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