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#balance resistor

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Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 18 \, \Omega \), \( R_2 = 36 \, \Omega \), \( R_3 = 12 \, \Omega \). What is \( R_4 \)

**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (18/36) = (12/R₄) . Solve: 0.5 = (12/R₄) ⇒ R₄ = (12/0.5) = 24 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 24 Ω,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 15 \, \Omega \), \( R_2 = 45 \, \Omega \), \( R_3 = 10 \, \Omega \). What is \( R_4 \)

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (15/45) = (10/R₄) . Solve: (1/3) = (10/R₄) ⇒ R₄ = 10 × 3 = 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,

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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