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#12 Ω resistor

2 public questions tagged with this topic.

Electrical Power, Energy and Heating Effect

A \( 12 \, \Omega \) resistor dissipates \( 48 \, \text{W} \) 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 = √((48/12)) = √(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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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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