#resistor power

Electrical Power, Energy and Heating Effect

A \( 10 \, \Omega \) resistor dissipates \( 25 \, \text{W} \) of power. What is the current through it?

**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). Power: P = I² R . Rearrange: I = √((P/R)) . Substitute: I = √((25/10)) = √(2.5) ≈ 1.58 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 1.58 A,

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

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Electrical Power, Energy and Heating Effect

A \( 6 \, \Omega \) resistor dissipates \( 24 \, \text{W} \) of power. What is the current through it?

**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. Power: P = I² R . Rearrange: I = √((P/R)) . Substitute: I = √((24/6)) = √(4) = 2 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 2.0 A,

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

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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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Electric Current, Drift Velocity and Mobility

A \( 5 \, \Omega \) resistor carries a current of \( 4 \, \text{A} \) for \( 15 \, \text{s} \). What is the energy dissi

**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 × 5 × 15 = 16 × 75 = 1200 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 1200 J,

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

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EMF, Internal Resistance and Cells Combination

A \( 15 \, \text{V} \) battery with negligible internal resistance is connected to a \( 3 \, \Omega \) and \( 12 \, \Ome

**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. Total resistance: R = 3 + 12 = 15 Ω . Current: I = (V/R) = (15/15) = 1 A . Power: P = I² R = 1² × 12 = 12 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

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

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EMF, Internal Resistance and Cells Combination

A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a \( 6 \, \Omega \) and \( 12 \, \Ome

**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Total resistance: R = 6 + 12 = 18 Ω . Current: I = (V/R) = (18/18) = 1 A . Power: P = I² R = 1² × 6 = 6 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

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

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EMF, Internal Resistance and Cells Combination

A \( 10 \, \text{V} \) battery with negligible internal resistance is connected to a \( 2 \, \Omega \) and \( 3 \, \Omeg

**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. Total resistance: R = 2 + 3 = 5 Ω . Current: I = (V/R) = (10/5) = 2 A . Power: P = I² R = 2² × 3 = 12 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

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

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