The direction of electric current is always opposite to
Conventional current opposite to electron flow, direction of positive charge motion.
Ref: NCERT Physics Class 12 Chapter 3 Current Electricity
#electron flow
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A copper wire of cross-sectional area \( 3 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 0.9 \, \text{A} \). I
**Conductivity** σ=1/ρ decreases with temperature for metals, σ = n e² τ/m, τ ∝1/T due to lattice vibrations. For semiconductors, n increases exponentially with T, so σ increases, opposite to metals, explaining why metallic resistance rises with temperature. Drift speed: v_d = (I/n e A) . Substitute: v_d = (0.9/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 3 × 10⁻⁷) . Calculate: v_d = (0.9/4.08 × 10³) ≈ 2.21 × 10⁻⁴ m/s . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and P = I²R, evaluation yields 2.21
Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity
A copper wire of cross-sectional area \( 4 \times 10^{-7} \, \text{m}^2 \) carries a current of \( 1.2 \, \text{A} \). I
**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 4 × 10⁻⁷) . Calculate: v_d = (1.2/5.44 × 10³) ≈ 2.21 × 10⁻⁴ m/s . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r
Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility
A copper wire of cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) carries a current of \( 1.5 \, \text{A} \). I
**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Drift speed: v_d = (I/n e A) . Substitute: v_d = (1.5/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1 × 10⁻⁶) . Calculate: v_d = (1.5/1.36 × 10⁴) ≈ 1.1 × 10⁻⁴ m/s . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε
Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility
Which of the following statements about the Z-scheme of electron transport is correct?
The Z-scheme describes the linear flow of electrons from water to NADP+ through Photosystem II and I.
Ref: NCERT Class 11 Biology Chapter 11: Photosynthesis in Higher Plants Light Reaction - Photosystems Electron Transport Photophosphorylation
