Part A If the internal resistance of the batteries is negligible, what power is
ID: 2136769 • Letter: P
Question
Part A If the internal resistance of the batteries is negligible, what power is delivered to the bulb? Part B Part B Part 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.) Part 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.) A typical small flashlight contains two batteries, each having an emf of 1.50V connected in series with a bulb having a resistance of 17. If the internal resistance of the batteries is negligible, what power is delivered to the bulb? If the batteries last for a time of 5.6h , what is the total energy delivered to the bulb? Ohm If the batteries last for a time of 5.6h , what is the total energy delivered to the bulb? 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.)Explanation / Answer
A) total voltage = 1.5 +1.5 = 3V
power delivered to bulb = V^2/ R = 3^2/ 17 =9/17 = .5294 W
b) t= 5x60+6 =366 min = 366x60 =21960 s
E=( V^2 / R) x t = .5294 x 21960 = 1.16 x10^4 J
c) new power = .5294 / 2 =.2647 let R1 be the combined internal resistance then
(3^2) / (R1+17) = .2647 solvinng we get R1= 17 ohm