#resistance calculation

Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 25 \, \Omega \), \( R_2 = 50 \, \Omega \), \( R_3 = 20 \, \Omega \). What is \( R_4 \)

**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. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (25/50) = (20/R₄) . Solve: 0.5 = (20/R₄) ⇒ R₄ = (20/0.5) = 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 = ε - I r and P = I²R, evaluation yields 40 Ω,

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Temperature Dependence of Resistance and Resistivity

A nichrome wire has a resistance of \( 80 \, \Omega \) at \( 25^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 80 [1 + 1.7 × 10⁻⁴ (225 - 25)] . Calculate: R_t = 80 [1 + 1.7 × 10⁻⁴ × 200] = 80 [1 + 0.034] = 80 × 1.034 = 82.72 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,

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Temperature Dependence of Resistance and Resistivity

A nichrome wire has a resistance of \( 50 \, \Omega \) at \( 20^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 50 [1 + 1.7 × 10⁻⁴ (380 - 20)] . Calculate: R_t = 50 [1 + 1.7 × 10⁻⁴ × 360] = 50 [1 + 0.0612] = 50 × 1.0612 = 53.06 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,

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 22 \, \Omega \), \( R_2 = 44 \, \Omega \), \( R_3 = 15 \, \Omega \). What is \( R_4 \)

**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. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (22/44) = (15/R₄) . Solve: 0.5 = (15/R₄) ⇒ R₄ = (15/0.5) = 30 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 = ε - I r and P = I²R, evaluation yields 30 Ω,

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Resistance, Resistivity and Ohm's Law

A wire of length \( 8 \, \text{m} \) and resistance \( 16 \, \Omega \) is stretched to \( 16 \, \text{m} \). What is the

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. 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 × 16 = 64 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 = ε - I r and P = I²R, evaluation yields 64 Ω, consistent

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Resistance, Resistivity and Ohm's Law

A nichrome wire has a resistance of \( 90 \, \Omega \) at \( 20^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 90 [1 + 1.7 × 10⁻⁴ (260 - 20)] . Calculate: R_t = 90 [1 + 1.7 × 10⁻⁴ × 240] = 90 [1 + 0.0408] = 90 × 1.0408 = 93.67 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 =

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Resistance, Resistivity and Ohm's Law

A nichrome wire has a resistance of \( 70 \, \Omega \) at \( 30^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 70 [1 + 1.7 × 10⁻⁴ (270 - 30)] . Calculate: R_t = 70 [1 + 1.7 × 10⁻⁴ × 240] = 70 [1 + 0.0408] = 70 × 1.0408 = 72.86 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 =

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Resistance, Resistivity and Ohm's Law

A nichrome wire has a resistance of \( 80 \, \Omega \) at \( 20^\circ \text{C} \) and \( \alpha = 1.7 \times 10^{-4} \,

**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. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: R_t = 80 [1 + 1.7 × 10⁻⁴ (300 - 20)] . Calculate: R_t = 80 [1 + 1.7 × 10⁻⁴ × 280] = 80 [1 + 0.0476] = 80 × 1.0476 ≈ 83.81 Ω .

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