Final answer:
In a small flashlight with two batteries connected in series, the power delivered to the bulb can be calculated. The current flowing through the circuit is found using Ohm's Law. The power delivered to the bulb is then determined by multiplying the current and voltage.
Step-by-step explanation:
In a small flashlight, two batteries are connected in series with a bulb. The batteries have an electromotive force (emf) of 1.5 V each and a negligible internal resistance. The bulb has a resistance of 16 ?. To calculate the power delivered to the bulb, we need to find the current flowing through the circuit first.
(a) current flowing through the circuit:
The total resistance of the circuit is the sum of the bulb resistance and the total internal resistance of the batteries:
Total resistance = bulb resistance + total internal resistance of batteries
Total resistance = 16 ? + 0 ? (negligible internal resistance)
Total resistance = 16 ?
Using Ohm's Law, we can find the current flowing through the circuit:
Current (I) = Voltage (V) / Resistance (R)
Current = 1.5 V / 16 ?
Current = 0.09375 A (or 93.75 mA)
(b) power delivered to the bulb:
Power (P) = Current (I) * Voltage (V)
Power = 0.09375 A * 1.5 V
Power = 0.140625 W (or 140.625 mW)
So, the power delivered to the bulb is approximately 0.140625 W (or 140.625 mW).