#load resistance

Electrical Power, Energy and Heating Effect

In a battery-powered circuit, if the external load resistance becomes very large, what happens to the current drawn from

**Electrical power** P = V I = I² R = V²/R (W), energy E = P t = I² R t (J), heating effect Joule's law H = I² R t. When internal r equals external R, total resistance 2R, I = ε/2R, power in external = I²R = ε²/4R, total = ε²/2R, fraction external = 1/2, illustrating maximum power transfer when R = r. Current I = ε / (R + r) . As external resistance R becomes very large, R + r ≈ R , so I ≈ ε / R , approaching zero as R to ∞ . Applying I = n

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

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Electric Current, Drift Velocity and Mobility

A \( 15 \, \text{V} \) battery with \( 3 \, \Omega \) internal resistance delivers a current of \( 2 \, \text{A} \) to a

**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. Terminal voltage: V = ε - I r = 15 - 2 × 3 = 9 V . Resistance: R = (V/I) = (9/2) = 4.5 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 4.5

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

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Resistance, Resistivity and Ohm's Law

A battery of emf \( 15 \, \text{V} \) and internal resistance \( 3 \, \Omega \) is connected to a resistor. If the termi

**Ohm's law deviation** at high fields occurs when τ or n vary with E, resistivity ρ = m/(n e² τ) changes, non-ohmic behaviour seen in semiconductors, electrolytes. At moderate fields, linear V-I holds, slope = R. Voltage drop: I r = ε - V = 15 - 12 = 3 V . Current: I = (3/r) = (3/3) = 1 A . Resistance: R = (V/I) = (12/1) = 12 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 12 Ω,

Ref: NCERT > Physics Book > Current Electricity > Resistance, Resistivity and Ohm's Law

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EMF, Internal Resistance and Cells Combination

A \( 20 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance delivers a current of \( 2.5 \, \text{A} \) to

**Cells combination** series ε_eq = Σ ε_i, r_eq = Σ r_i, parallel for identical cells ε_eq = ε, r_eq = r/n, n number of cells. Maximum current when external R = r_eq, power transfer theorem, explaining why matching resistances maximizes power. Terminal voltage: V = ε - I r = 20 - 2.5 × 2 = 15 V . Resistance: R = (V/I) = (15/2.5) = 6 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 6.0 Ω,

Ref: NCERT > Physics Book > Current Electricity > EMF, Internal Resistance and Cells Combination

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