Current Electricity MCQs

Ohm’s law states V = IR for a metallic conductor kept at a fixed temperature.

Re‑arranging gives V/I = R, showing the voltage‑current ratio is the resistance.

Since temperature is constant, R does not change, so the ratio remains a constant equal to resistance.

Mobility μ is defined as drift velocity v_d per unit electric field E, μ = v_d / E.

Substitute v_d = 2×10⁻⁴ m s⁻¹ and E = 5 V m⁻¹: μ = (2×10⁻⁴)/(5) = 0.4×10⁻⁴ m² V⁻¹ s⁻¹.

0.4×10⁻⁴ simplifies to 4×10⁻⁵ m² V⁻¹ s⁻¹.

Thus the mobility is 4×10⁻⁵ m² V⁻¹ s⁻¹.

In a series circuit the same charge passes sequentially through each resistor, so the flow of charge per unit time (current) is unchanged.

Since the resistors share a common path, there is no branching where current could split.

Thus the quantity that stays identical across every resistor is the current. Current.

At a junction the sum of currents entering equals the sum leaving because no charge can accumulate at a point. This follows from the principle that electric charge is conserved in any closed system. Hence Kirchhoff’s junction rule expresses conservation of charge. The correct answer is Charge.

Mobility μ = drift velocity / electric field = (m s⁻¹) / (V m⁻¹).

Divide the units: (m s⁻¹) × (m V⁻¹) = m² V⁻¹ s⁻¹.

Thus the SI unit of mobility is m² V⁻¹ s⁻¹.

For resistors in parallel, the reciprocal of the equivalent resistance equals the sum of reciprocals: 1/Req = 1/R + 1/R + … (n terms) = n/R.

Take reciprocal of both sides: Req = R/n.

Thus the equivalent resistance of n identical resistors each of R in parallel is R/n.

When cells are connected in parallel, their positive terminals are joined together and their negative terminals together, so the potential difference across the combination is the same as across each cell.

Since the cells are identical, each provides EMF E, and the parallel connection does not add voltages.

The net EMF of the parallel combination therefore remains E.

In a metre bridge a known resistor and an unknown resistor are placed in the two arms of a Wheatstone bridge.

When the galvanometer shows zero current, the bridge is balanced and the ratio of the resistances equals the ratio of the lengths of the wire.

Thus the condition of balance of a Wheatstone bridge is used to determine the unknown resistance.

Hence the principle is the balanced Wheatstone bridge.

Resistivity ρ is an intrinsic property of a material, defined by ρ = R·A/L, where R is resistance, A cross‑sectional area and L length.

Only the material’s nature (its atomic structure) and its temperature affect ρ; geometry (shape, area, length) influences resistance R, not ρ.

Thus the correct choice is nature of the material and temperature.

Current I = V/R and R ∝ length L, so doubling L halves the current for the same V. Drift velocity v_d = I/(n A e) is directly proportional to I, therefore v_d also halves. Hence the drift velocity becomes half.

For resistors in parallel, 1/Req = Σ(1/Ri).

Each resistor is 4 Ω, so 1/Req = 1/4 + 1/4 + 1/4 + 1/4 = 4/4 = 1.

Thus Req = 1 Ω, which matches the given answer.

Kirchhoff’s loop rule states that the algebraic sum of potential differences around any closed circuit is zero.

Potential difference is a form of energy per unit charge, so the rule ensures that energy gained equals energy lost in a loop.

Thus the loop rule expresses conservation of energy in an electrical circuit.

When a wire is stretched, its length doubles (L → 2L) and its cross‑sectional area halves because volume V = A L stays constant (A → A/2).

Resistance R = ρ L/A, so the new resistance R' = ρ(2L)/(A/2) = 4 ρL/A = 4R.

Thus the stretched wire has resistance 4R.

Increase in temperature gives lattice atoms larger vibrations, increasing electron scattering.

Higher scattering reduces the mean free path of electrons, so resistivity rises.

Since resistivity ∝ temperature for metals (ρ = ρ₀[1+αΔT]), the resistance also increases.

Thus the correct answer is Increases.

For two equal resistors R in parallel, equivalent resistance = (R·R)/(R+R)=R/2.

This combination is then in series with another resistor R, so total R = R + R/2.

Add the resistances: R + R/2 = (2R/2 + R/2) = 3R/2.

Hence the total resistance is 3R/2.

Sum of currents entering a junction equals sum leaving (Kirchhoff’s current law).

Incoming currents: 3 A + 2 A = 5 A.

Outgoing currents: 1 A + I, so 1 A + I = 5 A.

Thus I = 5 A − 1 A = 4 A. 4 A.

Use R = ρ L/A.

Substitute ρ = 2×10⁻⁸ Ω·m, L = 2 m, A = 1×10⁻⁶ m²: R = (2×10⁻⁸ × 2)/(1×10⁻⁶) = 4×10⁻⁸/10⁻⁶ = 4×10⁻² Ω.

4×10⁻² Ω equals 0.04 ohm, which matches the given answer.

Increase in temperature raises the number of charge carriers (electrons and holes) in a semiconductor.

Resistivity ρ = 1/(σ) and conductivity σ = nqμ, where n (carrier density) grows exponentially with T while mobility μ changes only slightly.

Thus σ increases with temperature, so ρ = 1/σ decreases.

Therefore the resistivity of a semiconductor decreases.

Potential difference across the terminals = emf minus the drop across internal resistance.

The internal drop is given by V_drop = I r (Ohm’s law for the internal resistor).

Hence V_terminal = E – I r, which matches the given answer E - Ir.

Current entering a junction is taken as positive and leaving as negative.

Applying conservation of charge, the net charge accumulation at the junction must be zero.

Therefore the algebraic sum of all currents meeting at the junction equals zero.

Hence the correct answer is Zero.