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#balance condition

3 public questions tagged with this topic.

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,

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

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 12 \, \Omega \), \( R_2 = 24 \, \Omega \), \( R_3 = 18 \, \Omega \). What is \( R_4 \)

**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. Balance condition: (R₁/R₂) = (R₃/R₄) . Substitute: (12/24) = (18/R₄) . Solve: 0.5 = (18/R₄) ⇒ R₄ = (18/0.5) = 36 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 36 Ω,

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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Wheatstone Bridge and Meter Bridge

A Wheatstone bridge has \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 25 \, \Omega \). 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: (30/60) = (25/R₄) . Solve: 0.5 = (25/R₄) ⇒ R₄ = (25/0.5) = 50 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 50 Ω,

Ref: NCERT > Physics Book > Current Electricity > Wheatstone Bridge and Meter Bridge

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