#electric power

Electrical Power, Energy and Heating Effect

A \( 7 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 25 \, \text{s} \). What is the energy dissi

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Energy: W = I² R t . Substitute: W = 2² × 7 × 25 = 4 × 175 = 700 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 700 J,

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

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Electrical Power, Energy and Heating Effect

A \( 12 \, \text{V} \) battery with \( 2 \, \Omega \) internal resistance is connected to a \( 10 \, \Omega \) resistor.

**Heating effect** depends on I² R t, explaining why high currents cause significant heating, need for thick wires, fuses. Energy supplied by battery ε I t = I²(R+r)t, split between external and internal as per resistances. Total resistance: Rtₒtₐl = 10 + 2 = 12 Ω . Current: I = (ε/Rtₒtₐl) = (12/12) = 1 A . Power: P = I² r = 1² × 2 = 2 W . 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 2.0 W,

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

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Electrical Power, Energy and Heating Effect

A \( 5 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 4.5 \, \Omega \) resisto

**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. Total resistance: Rtₒtₐl = 4.5 + 0.5 = 5 Ω . Current: I = (ε/Rtₒtₐl) = (5/5) = 1 A . Power: P = I² R = 1² × 4.5 = 4.5 W . Applying I = n e A

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

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Electric Current, Drift Velocity and Mobility

A \( 5 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 10 \, \text{s} \). How much energy is dissi

**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 dissipated: W = I² R t . Given: I = 2 A , R = 5 Ω , t = 10 s . Substitute: W = (2)² × 5 × 10 = 4 × 50 = 200 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

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

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