NTA will shortly release the JEE Main Syllabus 2025 at jeemain.nta.ac.in. Last Year, the complete syllabus of JEE Main underwent some major changes. According to the latest syllabus guidelines, the weightage of Physics has dropped to 22.5%. Due to these changes in the syllabus, certain shifts in chapter-wise weightage have been observed. Current Electricity are among such topics. It’s among the least weightage topics in JEE Main Physics Syllabus 2025. This chapter holds 1.39% weightage in the whole syllabus.

  • Current Electricity is a part of the General Physics section, and every year around 2-3 questions are asked from this chapter. Due to the drop in weightage, you can expect 1-2 questions in JEE Main 2025.
  • Current Electricity in JEE Mains Physics Syllabus 2025 covers all the basic concepts on units and measurements, fundamentals of units, dimensions etc.

Aspirants consider Current Electricity as one of the Easiest Chapters in Physics for JEE Main 2025. Even though this chapter has the least weightage in the syllabus, to develop a strong foundation in physics, this particular chapter plays a pivotal role. Hence the candidates need to develop a clear understanding of this chapter. The list of most asked questions from this chapter will help the students with their preparation.

Must Check News on JEE Main Physics:

Some of the most important topics in the Current Electricity chapter for the JEE Main exam include:

  • Ohm's Law: A fundamental concept in current electricity, Ohm's Law states that the current passing through a conductor is directly proportional to the voltage across it, as long as the temperature and other physical conditions remain constant
  • Kirchhoff's Laws: An important topic in the Current Electricity chapter
  • Grouping of resistors and cells: An important topic in the Current Electricity chapter
  • Wheatstone bridge: An important topic in the Current Electricity chapter
  • Metre bridge: An important topic in the Current Electricity chapter
  • Potentiometer: An important topic in the Current Electricity chapter

Current Electricity JEE Mains Questions -

List Of Most Asked Questions With Solutions

In the (JEE) Main, the topic of Physics and Measurement often features questions that have appeared multiple times in previous years. Many questions in this section are either repeated or follow similar patterns, with only minor changes, such as different numerical values or slight modifications in wording.Certain questions have been repeated as many as five times.

Here are some of the most frequently asked and repeated questions:

Question 1 :

A 220 V, 100 W bulb is connected across a 110 V supply. What is the power consumed by the bulb? (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 25 W

(B) 50 W

(C) 75 W

(D) 100 W

Question 2:

Three resistances of 1 ohm, 2 ohm, and 3 ohm are connected in series with a 12 V battery. Find the potential difference across the 2 ohm resistor. (This question has been asked thrice- JEE Main 2018,JEE Main 2023,JEE Main 2024)

(A) 2 V

(B) 4 V

(C) 6 V

(D) 8 V

Question 3:

A wire of length 2 m and resistance 10 ohms is uniformly stretched to a length of 10 m. What is the new resistance of the wire? (This question has been asked thrice- JEE Main 2019,JEE Main 2020,JEE Main 2021)

(A) 25 ohms

(B) 50 ohms

(C) 100 ohms

(D) 125 ohms

Question 4:

Two identical cells of emf 1.5 V each are connected in series across a 2 ohm resistor. The internal resistance of each cell is 0.5 ohms. What is the current through the resistor?

(This question has been asked thrice- JEE Main 2022,JEE Main 2023,JEE Main 2024)

(A) 0.5 A

(B) 1 A

(C) 1.5 A

(D) 2 A

Question 5:

The resistance of a wire is 10 ohms at 0°C and 11 ohms at 100°C. Find the temperature coefficient of resistance of the material. (This question has been asked thrice- JEE Main 2018,JEE Main 2022,JEE Main 2024)

(A) 0.005 /°C

(B) 0.01 /°C

(C) 0.02 /°C

(D) 0.03 /°C

Solution:1

The power consumed is given by P = (V^2 / R), where R is the resistance of the bulb.

First, find R using R = V^2 / P for the bulb's normal operation.

For 220 V, R = (220^2) / 100 = 484 ohms.

When connected to 110 V, the power consumed is P = (110^2) / 484 = 25 W.

Answer: (A) 25 W

Solution:2

The total resistance of the series combination is R_total = 1 + 2 + 3 = 6 ohms.

The total current is I = V / R_total = 12 / 6 = 2 A.

The potential difference across the 2 ohm resistor is V = I * R = 2 * 2 = 4 V.

Answer: (B) 4 V

Solution:3

The resistance of a wire is proportional to its length and inversely proportional to its cross-sectional area. When a wire is stretched, its length increases and its cross-sectional area decreases, so the resistance increases.

New resistance R_new = R * (l_new / l_old)^2 = 10 * (10 / 2)^2 = 250 ohms.

Answer: (D) 125 ohms

Solution:4

The total emf is 1.5 + 1.5 = 3 V. The total resistance is the sum of the external resistance and the internal resistances: R_total = 2 + 0.5 + 0.5 = 3 ohms.

The current is I = V / R_total = 3 / 3 = 1 A.

Answer: (B) 1 A

Solution:5

The temperature coefficient of resistance is given by alpha = (R2 - R1) / (R1 * delta_T), where delta_T is the temperature change.

Here, delta_T = 100°C - 0°C = 100°C, R1 = 10 ohms, and R2 = 11 ohms.

Substituting, alpha = (11 - 10) / (10 * 100) = 0.001 /°C.

Answer: (B) 0.01 /°C