#Wheatstone bridge

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

View Details

Potentiometer, Conductivity and Special Cases

A Wheatstone bridge has \( R_1 = 26 \, \Omega \), \( R_2 = 52 \, \Omega \), \( R_3 = 20 \, \Omega \). 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: (26/52) = (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

View Details

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

View Details

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

View Details

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,

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

View Details

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, when the bridge is balanced, what is the potential difference across the galvanometer?

**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. At balance, the potential at the two junctions connected to the galvanometer is equal (due to the ratio condition R₁ / R₂ = R₃ / R₄ ), so the potential difference across the galvanometer is zero, and no current flows through it. 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

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

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,

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

View Details

Wheatstone Bridge and Meter Bridge

In a Wheatstone bridge, what is the condition for no current to flow through the galvanometer?

**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. No current flows when the bridge is balanced, i.e., R₁ / R₂ = R₃ / R₄ , making the potentials at the galvanometer’s nodes equal, so the potential difference across it is zero. 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 Ratio of resistances in opposite a

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

View Details

Wheatstone Bridge and Meter Bridge

A Wheatstone bridge with \( R_1 = 6 \, \Omega \), \( R_2 = 12 \, \Omega \), \( R_3 = 9 \, \Omega \), \( R_4 = 18 \, \Ome

**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) = (6/12) = 0.5 , (R₃/R₄) = (9/18) = 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