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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?

Short Answer

Expert verified
The potential difference across the headlight bulbs is 12.6 V when the headlights are the only load on the battery, and it decreases to 9.6 V when the starter motor is also drawing current from the battery.

Step by step solution

01

Determine the Battery's output current in the first scenario

In the first scenario, only the headlights are drawing current from the battery. Using Ohm’s law (\(I = V / R\)), the current \(I\) delivered by the battery is \(I = 12.6 / (0.080 + 5.00) = 2.51 A\)
02

Calculate the Potential Difference across the headlights in the first scenario

The potential difference \(V\) across the headlights is given by \(V = I \cdot R\), where \(I\) is the current through the headlights and \(R\) is the total resistance of the headlights. Substituting the calculated current and given resistance, we have \(V = 2.51 \cdot 5.00 = 12.6 V\) which is the emf of the battery.
03

Determine the Battery's output current in the second scenario

In the second scenario, the starter motor is drawing an additional 35.0 A from the battery. So, the total current that the battery needs to deliver is \(I = 2.51 + 35.0 = 37.51 A\)
04

Calculate the Potential Difference across the headlights in the second scenario

The headlights are still in series with the internal resistance, so the potential difference across the headlights (which is equal to the battery's output voltage) is given by \( V = E - I \cdot r \), where \( E \) is the emf of the battery, \( I \) is the total current and \( r \) is the internal resistance. Substituting the calculated current and given values, we have \( V = 12.6 - (37.51 \cdot 0.080) = 9.6V \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
In electric circuits, Ohm's Law is an easy yet powerful tool used to determine how voltage, current, and resistance relate to each other. The law is formulated as: \( V = I \cdot R \), where:
  • \( V \) is the voltage in volts (V),
  • \( I \) is the current in amperes (A), and
  • \( R \) is the resistance in ohms (Ω).
This means the voltage across a conductor is directly proportional to the current flowing through it and the resistance of the conductor.
For example, in our exercise involving a car battery and headlights, Ohm’s Law was used to find the current when only the headlights were powered. The formula used was \( I = \frac{V}{R} \), where the total resistance included both the headlights and the internal resistance of the battery. This provided the current flowing through the circuit, leading to further calculations.
emf (electromotive force)
Electromotive force, often abbreviated as emf, is like the energy supply of a battery. It’s measured in volts and can be thought of as the "pressure" that pushes the electrical current around the circuit. Though termed as a force, it is actually not a force but a potential difference. It represents the total voltage the battery can supply when no current flows.
In our car battery example, the emf is given as \( 12.6 \mathrm{~V} \). This is the theoretical voltage available when nothing is connected to the battery yet. Once a load, like the headlights or the starter motor, is connected, the real voltage across the terminals might be less because of internal resistance causing voltage drop.
internal resistance
When you deal with real-world batteries, you can't ignore internal resistance, which is the resistance within the battery that opposes the flow of current. It plays a crucial role in determining how much voltage is available at the terminals of a battery when a current is flowing.
For instance, in the provided problem, the internal resistance of the car battery is \(0.080 \Omega\). This resistance causes some of the emf to be "used up" just to get the current through the battery itself before it even reaches the external circuit, such as the headlights. The effect becomes more noticeable when high currents are drawn, like when starting the car, leading to a lower potential difference across the components in the circuit.

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Most popular questions from this chapter

A lamp \((R=150 \Omega)\), an electric heater \((R=25 \Omega)\), and â fan \((R=50 \Omega)\) are connected in parallel across a \(120-V\) line. (a) What total current is supplied to the circuit? (b) What is the voltage across the fan? (c) What is the current in the lamp? (d) What power is expended in the heater?

A series combination of a \(12-\mathrm{k} \Omega\) resistor and an unknown capacitor is connected to a \(12-V\) battery, One second after the circuit is completed, the voltage across the capacitor is \(10 \mathrm{~V}\). Determine the capacitance of the capacitor.

A charged capacitor is connected to a resistor and a switch as in Active Figure \(18.17\), The circuit has a time constant of \(1.5 \mathrm{~s}\). After the switch is closed, the charge on the capacitor is \(75 \%\) of its initial charge. (a) Find the time it takes to reach this charge. (b) If \(R=250 \mathrm{k} \Omega\), what is the value of \(C ?\)

Consider the series RC-circuit shown in Active Figure \(18.16\) for which \(R=75.0 \mathrm{k} \Omega, C=25.0 \mu \mathrm{F}\), and \(\varepsilon=\) \(12.0 \mathrm{~V}\). Find (a) the time constant of the circuit and (b) the charge on the capacitor one time constant after the switch is closed.

An uncharged capacitor and a resistor are connected in series to a source of emf. If \(\varepsilon=9.00 \mathrm{~V}, C=20.0 \mu \mathrm{F}\), and \(R=100 \Omega\), find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, and (c) the charge on the capacitor after one time constant.

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