#resistor values

Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 8 \, \Omega \), \( R_2 = 16 \, \Omega \), \( R_3 = 12 \, \Omega \), \( R_4 = 24 \, \Om

**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. Check balance: (R₁/R₂) = (8/16) = 0.5 , (R₃/R₄) = (12/24) = 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,

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

View Details

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

View Details

Wheatstone Bridge and Meter Bridge

In a Wheatstone bridge with \( R_1 = 30 \, \Omega \), \( R_2 = 60 \, \Omega \), \( R_3 = 15 \, \Omega \), and \( R_4 = 3

**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. Apply Kirchhoff’s rules. Let currents be I₁ (AB), I₂ (AD), I_g (BD). Junction B: I₁ = I_g + I₄ , Junction D: I₂ = I_g + I₃ . Loop BADB: 30 I₁ + 10 I_g - 60 I₂ = 0 ⇒ 3 I₁ + I_g - 6 I₂ = 0 . Loop BCDB: 60 (I₁ - I_g) - 10 I_g - 15 (I₂ + I_g) = 0 ⇒ 4 I₁ - 2

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

View Details

Kirchhoff's Laws and Combination of Resistors

A circuit has a \( 24 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance and three resistors \( 6 \, \Ome

**Series combination** R_eq = R₁+R₂+..., same current I through each, voltage divides proportionally V_i = I R_i. Parallel combination 1/R_p = 1/R₁+1/R₂+..., same voltage V across each, current divides inversely, equivalent R_p = (R₁ R₂)/(R₁+R₂) for two resistors. Parallel resistance: (1/R_p) = (1/6) + (1/12) + (1/24) = (4 + 2 + 1/24) = (7/24) ⇒ R_p = (24/7) ≈ 3.43 Ω . Total resistance: Rtₒtₐl = 3 + 3.43 = 6.43 Ω . Total current: I = (ε/Rtₒtₐl) = (24/6.43) ≈ 3.73 A . Voltage across parallel: V = I R_p = 3.73 × 3.43 ≈ 12.79 V . Current through 12 Ω :

Ref: NCERT > Physics Book > Current Electricity > Kirchhoff's Laws and Combination of Resistors

View Details