An automobile battery has an emf of 12.6 V and an internal resistance of 0.080 Ω. The headlight

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 $0.080 \Omega$. The headlights have a total 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, taking an additional $35.0 \mathrm{~A}$ from the battery?

Key Concepts

Voltage Divider

A voltage divider is a configuration where the total voltage of a source is distributed among series components based on their resistances. This concept is key to calculating the voltage across a load when considering the additional voltage drop caused by the internal resistance of the source.

Ohm's Law

Ohm's Law is a fundamental principle in electrical circuits, stating that the current through a conductor is directly proportional to the voltage across it and inversely proportional to its resistance. This law is essential for determining current, voltage, and resistance values in series and parallel circuits.

Electromotive Force (EMF)

EMF is the voltage generated by a power source, such as a battery, when no current flows. It represents the ideal driving force for electrons in the circuit and sets the upper limit for the voltage available to any connected load.

Internal Resistance

Internal resistance is the inherent opposition to current flow within a power source like a battery. It causes a voltage drop inside the battery when current flows, meaning that the voltage across an external circuit is lower than the battery's EMF.

Transcript

00:01 We have an automobile battery that has a voltage or an emf of 12 .6 volts.

00:11 And it has an internal resistance inside of the battery, so we'll call it rv.

00:19 .08 oms.

00:21 And this is connected to the headlights, and the headlights have a total resistance we'll call rh of 5 oms.

00:32 So on part a, we want to calculate the voltage across the headlights.

00:39 So to do that, we use omslaw, which is voltage is equal to current times resistance.

00:48 So we already know the resistance of the headlights.

00:55 So we need to find the current that's passing through them.

01:01 And to do that, we use omsla again, but for the ems.

01:05 Of the battery so we want to find the current that the battery is producing which will go through the headlights and once we know that current we could plug it into our equation and solve for the voltage across the headlights now part b now there's a starter motor and it's going to draw an additional current of 35 amps and because of that we want to calculate the new voltage across the headlights so we have to find the voltage that that the starter the starter motor has once you know that voltage we subtract it from the voltage across the headlights and we get a new voltage across the headlights due to this new starter motor so in part a we start off by om's law for the battery so we have the emf of the battery is equal to the current that it's going to give times the resistance equivalent resistance so the sum total of the resistance inside of the battery their internal resistance and the resistance of the headlights and we can just add those up which will have resistance of the battery plus the resistance of the headlights and we have for the battery 0 .08 headlights 5 so we get our equivalent of 5 .08 and with that we can calculate the current that the battery is going to provide.

02:47 So we have current is equal to, and then dividing both sides by our equivalent, you get the emf, the battery divided by our equivalent.

02:58 So we just plug in our values.

03:01 We've 12 .6 for the battery, and our equivalent we found is 5 .08.

03:07 So we get the current that the battery will provide is 2 .48 amps.

03:13 And this is the current that's going to flow through the headlights.

03:17 And we use the oms law for the headlights...

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