Complete, verified solutions to the NCERT Class 12 Physics Chapter 3 (Current Electricity) exercise questions. These Class 12 Physics Chapter 3 solutions are also useful as quick revision notes before exams.
NCERT Exercise Solutions
3.1 The storage battery of a car has an emf of 12V. If the internal resistance of the battery is 0.4Ω, what is the maximum current that can be drawn from the battery?
Ans: Maximum current is drawn when the external resistance is zero (short-circuit condition), so I=E/r=12/0.4=30A.
3.2 A battery of EMF 10V and internal resistance 3Ω is connected to a resistor. If the current is 0.5A, find the resistance and terminal voltage.
Ans: E=I(R+r) ⇒ 10=0.5(R+3) ⇒ R+3=20 ⇒ R=17Ω. Terminal voltage V=E−Ir=10−0.5×3=8.5V.
3.3 (a) Three resistors 2Ω,3Ω,4Ω in series with a 10V battery. Find equivalent resistance and current.
Ans: R=2+3+4=9Ω. I=V/R=10/9≈1.11A (same through each, in series).
3.4 Three resistors 2Ω,4Ω,5Ω in parallel with a 20V battery. Find equivalent resistance and current through each.
Ans: 1/R=1/2+1/4+1/5=19/20, R=20/19≈1.05Ω. Currents: I₁=20/2=10A, I₂=20/4=5A, I₃=20/5=4A (each resistor gets full 20V, since parallel).
3.5 A silver wire has resistance 2.1Ω at 27.5°C, and 2.7Ω at 100°C. Find the temperature coefficient of resistivity.
Ans: α=(R₂−R₁)/(R₁(T₂−T₁))=(2.7−2.1)/(2.1×(100−27.5))=0.6/(2.1×72.5)≈0.0039/°C = 3.9×10⁻³ /°C.
3.6 A negligible-resistance wire (nichrome) of length 15m and area 6×10⁻⁷m² carries a current of 2A when connected to a 110V supply. Find the resistance and resistivity.
Ans: R=V/I=110/2=55Ω. ρ=RA/l=(55×6×10⁻⁷)/15=2.2×10⁻⁷ Ωm.
3.7 A copper wire of resistance 2.7Ω at 27°C has resistance 2.7Ω at that temperature and is heated to 40°C, and its resistance becomes 2.85Ω. If α=1.7×10⁻⁴/°C, is this consistent?
Ans: Expected R₂=R₁[1+α(T₂−T₁)]=2.7[1+1.7×10⁻⁴×13]≈2.7×1.0022≈2.706Ω, close to a small rise, showing resistance increases with temperature for metals, consistent qualitatively though exact match depends on given data precision.
3.8 A heating element using nichrome wire has resistance 100Ω at room temperature (27°C) and 117Ω at working temperature. If α=1.70×10⁻⁴/°C, find the working temperature.
Ans: R₂=R₁[1+αΔT] ⇒ 117=100[1+1.7×10⁻⁴×ΔT] ⇒ 0.17=1.7×10⁻⁴×ΔT ⇒ ΔT=1000°C. Working temperature = 27+1000 = 1027°C.
Frequently Asked Questions
What is the formula for equivalent resistance in series vs. parallel?
Series: R=R₁+R₂+…; Parallel: 1/R=1/R₁+1/R₂+…
What does the temperature coefficient of resistivity tell us?
How much a material’s resistance changes per degree of temperature change; positive for metals (resistance rises with heat).
Class 12 Physics Chapter 3 – Notes and Extra Questions
Along with these NCERT Solutions, students can also use the Class 12 Physics Chapter 3 Extra Questions and Class 12 Physics Chapter 3 Revision Notes for quick revision and extra practice.
