#Joule heating

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

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Electrical Power, Energy and Heating Effect

A \( 8 \, \Omega \) resistor carries a current of \( 2.5 \, \text{A} \) for \( 10 \, \text{s} \). What is the energy dis

**Power dissipation** in resistor converts electrical energy to heat, P = V²/R inversely proportional to R for fixed V, directly proportional for fixed I. For battery with internal r, power wasted internally = I² r, useful power = I² R, efficiency η = R/(R+r). Energy: W = I² R t . Substitute: W = (2.5)² × 8 × 10 = 6.25 × 80 = 500 J . 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 500 J,

Ref: NCERT > Physics Book > Current Electricity > Electrical Power, Energy and Heating Effect

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Electric Current, Drift Velocity and Mobility

What is the primary source of energy dissipation in a resistor carrying current?

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Energy dissipation in a resistor occurs as electrons collide with lattice ions, transferring kinetic energy gained from the electric field into thermal energy (heat) via lattice vibrations. 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 Collisions with

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

A \( 6 \, \Omega \) resistor carries a current of \( 4 \, \text{A} \) for \( 5 \, \text{s} \). What is the energy dissip

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Energy: W = I² R t . Substitute: W = 4² × 6 × 5 = 16 × 30 = 480 J . 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 480 J,

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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