6 public questions tagged with this topic.
**Transformer** works on mutual induction, V_s/V_p = N_s/N_p, I_s/I_p = N_p/N_s for ideal (power conserved V_p I_p = V_s I_s), step-up N_s>N_p V_s>V_p I_s
Ref: NCERT > Physics Book > Alternating Currents > Transformer, AC Generator and LC Oscillations
**Applications** include charge sharing for voltage division, Van de Graaff generator accumulates charge on spherical dome to high potential V = k Q/R, up to MV, using belt to transport charge, capacitance of sphere C=4π ε₀ R ≈10 pF for R=0.1 m, so Q= C V ≈10⁻⁸ C for 1000 V. Initial charge: Q = 12 × 10⁻⁶ × 30 = 3.6 × 10⁻⁴ C . Total capacitance: 12 + 12 = 24 μF . Final voltage: V = (Q/C) = (3.6 × 10⁻⁴/24 × 10⁻⁶) = 15 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq
Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Sharing of Charges, Common Potential and Applications
**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. Total resistance: Rtₒtₐl = 6 + 2 = 8 Ω . Current: I = (ε/Rtₒtₐl) = (8/8) = 1 A . Terminal voltage: V = ε - I r = 8 - 1 × 2 = 6 V . 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 =
Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity
**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Current: I = (ε/R + r) = (6/4 + 2) = 1 A . Terminal voltage: V = ε - I r = 6 - 1 × 2 = 4 V . 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 4 V,
Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity
**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Power: P = (V²/R) . Rearrange: V = √(P R) . Substitute: V = √(60 × 15) = √(900) = 30 V . 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 30 V,
Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect
**Electrical power** P = V I = I² R = V²/R (W), energy E = P t = I² R t (J), heating effect Joule's law H = I² R t. When internal r equals external R, total resistance 2R, I = ε/2R, power in external = I²R = ε²/4R, total = ε²/2R, fraction external = 1/2, illustrating maximum power transfer when R = r. Power: P = (V²/R) . Rearrange: V = √(P R) . Substitute: V = √(48 × 12) = √(576) = 24 V . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0,
Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect
