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A \(10.0-\mathrm{V}\) battery is connected to a \(120-\Omega\) resistor. Ignoring the internal resistance of the battery, calculate the power delivered to the resistor.

Short Answer

Expert verified
The power delivered to the resistor is 0.833 watts.

Step by step solution

01

Identify the Relevant Formula

To calculate the power delivered to the resistor, use the formula for electrical power, which is given by P = VI, where P is power, V is voltage, and I is current. Since we don't have the current, we can use Ohm's law, V = IR, to find the current, and then substitute it back into the power equation.
02

Calculate the Current

Ohm’s law states that I = V / R, where I is the current in amperes, V is the voltage in volts, and R is the resistance in ohms. Plug in the given values to find the current: I = (10.0 V) / (120 Ω) = 0.0833 A.
03

Calculate the Power

Now, use the power formula P = VI, substituting the voltage and the current calculated in the previous step: P = (10.0 V) × (0.0833 A) = 0.833 W.

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Key Concepts

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

Ohm's Law
Understanding Ohm's law is crucial when dealing with electrical circuits. This fundamental principle defines how voltage (\textbf{V}), current (\textbf{I}), and resistance (\textbf{R}) are related. The law is usually presented in a simple equation: \( I = \frac{V}{R} \). It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. To put it simply, if you increase the voltage, the current will also increase, provided that the resistance remains constant. Conversely, if you increase the resistance, the current will decrease for a given voltage.

For instance, in the case of a \(10.0-\mathrm{V}\) battery connected to a \(120-\Omega\) resistor, Ohm's law allows us to calculate the current by rearranging the formula to solve for \(I\) yielding \(I = \frac{10.0 \mathrm{V}}{120 \Omega} \approx 0.0833 \mathrm{A}\). It's this simplicity and ease of use that make Ohm's law a core concept in understanding how electrical circuits operate.
Electrical Resistance
Electrical resistance is a measure of how much an object opposes the flow of electric current through it. It's often compared to hydraulic resistance, which measures how much a pipe restricts the flow of water. Resistance is measured in ohms (\(\Omega\)) and depends on the material, geometry, and temperature of the conductor.

Materials can generally be classified into three categories:

In the textbook example where we have a \(120-\Omega\) resistor, this resistor limits the current that can flow from the \(10.0-\mathrm{V}\) battery, providing a safe and controlled flow of electrical energy. The value of the resistance plays a pivotal role in determining both the current and the power in a circuit.
Current and Voltage Relationship
The relationship between current (\(I\)) and voltage (\(V\)) is fundamental to understanding electrical circuits. Current, measured in amperes, is the flow of electric charge, while voltage, measured in volts, is the electric potential that causes current to flow.

Imagine voltage as the 'pressure' pushing the current through a conductor, similar to water pressure pushing water through a hose. Without enough voltage, current will not flow; without a path (conductor with low resistance), voltage cannot push the current through. This relationship is directly described by Ohm's law, which we've already explored.

Now, referring back to our example, after using Ohm's law to find the current, we can determine the power delivered to the resistor. Power, measured in watts (\(W\)), is the rate at which energy is transferred or converted. It is calculated using the formula \(P = VI\), where \(P\) is power, \(V\) is voltage, and \(I\) is current. By substituting the current we calculated earlier into the power formula, we find \(P = 10.0 \mathrm{V} \times 0.0833 \mathrm{A} = 0.833 \mathrm{W}\). This ultimately tells us how much energy per second is being dissipated by the resistor.

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Most popular questions from this chapter

The current in a resistor decreases by 3.00 A when the voltage applied across the resistor decreases from 12.0 V to 6.00 V. Find the resistance of the resistor.

One lightbulb is marked ‘25 W 120 V,’ and another ‘100 W 120 V’; this means that each bulb has its respective power delivered to it when plugged into a constant 120-V potential difference. (a) Find the resistance of each bulb. (b) How long does it take for 1.00 C to pass through the dim bulb? Is the charge different in any way upon its exit from the bulb versus its entry? (c) How long does it take for 1.00 J to pass through the dim bulb? By what mechanisms does this energy enter and exit the bulb? (d) Find how much it costs to run the dim bulb continuously for 30.0 days if the electric company sells its product at \(\$ 0.0700\) per kWh. What product does the electric company sell? What is its price for one SI unit of this quantity?

A close analogy exists between the flow of energy by heat because of a temperature difference (see Section 20.7\()\) and the flow of electric charge because of a potential difference. The energy \(d Q\) and the electric charge \(d q\) can both be transported by free electons in the conducting material. Consequently, a good electrical conductor is usually a good thermal conductor as well. Consider a thin conducting slab of thickness \(d x,\) area \(A,\) and electrical conductivity \(\sigma,\) with a potential difference \(d V\) between opposite faces. Show that the current \(I=d q / d t\) is given by the equation on the left below: \(\begin{array}{ll}{} & {\text { Thermal Conduction }} \\ {\text { Charge Conduction }} & {(\text { Eq. } 20.14)} \\ {\frac{d q}{d t}=\sigma A\left|\frac{d V}{d x}\right|} & {\frac{d Q}{d t}=k A\left|\frac{d T}{d x}\right|}\end{array}\) In the analogous thermal conduction equation on the right, the rate of energy flow \(d Q / d t\) (in SI units of joules per second) is due to a temperature gradient \(d T / d x,\) in a material of thermal conductivity \(k\) . State analogous rules relating the direction of the electric current to the change in potential, and relating the direction of energy flow to the change in temperature.

A lightbulb has a resistance of 240\(\Omega\) when operating with a potential difference of 120 \(\mathrm{V}\) across it. What is the current in the lightbulb?

A high-voltage transmission line with a diameter of 2.00 \(\mathrm{cm}\) and a length of 200 \(\mathrm{km}\) carries a steady current of 1000 \(\mathrm{A}\) . If the conductor is copper wire with a free charge density of \(8.49 \times 10^{28}\) electrons/m \(^{3}\) , how long does it take one electron to travel the full length of the line?

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