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Current Electricity — NEET & JEE MCQs with answers
This chapter studies steady electric currents in conductors. It covers the definition of current and drift velocity, Ohm's law and resistivity, temperature dependence of resistance, electrical power, cells (EMF and internal resistance), combinations of cells, and Kirchhoff's rules applied to circuit analysis including the Wheatstone bridge.
Topics in this chapter
- 3.1 Introduction
- 3.2 Electric Current
- 3.3 Electric Currents in Conductors
- 3.4 Ohm's Law
- 3.5 Drift of Electrons and the Origin of Resistivity
- 3.6 Limitations of Ohm's Law
- 3.7 Resistivity of Various Materials
- 3.8 Temperature Dependence of Resistivity
- 3.9 Electrical Energy and Power
- 3.10 Cells, EMF, and Internal Resistance
- 3.11 Cells in Series and in Parallel
- 3.12 Kirchhoff's Rules
- 3.13 Wheatstone Bridge
Free practice questions
A cell of emf e and internal resistance r delivers maximum power to an external resistance R when R = r. Using the NCERT relations, the current at this condition is
- A. e/(2r)
- B. 2e/r
- C. e/(4r)
- D. e/r
Show answer
Answer: A — With R = r, I = e/(R + r) = e/(r + r) = e/(2r). (Using I = e/(R + r) from NCERT Eq. 3.40.)
NCERT Class 12 Physics Ch 3.10
At a given time, in a field-free conductor, the velocities of the free electrons have:
- A. A direction fixed by the nearest ion
- B. No preferential direction
- C. Zero magnitude
- D. A single common direction
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Answer: B — NCERT: 'At a given time, there is no preferential direction for the velocities of the electrons.'
NCERT Class 12 Physics Ch 3.3
The introduction compares charges flowing in a steady manner in everyday devices to:
- A. Water flowing smoothly in a river
- B. Waves crashing on a shore
- C. A whirlpool spinning in place
- D. Rain falling in scattered drops
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Answer: A — The text compares steady flow of charge to 'water flowing smoothly in a river' — a helpful analogy for a constant, uniform current.
NCERT Class 12 Physics Ch 3.1
A steady current of 3.2 A flows through a wire for 5 s. The number of electrons crossing a cross-section in this time is (e = $1.6 \times 10^{-19}$ C):
- A. $1.6 \times 10^{20}$
- B. $3.2 \times 10^{19}$
- C. $1 \times 10^{20}$
- D. $1 \times 10^{19}$
Show answer
Answer: C — Charge $q = I t = 3.2 \times 5 = 16$ C. Number of electrons $n = q/e = \dfrac{16}{1.6 \times 10^{-19}} = 1 \times 10^{20}$.
NCERT Class 12 Physics Ch 3.2
Two cells with $\varepsilon_1=8$ V, $r_1=2$ Ω and $\varepsilon_2=4$ V, $r_2=2$ Ω are connected in parallel. The equivalent emf of the combination is:
- A. 6 V
- B. 3 V
- C. 4 V
- D. 12 V
Show answer
Answer: A — $\varepsilon_{eq}=\dfrac{\varepsilon_1 r_2+\varepsilon_2 r_1}{r_1+r_2}=\dfrac{8\times2+4\times2}{2+2}=\dfrac{16+8}{4}=6$ V (Eq. 3.54).
NCERT Class 12 Physics Ch 3.11
The galvanometer G in a Wheatstone bridge is connected between vertices:
- A. B and C
- B. A and D
- C. B and D (the galvanometer arm)
- D. A and C (the battery arm)
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Answer: C — NCERT: 'Between the other two vertices, B and D, a galvanometer G ... is connected. This line, shown as BD in the figure, is called the galvanometer arm.'
NCERT Class 12 Physics Ch 3.13
The electrical power dissipated in a conductor carrying current I across a potential difference V is given by:
- A. $P = IV$
- B. $P = I^2V$
- C. $P = I/V$
- D. $P = V/I$
Show answer
Answer: A — The energy dissipated per unit time is the power. NCERT gives P = ΔW/Δt = IV (Eq. 3.32), where the heat energy dissipated in time Δt is ΔW = IVΔt.
NCERT Class 12 Physics Ch 3.9
In the relation $\rho_T = \rho_0[1 + \alpha(T - T_0)]$, the quantity $\alpha$ is called the
- A. temperature co-efficient of resistivity
- B. drift velocity
- C. relaxation time
- D. mobility of electrons
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Answer: A — As per NCERT, $\alpha$ is called the temperature co-efficient of resistivity.
NCERT Class 12 Physics Ch 3.8
In a copper wire carrying a steady current, if the applied electric field is doubled (other factors unchanged), the drift velocity of the electrons will
- A. remain unchanged
- B. become four times as large
- C. become twice as large
- D. become half as large
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Answer: C — $v_d = eE\tau/m$, so $v_d \propto E$ at constant $\tau$. Doubling $E$ doubles $v_d$.
NCERT Class 12 Physics Ch 3.5
A two-loop network has junction relation I1 = I2 + I3. The loop equations are 2I2 − 3I3 = 0 and 2I2 + 4I1 = 24 (SI units). The current I1 is:
- A. 6 A
- B. ≈ 2.77 A
- C. 3 A
- D. ≈ 4.62 A (60/13 A)
Show answer
Answer: D — From 2I2 − 3I3 = 0, I3 = (2/3)I2. Junction: I1 = I2 + (2/3)I2 = (5/3)I2. Substitute into 2I2 + 4I1 = 24: 2I2 + 4(5/3)I2 = 24 → 2I2 + (20/3)I2 = 24 → (6/3 + 20/3)I2 = 24 → (26/3)I2 = 24 → I2 = 72/26 = 36/13 ≈ 2.77 A. Then I1 = (5/3)(36/13) = 60/13 ≈ 4.62 A.
NCERT Class 12 Physics Ch 3.12
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