohm's Law in the LabL3 Apply
Materials: battery (1.5 V cells), rheostat, voltmeter, ammeter, resistance wire, key.
Procedure:
- Connect the resistance wire in series with the rheostat, ammeter and key.
- Connect the voltmeter in parallel across the resistance wire.
- Vary the rheostat to obtain at least six pairs of (V, I) readings.
- Plot V (x-axis) versus I (y-axis).
Predict: What shape will the V–I plot have for a metallic wire? What will its slope represent?
Observation: The plot is a straight line through the origin, confirming \(V \propto I\). Slope = 1/R.
Conclusion: The wire is ohmic at room temperature. If we repeated with a torch bulb filament (whose temperature changes with current), the line would curve, showing non-ohmic behaviour.
Competency-Based Questions
A student is studying current flow in a copper wire (length 1 m, area 1 mm²) connected to a 1.5 V cell. She measures a current of 0.5 A and notes that the bulb glows immediately even though electrons drift at only ~0.04 mm/s. She plots V–I for the wire and also for a junction diode for comparison.
Q1. The SI unit of electric current and its definition is: L1 Remember
Answer: (c) 1 A = 1 C/s.
Q2. Even though electrons drift at less than a mm/s, why does the bulb light up almost instantly when the switch is closed? L2 Understand
Answer: When the switch is closed, the electric field is established along the entire length of the wire almost instantaneously (at nearly the speed of light). All free electrons throughout the wire begin drifting at once, so current starts flowing in the bulb's filament immediately even though each individual electron is sluggish.
Q3. The student doubles the length of the wire while keeping V constant. The current will: L3 Apply
Answer: (b). R = ρℓ/A. Doubling ℓ doubles R, so I = V/R halves.
Q4. True/False: The V–I graph of a junction diode passes through the origin and is a straight line. Justify. L4 Analyse
FALSE. A diode is a non-ohmic device. Although V–I passes through the origin, it is highly non-linear: very small current for negative bias (reverse), and a sharply rising current only after the threshold voltage in forward bias. Therefore Ohm's law does not apply.
Q5. HOT: Suppose the relaxation time τ in a metal could be doubled by lowering the temperature. Predict the effect on (i) drift velocity at the same applied field, (ii) resistivity, (iii) current density. L6 Create
(i) v_d = eEτ/m doubles. (ii) ρ = m/(ne²τ) halves. (iii) J = nev_d also doubles. Lower temperature → fewer collisions → larger τ → smaller resistivity → larger current. This is qualitatively why metals conduct better when cool.
Assertion–Reason Questions
Choose: (A) Both A and R true; R correctly explains A. (B) Both true; R does not explain A. (C) A true, R false. (D) A false, R true.
Assertion (A): The conventional current in a metallic wire flows opposite to the direction of electron drift.
Reason (R): The conventional direction of current is taken as the direction in which positive charges would flow.
(A) Both true and R correctly explains A. Since electrons are negative, they drift opposite to the conventional current direction.
Assertion (A): Resistivity of a conductor depends on its length.
Reason (R): Resistance R = ρℓ/A increases with length.
(D). A is FALSE — resistivity is a material property, independent of length and area. R (resistance) does increase with length, which is true.
Assertion (A): Drift velocity of electrons in a copper wire is very small (~10⁻⁴ m/s) yet a large current can flow.
Reason (R): The number density of free electrons in copper is extremely large (~10²⁹ m⁻³).
(A). Both true and R correctly explains A. I = nAev_d — even tiny v_d produces ampere-scale currents thanks to enormous n.
Frequently Asked Questions - Electric Current Ohms Law
What is the main concept covered in Electric Current Ohms Law?
In NCERT Class 12 Physics Chapter 3 (Current Electricity), "Electric Current Ohms Law" covers the core principles and equations students need for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Electric Current Ohms Law useful in real-life applications?
Real-life applications of "Electric Current Ohms Law" from NCERT Class 12 Physics Chapter 3 include electronics, communication systems, medical imaging, solar energy, semiconductor devices, and modern technology. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Electric Current Ohms Law?
Key formulas in "Electric Current Ohms Law" (NCERT Class 12 Physics Chapter 3 Current Electricity) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 3?
NCERT Class 12 Physics Chapter 3 (Current Electricity) is structured so each part builds on the previous one. "Electric Current Ohms Law" connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Electric Current Ohms Law?
CBSE board questions from "Electric Current Ohms Law" typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Electric Current Ohms Law" lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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