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Question

A flashlight consists of two 1.5-V batteries connected in series to a bulb with resistence R = 10 Ω.

(a) Which is the power delivered to the bulb? (0.9)

(b) Batteries run down when they acquire an internal resistance. How large is the additional resistance

if the power delivered to the bulb has decreased by one-third of its initial value?

Answer 1

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Power = V²/R = (1.5+1.5)²/10Ω = 0.9W <----- (a)

2/3 * 0.9 = 0.6W = i²*10 -----> i² = 0.06 -----> i = 0.245A

0.3W = i²*Ri -----> 0.3/0.245² = Ri = 5 Ω <----- (b)