Potentiometer, Conductivity and Special Cases

Potentiometer, Conductivity and Special Cases

Why does the power dissipated in a resistor increase quadratically with current?

**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. Power P = I² R . Since power depends on the square of the current ( I² ), doubling the current quadruples the power, assuming resistance remains constant. 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 Power depends on current squared,

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

View Details

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 conductor has a resistivity of \( 8 \times 10^{-8} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\te

**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. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 8 × 10⁻⁸ [1 + 4 × 10⁻³ (70 - 20)] . Calculate: rho_t = 8 × 10⁻⁸ [1 + 0.2] = 8 × 10⁻⁸ × 1.2 = 9.6 × 10⁻⁸ Ω m . Applying I = n e A v_d, R

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

View Details

Potentiometer, Conductivity and Special Cases

A wire of length \( 6 \, \text{m} \) and cross-sectional area \( 5 \times 10^{-6} \, \text{m}^2 \) has a resistance of \

**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. Resistance: R = (rho l/A) . Rearrange: rho = (R A/l) . Substitute: rho = (12 × 5 × 10⁻⁶/6) = 10 × 10⁻⁶ = 1.0 × 10⁻⁵ 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 - I r and P = I²R, evaluation yields 1.0 ×

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

View Details

Potentiometer, Conductivity and Special Cases

What is the physical significance of the temperature coefficient of resistivity being positive for metals?

**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. A positive temperature coefficient ( α ) means resistivity ( rho_t = rho₀ [1 + α (T - T₀)] ) increases with temperature, as increased lattice vibrations reduce the mean free time between collisions, raising 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 P =

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

View Details

Potentiometer, Conductivity and Special Cases

A copper wire carries \( 2 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \tim

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (2/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁵) . Calculate: A = (2/1.088 × 10⁵) ≈ 1.84 × 10⁻⁵ m² . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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

View Details

Potentiometer, Conductivity and Special Cases

A \( 15 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance delivers a current of \( 2.5 \, \text{A} \) to

**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. Terminal voltage: V = ε - I r = 15 - 2.5 × 1 = 12.5 V . Resistance: R = (V/I) = (12.5/2.5) = 5 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 5.0 Ω,

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

View Details

Potentiometer, Conductivity and Special Cases

A circuit has a \( 10 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance and two resistors \( 5 \, \Omega

**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. Parallel resistance: (1/R_p) = (1/5) + (1/10) = (2 + 1/10) = (3/10) ⇒ R_p = (10/3) ≈ 3.33 Ω . Total resistance: Rtₒtₐl = 1 + 3.33 = 4.33 Ω . Current: I = (ε/Rtₒtₐl) = (10/4.33) ≈ 2.31 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

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

View Details

Potentiometer, Conductivity and Special Cases

A copper wire carries \( 2.72 \, \text{A} \) with a drift speed of \( 8 \times 10^{-5} \, \text{m/s} \). If \( n = 8.5 \

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (2.72/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 8 × 10⁻⁵) . Calculate: A = (2.72/1.088 × 10⁵) ≈ 2.5 × 10⁻⁵ m² . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT],

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

View Details

Potentiometer, Conductivity and Special Cases

In a circuit with a battery, why does the internal resistance of the battery affect the maximum power delivered to an ex

**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. Maximum power transfer occurs when the external resistance equals the internal resistance ( R = r ), as P = I² R = (ε / (R + r))² R . Internal resistance limits current, influencing the power distribution. 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

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

View Details

Potentiometer, Conductivity and Special Cases

A copper wire carries \( 4.5 \, \text{A} \) with a drift speed of \( 1.8 \times 10^{-4} \, \text{m/s} \). If \( n = 8.5

**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. Drift speed: v_d = (I/n e A) . Rearrange: A = (I/n e v_d) . Substitute: A = (4.5/8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.8 × 10⁻⁴) . Calculate: A = (4.5/2.448 × 10⁵) ≈ 1.84 × 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

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