An automobile battery has an emf of 12.6 V and an internal resistance of 0.0800Ω. The headlights

An automobile battery has an emf of 12.6 V and an internal...

An automobile battery has an emf of 12.6 $\mathrm{V}$ and an internal resistance of 0.0800$\Omega$ . The headlights together have an equivalent resistance of 5.00$\Omega$ (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and (b) when the starter motor is operated, requiring an additional 35.0 A from the battery?

An automobile battery has an emf of 12.6 $\mathrm{V}$ and an internal resistance of 0.0800$\Omega$ . The headlights together have an equivalent resistance of 5.00$\Omega$ (assumed constant). What is the potential difference across the headlight bulbs (a) when they are the only load on the battery and
(b) when the starter motor is operated, requiring an additional 35.0 A from the battery?

Key Concepts

Ohm's Law

This fundamental law in electrical circuits states that the current through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. It provides the basis for calculating voltage drops, currents, and resistances in circuits.

Electromotive Force (EMF)

This is the voltage provided by the battery when no current is flowing. It represents the energy per unit charge generated by the chemical reactions within the battery and forms the basis for understanding how batteries drive current through circuits.

Internal Resistance

Every battery has a small resistance that opposes the flow of current. This resistance causes a voltage drop within the battery itself when current flows, meaning the terminal voltage available to the external circuit is lower than the EMF.

Series Circuits

In a series circuit, components are connected end-to-end so that the same current flows through each component. In the context of internal resistance and load, the battery’s internal resistance and the external resistance are in series, which means the total resistance is the sum of the individual resistances.

Voltage Division

This concept explains how the total voltage in a series circuit is divided among the series elements according to their resistances. It is relevant for calculating the potential difference across any individual component (like the load) in the circuit by accounting for the voltage drop across the internal resistance.

Transcript

00:01 Okay, so in this problem, we have a battery that has an emf of 12 .6 volts.

00:07 It has an internal resistance of 0 .08 oms, and the light bulbs attached to it have a combined resistance that's static of 5 oms.

00:17 The two questions that were asked are, what is the voltage across the light bulbs when they are the only thing connected to the battery? so our first situation looks like this.

00:29 Here's our battery or voltage source.

00:32 Here's our lights.

00:34 And this has the emf of the 12 .6.

00:39 And i'm going to call this rl for the lights.

00:44 And so if we need to figure out what the voltage difference is across these, what we need to do is figure out how much voltage drop there is from battery internally.

00:55 And the only way to do that, or the easiest way to do that, is oms law figuring out the current.

01:01 If i know the current flow, then i can figure out the voltage.

01:07 So, based off the information that we have and omslaw, v is equal to i times r, we can try and figure out the current.

01:20 So let's do that.

01:21 The current is going to be equal to the total voltage divided by the resistance.

01:27 Now if we look at the total voltage supplied to the entire circuit, including the internal battery, that's our e.

01:36 And then this r down here is total.

01:39 So this will be rl, the lights, plus the batteries internal, because they're in series.

01:47 So 12 .6 divided by 5 plus 0 .08.

01:56 12 .6 divided by 5 plus 0 .08...

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