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Question

A \( 6 \, \Omega \) resistor dissipates \( 24 \, \text{W} \) of power. What is the current through it?

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Explanation

**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 = I² R . Rearrange: I = √((P/R)) . Substitute: I = √((24/6)) = √(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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