Class 12 Physics Chapter 3 Current Electricity – Extra Questions with Answers

Extra practice questions for Class 12 Physics Chapter 3 (Current Electricity), beyond the textbook. These Class 12 Physics Chapter 3 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. State Ohm’s law.
Ans: V=IR, at constant temperature for an ohmic conductor.

Q2. Write the formula for equivalent resistance of two resistors in parallel.
Ans: 1/R=1/R₁+1/R₂.

Q3. What is the balance condition of a Wheatstone bridge?
Ans: P/Q=R/S.

Q4. State Kirchhoff’s junction rule.
Ans: Sum of currents entering a junction equals sum of currents leaving it.

Q5. What is the SI unit of resistivity?
Ans: Ohm-metre (Ω·m).

Short Answer Questions (2–3 marks)

Q6. Two resistors of 4Ω and 6Ω are connected in series to a 10V battery. Find the current.
Ans: R=4+6=10Ω. I=V/R=10/10=1A.

Q7. A wire of resistance 10Ω is stretched to double its length. Find the new resistance.
Ans: R∝l²/V(volume constant), so R′=R×2²=10×4=40Ω.

Q8. Find the equivalent resistance of three 6Ω resistors all in parallel.
Ans: 1/R=1/6+1/6+1/6=3/6=1/2, so R=2Ω.

Higher-Order Thinking / Application Questions

Q9. Using Kirchhoff’s laws, find the current through each branch of a circuit with two batteries: a 10V battery with internal resistance 1Ω and a 6V battery with internal resistance 1Ω, both connected to a common external resistor of 4Ω in a single loop opposing each other.
Ans: Since the batteries oppose each other in the loop, the net EMF driving the current is the difference: 10−6=4V. The total resistance in the loop is the sum of both internal resistances plus the external resistor: R_total=1+1+4=6Ω. By Kirchhoff’s voltage law (or simply Ohm’s law applied to the net EMF and total resistance), the current is I=V_net/R_total=4/6=2/3 A≈0.67 A, flowing in the direction of the stronger (10V) battery. This demonstrates how Kirchhoff’s loop rule correctly handles opposing EMFs by taking their algebraic (signed) sum around the loop, rather than simply adding magnitudes.

Q10. In a meter bridge experiment, the null point is obtained at 40cm from the left end when an unknown resistance X is in the left gap and a known resistance of 15Ω is in the right gap. Find X, and explain why the meter bridge is more accurate near the midpoint (50cm).
Ans: In a meter bridge, the balance condition is X/R=l/(100−l), where l is the balance length from the end with the unknown resistance. Here l=40cm, R=15Ω: X/15=40/(100−40)=40/60=2/3. So X=15×2/3=10Ω. Regarding accuracy: the meter bridge is most accurate when the null point is near the midpoint (50cm) because the bridge wire has uniform resistance per unit length, and the percentage error in reading the balance length (a fixed absolute error, e.g. ±0.1cm from parallax) has the LEAST relative impact on the calculated resistance ratio when l is close to 50cm (since the sensitivity of the bridge to small changes in resistance is maximized there, following from the mathematics of the l/(100−l) ratio, whose derivative with respect to l is most favorable to error-cancellation near the center). This is why experimentalists choose the value of the known resistance R to bring the balance point close to the 50cm mark.

Written by Satish

NCERTBooks.org is an independent educational resource run by a small team focused on making official NCERT textbooks easy to find, read, and download for students, parents, and teachers across India. We are not affiliated with NCERT or the Ministry of Education -- we organise publicly available NCERT content by class and subject, verify links against official sources, and build tools (like our in-browser reader) that make studying more convenient. Every guide we publish is written and reviewed by our team based on the actual NCERT curriculum and syllabus.