Practice question
Question
A \( 18 \, \text{V} \) battery with negligible internal resistance is connected to a \( 6 \, \Omega \)
and \( 12 \, \Omega \) resistor in series. What is the power dissipated in the \( 6 \, \Omega \)
resistor?
Explanation
**Internal resistance** causes voltage drop I r inside battery, so V = ε - I r decreases with I. For 16 V battery, r=2 Ω, I=2 A, V=16-4=12 V, external R = V/I =6 Ω. Measurement of V and I yields r = (ε - V)/I. Total resistance: R = 6 + 12 = 18 Ω . Current: I = (V/R) = (18/18) = 1 A . Power: P = I² R = 1² × 6 = 6 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
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