A typical small flashlight contains two batteries, each having an emf of 1.5 V, connected in series with a bulb having resistance 17 Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb?

Video transcript

Hey everyone in this problem, we have a vehicle that has 212 volt batteries connected in series the combined batteries drive a starter motor of resistance 5.4 if the batteries are ideal, were asked to determine the power supply to the motor. Now recall that the power P is equal to the voltage V times the current. I Now we're giving information about the voltage and were given information about the resistance but were not given information about the current. I well recall that we can relate voltage current and resistance through OEMs Law. And so we can write I as V over R. Which means that our power can be written as V squared over. R our voltage while we have 12 volt batteries, ideal batteries that are connected in series. This tells us that our total voltage is going to be 24 volts and our resistance is 5.4 ohms. So we get 24 volts squared divided by 5.4 ohms. Which gives us a power of 106. watts. Now, if we look at our answer choices and we round to the nearest wat, we see the answer choice. C is going to be the correct answer. The power is 107 watts. That's it for this one. Thanks everyone for watching. See you in the next video