Wheatstone Bridge and Meter Bridge

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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Wheatstone Bridge and Meter Bridge

A nichrome wire has a resistance of \( 50 \, \Omega \) at \( 25^\circ \text{C} \) and \( 55 \, \Omega \) at a higher tem

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Use: R_t = R₀ [1 + α (T - T₀)] . Substitute: 55 = 50 [1 + 1.7 × 10⁻⁴ (T - 25)] . Solve: 55 = 50 + 50 × 1.7 × 10⁻⁴ (T - 25) ⇒ 5 = 8.5 × 10⁻³ (T - 25) . T - 25 = (5/8.5 × 10⁻³) ≈ 588 ⇒ T ≈ 613° C . Applying I = n e A v_d, R

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Wheatstone Bridge and Meter Bridge

Why does the resistivity of an insulator decrease dramatically when impurities are added?

**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. Impurities in insulators (e.g., semiconductors) introduce additional charge carriers by creating energy levels within the bandgap, allowing more electrons or holes to conduct, reducing resistivity. 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 Charge carrier density increases,

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Wheatstone Bridge and Meter Bridge

A copper wire carries \( 4 \, \text{A} \) with a drift speed of \( 1.0 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5 \t

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (4/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁴) . Calculate: A = (4/1.36 × 10⁵) ≈ 2.94 × 10⁻⁵ m² . 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

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Wheatstone Bridge and Meter Bridge

A copper wire of length \( 2 \, \text{m} \) and cross-sectional area \( 2 \times 10^{-6} \, \text{m}^2 \) carries a curr

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Drift speed is given by v_d = (I/n e A) . Given: I = 2 A , n = 8.5 × 10²⁸ m⁻³ , e = 1.6 × 10⁻¹⁹ C , A = 2 × 10⁻⁶ m² . Substitute: v_d = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 2 × 10⁻⁶) . Calculate: v_d = (2/2.72 × 10⁴) = 7.35 × 10⁻⁵ m/s . Applying I = n e

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Wheatstone Bridge and Meter Bridge

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 3 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 12 - 3 × 2 = 6 V . Resistance: R = (V/I) = (6/3) = 2 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 2.0 Ω,

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Wheatstone Bridge and Meter Bridge

A conductor of length \( 1 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Resistance is given by R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 2 Ω , A = 1 × 10⁻⁶ m² , l = 1 m . Substitute: rho = (2 × 1 × 10⁻⁶/1) = 2 × 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

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Wheatstone Bridge and Meter Bridge

Why does a superconductor exhibit zero resistance below its critical temperature?

**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. Below the critical temperature, electrons in a superconductor form Cooper pairs, which move without scattering off lattice ions, eliminating resistance as there’s no energy loss to collisions. 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 No collisions with lattice,

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Wheatstone Bridge and Meter Bridge

Why does the equivalent resistance of two resistors in parallel always lie between zero and the smallest individual resi

**Meter bridge** uses uniform wire of length 1 m, balance length l gives R_unknown = R_known·l/(100-l). Principle same as Wheatstone, with wire resistances proportional to lengths, allowing unknown resistance determination from length ratio. For parallel resistors, 1/Rₑq = 1/R₁ + 1/R₂ , so Rₑq = R₁ R₂ / (R₁ + R₂) . This value is less than the smaller resistance (e.g., if R₁ < R₂ , Rₑq < R₁ ) but greater than zero, as additional paths reduce 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

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Wheatstone Bridge and Meter Bridge

A copper wire carries a current of \( 3 \, \text{A} \) with a drift speed of \( 1.2 \times 10^{-4} \, \text{m/s} \). If

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (3/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.2 × 10⁻⁴) . Calculate: A = (3/1.632 × 10⁶) ≈ 1.84 × 10⁻⁶ m² . 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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 7 \, \Omega \), \( R_2 = 14 \, \Omega \), \( R_3 = 21 \, \Omega \), \( R_4 = 42 \, \Om

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Check balance: (R₁/R₂) = (7/14) = 0.5 , (R₃/R₄) = (21/42) = 0.5 . Bridge is balanced. Since balanced, I_g = 0 A . 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 0 A,

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Wheatstone Bridge and Meter Bridge

Why does the resistance of a conductor increase when its cross-sectional area is reduced?

**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. Resistance R = rho l / A . Reducing A increases R inversely, as fewer charge carriers can pass through a smaller area, increasing opposition to current flow. 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 Fewer paths for current,

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