Question

(6.9)

A typical small flashlight contains two batteries, each having an emf of 1.50 V connected in series with a bulb having a resistance of 18 Ω

A) If the internal resistance of the batteries is negligible, what power is delivered to the bulb?

B) If the batteries last for a time of 6.0 h , what is the total energy delivered to the bulb?

C) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

Answer 1

Part (A): It is given that the two batteries are connected in series, -1.50+1.50 P=VI since l=- R. R. 18 Thus, the power delivered in the bulb is[0.5 W Part (B) Consider the equation for total energy delivered Energy Powerxtime Substitute, 0.5W for 'P' and 6h for't' in (1) Energy,oal-0.5x 6x3600 -10800J Thus, the total energy delivered to the bulb is 10800 J Part (C): Let (r+R) be the intemal resistance when the power developed is P and 'r' be the combined internal resistance ofthe batteries

1(v) IR- 2 R 2 R (r+R2 (r+R) 18(5-1) -18V2-1 7.4558 Ω Thus, the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value is 7.4558 Ω