Class 12 Physics, Chapter 3 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
Electrons in a uniform ohmic metal drift towards -x. What are the directions of conventional current density j and electric field E?
(d) Both towards +x
Because electrons carry negative charge, j=-ne v_d. In the ohmic metal j is parallel to E, so both oppose the electron drift.
Across a fixed section of an ionised gas each second, positive ions carry +6 C towards +x while electrons carry -3 C towards -x. What is the conventional current?
(d) 9 A towards +x
Negative charge moving towards -x contributes conventional current towards +x, as does positive charge moving towards +x. Add their current magnitudes.
A uniform specimen has resistance R, length L and cross-sectional area A. An otherwise identical specimen is twice as long with the same area and temperature. Which pair describes the second specimen?
(d) Resistance 2R; resistivity RA/L
For the second specimen rho=(2R)A/(2L)=RA/L. The factor from greater resistance cancels its greater length.
Two wires of the same metal at the same temperature have equal cross-sectional areas; wire B has 2 times the length of A. What is rho_B:rho_A?
(c) 1:1
Resistivity belongs to the material at this temperature. Length changes resistance, but does not change resistivity.
A metal at uniform temperature has rapid random electron motion but no applied field or concentration gradient. What is its macroscopic net current?
(c) Zero; the random velocities cancel in the average
Thermal motion has no preferred direction, giving zero mean velocity. A directed drift is needed for macroscopic net conduction current.
The voltage across an ohmic wire at fixed temperature is increased by a factor 4, with dimensions unchanged. What happens to mobility μ and drift-speed magnitude vd?
(a) μ unchanged; vd multiplied by 4
E=V/L increases by the given factor. Since μ=e tau/m is unchanged at fixed conditions, vd=μE grows by that factor.
A student measures a diode at many voltages, defines R=V/I separately at each nonzero reading, and concludes that the diode obeys Ohm law. Which correction is needed?
(c) Ohm law requires the same V/I across the relevant readings, not a separately defined value at each point
The equation can define an operating-point resistance even for a nonlinear device. Ohmic behaviour requires proportionality and a constant ratio under fixed conditions.
Why does the resistivity of an ordinary metal generally increase on heating over its normal operating range?
(d) The carrier density changes little while increased scattering reduces relaxation time
For the metal rho=m/(ne²tau); n is nearly unchanged and tau falls, raising resistivity.
A load R dissipates power P across an ideal constant-voltage source. It is replaced by 3R across the same source. What is the new power?
(a) P/3
At fixed voltage P=V²/R. Multiplying resistance by the stated factor divides the power by it.
2 identical cells, each of emf E and internal resistance r, are connected in parallel with matching polarities. What are their equivalent emf and internal resistance?
(b) E and r/2
Identical cells connected in parallel with matching polarities give equivalent emf E.
Their 2 internal resistances are in parallel, so 1/r_eq = 2/r.
Therefore r_eq = r/2, with equivalent emf E.
Therefore, the requested answer is E and r/2.
A rechargeable battery of emf 8 V and internal resistance 1 Ω is charged by current 2 A entering its positive terminal. What is the voltage across its terminals?
(b) 10 V
During charging the external terminal voltage supplies both emf and the internal resistive drop: V=E+Ir.
Two same-material wires of equal radius have lengths in the ratio 2:3 and are connected in parallel. Total current is 20 A. What current passes through the shorter wire?
(b) 12 ampere
The resistance ratio is 2:3; the currents divide in ratio 3:2. The shorter wire gets three fifths of the total.
At the steady junction P in the diagram, two labelled currents enter and current I leaves. What is the magnitude of I?
(d) 14 ampere
Charge conservation adds incoming currents algebraically, irrespective of the angle between the wires. The outgoing current is their sum.
A steady node has one incoming current 10 A and two outgoing currents. One outgoing current is 6 A. The two outgoing wires meet at 90°. What is the other outgoing current?
(a) 4 ampere
Charge conservation gives 10=6+I in amperes. Subtract the known outgoing current from the incoming current; wire angles do not enter the sum.
A Wheatstone bridge has two series paths A-P-B and A-Q-B across a nonzero supply. Their resistances are (2R,3R) and (3R,2R), respectively. A galvanometer joins P and Q. Is it balanced?
(c) No; the midpoint voltage division ratios differ
Test whether zero galvanometer current is possible. If it were zero, each arm pair would act as a series voltage divider across the supply.
Take and . The A–P–B path would give , while A–Q–B would give .
These midpoint potentials differ, contradicting the zero-current balance condition. Equivalently, the arm ratios and are unequal.
The bridge is not balanced. Equal path resistance sums alone do not justify assuming equal branch currents in the connected, unbalanced bridge.
Wrong
Two resistors in series have identical voltage drops because they carry the same current.
Why it feels right. I treat current and potential difference as interchangeable measures of electricity.
Right
Series resistors carry the same current. Each drop is set by V = IR for that resistor.
Worked example
