Practice question
Question
A \( 12 \, \Omega \) resistor dissipates \( 48 \, \text{W} \) of power. What is the current through it?
Explanation
**Conductivity** σ = 1/ρ = n e² τ/m (S/m), τ relaxation time, m electron mass. Resistivity deviation at high fields occurs when τ depends on E or n changes due to impact ionization, breaking Ohm's law, seen in varistors, gas discharge. 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,
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.