Practice question
Question
A \( 24 \, \text{V} \) battery with negligible internal resistance is connected to a \( 4 \, \Omega \)
and \( 8 \, \Omega \) resistor in series. What is the power dissipated in the \( 4 \, \Omega \)
resistor?
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. Total resistance: R = 4 + 8 = 12 Ω . Current: I = (V/R) = (24/12) = 2 A . Power: P = I² R = 2² × 4 = 16 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 16 W,
Discussion
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