- #1
moenste
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Homework Statement
A bulb is used in a torch which is powered by two identical cells in series each of EMF 1.5 V. The bulb then dissipates power at the rate of 625 mW and the PD across the bulb is 2.5 V. Calculate (i) the internal reistance of each cell and (ii) the energy dissipated in each cell in one minute.
Answers: (i) 1 Ω, (ii) 3.75 J.
2. The attempt at a solution
(i) I looked for current. P = V I → I = P / V = 625 * 10-3 / 2.5 = 0.25 A. I found the resistance of the lamp: P = V2 / R → R = V2 / P = 2.52 / 625 * 10-3 = 10 Ω. Then plug in E = I (r + R) → 2.5 = 0.25 (r + 10) → r = 0 Ω. I also tried (2.5 + 1.5 + 1.5) = 0.25 (r + 10) → r = 12 Ω / 3 = 4 Ω.
(ii) W = V I t → W = (2.5 + 1.5 + 1.5) * 0.25 * 60 - 82.5 J.
What's wrong with every part?