A typical small flashlight contains two batteries, each having an emf of 1.50 V

Question

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 14 Ohm If the internal resistance of the batteries is negligible, what power is delivered to the bulb? If the batteries last for a time of 6.0 h. 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) P = EI = E^2/R = (2V)^2/R = 3^2/14 = 0.64 W
B) Energy = Pt = 0.64*6*3600 = 13824 J = 1.4*10^4 J

C) half power = 3²/28
Voltage at half power
P = V²/R
3²/28 = V²/14 ======> V = (14 x 3²/28) = 2.121
The current at this voltage is V = IR =====> 2.121 = I x 14 =====> I = 2.121/14 = 0.15152

The voltage across the internal resistance is Vi = 3 - 2.121 = 0.87868
Thus the internal resistance is
Vi = IRi =====> 0.87868 = 0.15152 x Ri
Ri = 0.87868 / 0.15152 = 5.8 ohms

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