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#resistance change

3 public questions tagged with this topic.

Electric Current, Drift Velocity and Mobility

What happens to the total resistance of a circuit when an additional resistor is added in parallel to an existing resist

**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. For resistors in parallel, 1/Rtₒtₐl = 1/R₁ + 1/R₂ . Adding another resistor increases the sum of reciprocals, reducing Rtₒtₐl below the smallest individual resistance. 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 It

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and resistance \( 10 \, \Omega \) is stretched to \( 8 \, \text{m} \). What is the

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Volume constant: l A = l' A' ⇒ A' = (A/2) . New resistance: R' = (rho l'/A') = (rho (2l)/(A/2)) = 4 (rho l/A) = 4R = 4 × 10 = 40 Ω . 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 =

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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Resistance, Resistivity and Ohm's Law

A wire has a resistance of \( 30 \, \Omega \) at \( 25^\circ \text{C} \) and \( 31.5 \, \Omega \) at \( 75^\circ \text{C

**Resistivity** depends on material and temperature, not geometry. For metallic conductor, ρ ≈10⁻⁸ Ω·m for copper. Given ρ=4×10⁻⁸ Ω·m, l=2 m, A=π r², R calculation uses R=ρ l/A. Stretching wire conserves volume V = l A = l' A', so A' = A l/l', new R' = ρ l'²/V. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 31.5 = 30 [1 + α (75 - 25)] . Solve: 31.5 = 30 + 1500α ⇒ 1500α = 1.5 ⇒ α = (1.5/1500) = 1.0 × 10⁻³ °C⁻¹ . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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