Electrical Power, Energy and Heating Effect

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,

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Electrical Power, Energy and Heating Effect

A \( 8 \, \Omega \) resistor dissipates \( 32 \, \text{W} \) 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 = √((32/8)) = √(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,

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Electrical Power, Energy and Heating Effect

A \( 5 \, \Omega \) resistor carries a current of \( 3 \, \text{A} \) for \( 20 \, \text{s} \). 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 = 3² × 5 × 20 = 9 × 100 = 900 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 900 J,

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Electrical Power, Energy and Heating Effect

A \( 6 \, \Omega \) resistor carries a current of \( 3 \, \text{A} \) for \( 30 \, \text{s} \). What is the energy dissi

**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. Energy: W = I² R t . Substitute: W = 3² × 6 × 30 = 9 × 180 = 1620 J . Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0,

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Electrical Power, Energy and Heating Effect

A \( 10 \, \Omega \) resistor dissipates \( 40 \, \text{W} \) of power. What is the voltage across 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 = (V²/R) . Rearrange: V = √(P R) . Substitute: V = √(40 × 10) = √(400) = 20 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 20 V,

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Electrical Power, Energy and Heating Effect

A \( 9 \, \text{V} \) battery with \( 1 \, \Omega \) internal resistance is connected to a \( 8 \, \Omega \) resistor. W

**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 + 1 = 9 Ω . Current: I = (ε/Rtₒtₐl) = (9/9) = 1 A . Power: P = I² r = 1² × 1 = 1 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

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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,

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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

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Electrical Power, Energy and Heating Effect

A \( 8 \, \Omega \) resistor carries a current of \( 2.5 \, \text{A} \) for \( 10 \, \text{s} \). What is the energy dis

**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). Energy: W = I² R t . Substitute: W = (2.5)² × 8 × 10 = 6.25 × 80 = 500 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 500 J,

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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

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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

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Electrical Power, Energy and Heating Effect

A \( 8 \, \Omega \) resistor carries a current of \( 2 \, \text{A} \) for \( 15 \, \text{s} \). 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 × 8 × 15 = 4 × 120 = 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,

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