A typical small flashlight contains two batteries, each...

A typical small flashlight contains two batteries, each having an emf of $1.5 \mathrm{~V},$ connected in series with a bulb having resistance $17 \Omega .$ (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for $5.0 \mathrm{~h}$, 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.

Key Concepts

Series Connection

A series circuit is one where components are connected end-to-end, ensuring the same current flows through each element. In series circuits, the total voltage is the sum of the individual emfs, and the total resistance is the sum of all resistances, which is crucial for analyzing circuit performance.

Electromotive Force (EMF)

Electromotive force, or emf, represents the voltage generated by a battery or a power source when no current is flowing. In a series configuration, individual battery emfs add up, providing the net voltage available for driving current through the circuit.

Power Calculation in Circuits

The power delivered to a component, such as a bulb, can be calculated using formulas like P = V²/R or P = I²R, where V represents voltage, I represents current, and R is resistance. This concept is essential for determining how electrical energy is converted into other forms, such as light or heat.

Energy Delivered Over Time

Energy in electrical circuits is the product of power and time, typically calculated with E = P × t. This concept relates the instantaneous power consumption to the total energy used over a period, which is important for evaluating the performance of devices such as flashlights.

Internal Resistance of Batteries

Internal resistance is an inherent property of batteries that causes a voltage drop when current flows, affecting the efficiency and performance of the power source. Understanding internal resistance helps explain why the effective voltage delivered to a load can decrease under load conditions.

Impact of Internal Resistance on Circuit Performance

The presence of internal resistance in batteries introduces additional resistance in the circuit, which can reduce the power delivered to the load. Analyzing how internal resistance affects voltage distribution and power dissipation is key to understanding the behavior of real-world circuits, especially in cases where the performance declines as the batteries discharge.

Transcript

00:01 The things we know from the problem is that there are two batteries that are connected in series, and each battery is 1 .5 volts.

00:10 And connected to these batteries is a light bulb that has a resistance of 17 oms.

00:18 In part a of the problem, we're asked to find what power is delivered to the bulb, assuming that we can neglect the internal resistance, of the batteries.

00:34 Right away, we can use the equation p is equal to v squared over r since we have actually the emf of the batteries, and if we ignore the resistance, then this is also the terminal voltage, so we can use it directly in here, and we're given resistance.

01:01 So there are two batteries, so that's 2 times 1 .5 volt squared over 17 oms.

01:15 And the power that's dissipated here, or that's the levered to the ball is 0 .53.

01:30 And now we move on to part b of the problem.

01:34 And in part b of the problem, we're asked that if the battery lasts for five hours, so we have d equals the five hours.

01:47 What is the total energy delivered to the bulb? okay, so i'm going to convert this immediately to seconds.

01:59 So if we multiply by 60 minutes in one hour, and then 60 seconds in one minute, this will give us the time.

02:16 In seconds and then so i won't i'll just save this part of the computation for last so we want to know what energy is the levered so energy is power times time so we have found the power in part a and then we have written out the time here in this long expression so you e which is the electrical energy delivered to the bulb is 0 .53 watts times 5 times 60 seconds.

03:08 And remember that watts is actually joules per second, so the seconds will cancel out and just giving me dimensions of joules, which is the dimension of energy.

03:20 So when you put all these numbers in your calculator, you find that you have 9 ,540 jules.

03:36 And i urge you to actually put in the numbers yourself and make sure that i haven't made any mistakes.

03:46 Okay, and now we're going to move on to part c of the problem.

03:51 So let's read part c.

03:52 In part c, it says the resistance of real batteries increase as they run down.

04:00 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? and we are to assume that the resistance of the bulb is constant.

04:16 So when the power changes, we're going to use the same resistance of the bulb, that same 17 watts, i mean 17 oms, sorry.

04:29 So let's first compute what half the power is...

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