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#external resistor

2 public questions tagged with this topic.

Electrical Power, Energy and Heating Effect

A \( 9 \, \text{V} \) battery with \( 0.5 \, \Omega \) internal resistance is connected to a \( 8.5 \, \Omega \) resisto

**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). Total resistance: Rtₒtₐl = 8.5 + 0.5 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Power: P = I² R = 1² × 8.5 = 8.5 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

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

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Temperature Dependence of Resistance and Resistivity

A cell of emf \( 6 \, \text{V} \) and internal resistance \( 2 \, \Omega \) is connected to an external resistor \( 4 \,

**Resistivity temperature variation** ρ_t = ρ₀[1+α(T-T₀)], α ≈4×10⁻³ /°C for copper, 1.7×10⁻⁴ /°C for nichrome. Given R=60 Ω at 30°C, α=1.7×10⁻⁴ /°C, T=330°C, ΔT=300°C, R_t=60[1+1.7×10⁻⁴×300]=60×1.051=63.06 Ω, modest increase for nichrome due to small α. Current: I = (ε/R + r) = (6/4 + 2) = 1 A . Terminal voltage: V = ε - I r = 6 - 1 × 2 = 4 V . 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 V,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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