#resistivity calculation

Potentiometer, Conductivity and Special Cases

A conductor has a resistivity of \( 8 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 8 × 10⁻⁸ [1 + 4 × 10⁻³ (70 - 20)] . Calculate: rho_t = 8 × 10⁻⁸ [1 + 0.2] = 8 × 10⁻⁸ × 1.2 = 9.6 × 10⁻⁸ Ω m . Applying I = n e A v_d, R

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Potentiometer, Conductivity and Special Cases

A wire of length \( 9 \, \text{m} \) and cross-sectional area \( 1.5 \times 10^{-6} \, \text{m}^2 \) has a resistance of

**Potentiometer** measures potential difference without drawing current, using null deflection, principle V ∝ l, l balance length, accurate because no I r drop. Potential drop across resistor V = I R arises because electric field does work on charges, energy converted to heat, maintaining E = -dV/dx along wire. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (18 × 1.5 × 10⁻⁶/9) = 3 × 10⁻⁶ Ω m . 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,

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Wheatstone Bridge and Meter Bridge

A wire has a resistance of \( 10 \, \Omega \) at \( 20^\circ \text{C} \) and \( 12 \, \Omega \) at \( 100^\circ \text{C}

**Wheatstone bridge balance** condition R₁/R₂ = R₃/R₄, R₄ = R₂ R₃/R₁, when galvanometer current zero, potentials at midpoints equal. At balance, no current through galvanometer, enabling precise resistance measurement independent of source voltage. Use: R_t = R₀ [1 + α (T - T₀)] . Given: R₀ = 10 Ω , R_t = 12 Ω , T = 100° C , T₀ = 20° C . Substitute: 12 = 10 [1 + α (100 - 20)] . Solve: 12 = 10 + 80α ⇒ 80α = 2 ⇒ α = (2/80) = 0.025 × 10⁻² = 2.5 × 10⁻⁴ °C⁻¹ . Applying I = n e

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Resistance, Resistivity and Ohm's Law

A wire of length \( 4 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**Resistance** R = ρ l/A, ρ resistivity (Ω·m), l length (m), A area (m²), ρ = m/(n e² τ) from Drude model, τ average collision time. Ohm's law V = I R holds when ρ constant, independent of V. Volume constant stretching l→2l implies A→A/2, so R' = ρ·2l/(A/2)=4R, resistance quadruples when length doubles at constant volume. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (8 × 2 × 10⁻⁶/4) = 4 × 10⁻⁶ Ω m . 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 of length \( 3 \, \text{m} \) and cross-sectional area \( 3 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**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. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 5 Ω , A = 3 × 10⁻⁶ m² , l = 3 m . Substitute: rho = (5 × 3 × 10⁻⁶/3) = 5 × 10⁻⁶ Ω m . Applying I = n e A v_d, R = ρ l/A,

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

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