Read and download the CBSE Class 12 Physics Electric Circuits Important Questions. Designed for 2025-26, this advanced study material provides Class 12 Physics students with detailed revision notes, sure-shot questions, and detailed answers. Prepared by expert teachers and they follow the latest CBSE, NCERT, and KVS guidelines to ensure you get best scores.
Advanced Study Material for Class 12 Physics Chapter 1 Electric Charges and Fields
To achieve a high score in Physics, students must go beyond standard textbooks. This Class 12 Chapter 1 Electric Charges and Fields study material includes conceptual summaries and solved practice questions to improve you understanding.
Class 12 Physics Chapter 1 Electric Charges and Fields Notes and Questions
CBSE Class 12 Physics Electric Circuits Important Questions. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.
Comprehension-1
Two persons are pulling a square of side a along one of the diameters horizontally to make it rhombus. Plane of rhombus is always vertical and uniform magnetic field B exist perpendicular to plane. They start pulling at t = 0 and with constant velocity v.
1. The induced emf in the frame when angle at corner being pulled is 60°
(a) Bav (b) 2Bav (c) 2B/av (d) 4B/av
2. If the resistance of the frame is R the current induced is
(a) Bav/2R (b)2Ba v/R (c) Bav/ R (d) Bav/4R
3. Finally square frame reduces to straight wire. The total charge flown is
(a)a2B/ R (b) a2B/2 R (c) 2a2B/R (d) 4a2B/R
Comprehension-2
In case of analysis of circuits, containing cells, resistences and inductances two things are very important one is conservation of charge which leads to the fact that at any junction of circuit incoming current is equal to out going current. The other thing is that sum of voltage drop in a closed loop is equal to zero. Inductors have a unique property by which they oppose the change in magnetic flux linked to them. The voltage drop across resistor is
VR = IR and across inductor is L dt /dI
. In the steady state current through inductor becomes constant which leads to zero voltage drop across inductor. ie. it behaves like short circuit. Refer to circuit given in the figure. E = 10 V, R1 = 2Ω, R2 = 3Ω, R3 = 6Ω and L = 5H.
Now answer the following questions
4. The current I1 just after pressing the switch S is
(a) 10/8 A (b) 10/5 A (c) 10/12 A (d) 10/6A
5. The current I1 long after pressing the switch S is
(a) 10/4 A (b) 10/5 A (c) 10/12 A (d) 10/6 A
6. The current I2 long after pressing the switch S is
(a) 10/4 A (b) 10/5 A (c) 10/12 A (d) 10/6A
7. The current through R2 just after releasing the switch S is
(a) 10/4A (b) 10/5 A (c) 10/6 A (d) 10/12 A
Each of the questions given below consists of two statements, an assertion (A) and reason (R).Select the number corresponding to the appropriate alternative as follows
(a) If both A and R are true and R is the correct explanation of A
(b) If both A and R are true but R is not the correct explanation of A
(c) If A is true but R is false
(d) If A is false but R is true
8. A: We prefer a potentiometer with a longer bridge wire.
R: By doing this, the sensitivity of the bridge is increased because for longer wire potential drop per unit length will be small.
9. A: In steady state the branch of the circuit containing a capacitor acts as a open circuit.
R: In steady state, inductor acts as a conductor of zero resistance.
10. A: Ohm’s law and Kirchhoff’s laws are contradictory.
R: Electrons in a conductor have no motion in the absence of a potential difference across it.
11. A: Current is a vector quantity.
R: It obeys vector addition.
12. A: The wire of a potentiometer should be of uniform area of cross section.
R: The resistance per unit length and hence fall of potential per unit of length in case of a uniform wire is constant.
13. A: Fuse wire should have high resistance but low melting point.
R: It is used to control the current.
14. A: Terminal voltage of a cell is greater than e.m.f. of the cell, during charging of the cell.
R: For charging, current must flow from positively to the negative plate inside the cell.
15. A: The element of a heater is very hot, while the conducting wire carrying the current are cold.
R: It happens because the resistance of connecting wires is high as compared to that of element of heater.
16. A: Higher the range greater is the resistance of ammeter.
R: To increase the range of ammeter shunt has to be increased.
17. A: The range of an ammeter can be decreased.
R: The value of shunt, S =− Ig-G/I-Ig . S can not be negative. Therefore range of ammeter can not be decreased.
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Important Practice Resources for Class 12 Physics
CBSE Class 12 Physics Chapter 1 Electric Charges and Fields Study Material
Students can find all the important study material for Chapter 1 Electric Charges and Fields on this page. This collection includes detailed notes, Mind Maps for quick revision, and Sure Shot Questions that will come in your CBSE exams. This material has been strictly prepared on the latest 2026 syllabus for Class 12 Physics. Our expert teachers always suggest you to use these tools daily to make your learning easier and faster.
Chapter 1 Electric Charges and Fields Expert Notes & Solved Exam Questions
Our teachers have used the latest official NCERT book for Class 12 Physics to prepare these study material. We have included previous year examination questions and also step-by-step solutions to help you understand the marking scheme too. After reading the above chapter notes and solved questions also solve the practice problems and then compare your work with our NCERT solutions for Class 12 Physics.
Complete Revision for Physics
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