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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.

Short Answer

Expert verified
The resistance of the resistor is 2.00 \( \Omega \).

Step by step solution

01

Understand Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, with the constant of proportionality being the resistance. It is given by the equation: \( V = IR \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance.
02

Calculate the Change in Voltage

Find the change in voltage (\( \Delta V \)) by subtracting the final voltage from the initial voltage: \( \Delta V = V_{initial} - V_{final} = 12.0\,V - 6.00\,V = 6.00\,V \).
03

Relate the Change in Current to Resistance

Using Ohm's Law, the change in voltage is also equal to the change in current times resistance (\( \Delta V = \Delta I \times R \)). We're given that the change in current (\( \Delta I \)) is 3.00 A, so we can write the equation as: \( 6.00\,V = 3.00\,A \times R \).
04

Calculate the Resistance

Solve for resistance (\( R \)) by dividing the change in voltage by the change in current: \( R = \frac{\Delta V}{\Delta I} = \frac{6.00\,V}{3.00\,A} = 2.00 \Omega \).

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Key Concepts

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

Understanding Electrical Resistance
When we talk about electrical resistance, we're referring to the measure of how much an object opposes the flow of electric current through it. You can liken it to a water pipe: just as a narrow pipe restricts water flow, a material with high resistance restricts electric current.

Resistance is measured in ohms, symbolized by the Greek letter Omega (\( \text{\textOmega} \)). The higher the resistance, the more the material inhibits the current. Various factors affect resistance, including the material's intrinsic properties, its temperature, length, and cross-sectional area.

For example, metals like copper or aluminum are considered good conductors because they have low resistance, permitting current to flow easily. On the other hand, materials like rubber or glass offer high resistance and are used as insulators to prevent unwanted current flow.

Understanding resistance is crucial not only for solving physics problems but also for practical applications in electric circuits and electronics design.
Exploring the Voltage-Current Relationship
The voltage-current relationship is a cornerstone of electrical engineering and physics, described by the famous Ohm's Law. This principle illustrates the direct relationship between voltage, current, and resistance in an electrical circuit.

According to Ohm's Law, the current (\( I \) flowing through most conductors is proportional to the voltage (\( V \) across it. This implies that if you increase the voltage, the current will increase, provided the resistance remains constant. The relationship can be written as the equation: \( V = IR \).

An increase in voltage means more 'electrical pressure' pushing the electrons through a circuit, thus leading to a higher current. Conversely, a decrease in voltage results in a reduced flow of current. This interdependence is fundamental for designing circuits and understanding how different components will affect overall performance.

For instance, if a bulb's brightness depends on the current, adjusting the voltage will result in changing how bright the light is. This simple yet powerful understanding allows for sophisticated control over complex electrical systems.
Approaching Physics Problem Solving
Solving problems in physics, especially those involving circuits and electrical components, requires a systematic approach. Learning to navigate through these challenges improves critical thinking and analytical skills, which are beneficial beyond just physics class.

One effective strategy is to start by identifying the known variables in a problem, like in the given exercise where the initial and final voltages, as well as the change in current, were provided. The next step is to apply relevant principles or formulas, such as Ohm's Law in this case. After setting up the correct equation, it's simply a matter of algebraic manipulation to solve for the unknown quantity—in this instance, resistance.

Always ensure to check that the units are consistent and that the answer makes sense in the context of the problem. A calculated resistance of 2.00 ohms, for example, implies a low level of opposition to current flow, which could be expected in a simple circuit using a standard resistor.

Physics problem-solving is a skill that develops with practice and patience. Each problem, while unique, is an opportunity to apply and reinforce fundamental concepts, enhancing your comprehension and proficiency in the subject.

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Most popular questions from this chapter

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.

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?

Suppose that a voltage surge produces 140 V for a moment. By what percentage does the power output of a 120-V, 100-W lightbulb increase? Assume that its resistance does not change.

A toaster is rated at 600 W when connected to a 120-V source. What current does the toaster carry, and what is its resistance?

An electric utility company supplies a customer's house from the main power lines \((120 \mathrm{V})\) with two copper wires, each of which is 50.0 \(\mathrm{m}\) long and has a resistance of 0.108\(\Omega\) per 300 \(\mathrm{m}\) (a) Find the voltage at the customer's house for a load current of 110 \(\mathrm{A}\) . For this load current, find \((\mathrm{b})\) the power the customer is receiving and \((\mathrm{c})\) the electric power lost in the copper wires.

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