Calculating Electrical Power In A Circuit Resistor A Step-by-Step Guide

Hey everyone! Let's dive into a common physics problem involving electrical circuits. We're going to break down how to calculate the electrical power dissipated by a resistor in a circuit. This is a fundamental concept in electronics, and once you grasp it, you'll be able to tackle a wide range of circuit analysis problems. Let's jump right in!

Understanding the Problem

In this scenario, electrical resistance is a key factor. We have a resistor with a resistance of 25 ohms (Ω). This resistance is like a bottleneck in the circuit, impeding the flow of electrical current. Think of it like a narrow pipe restricting water flow – the higher the resistance, the harder it is for the current to pass through. The problem also states that there's an electrical current of 2 amperes (A) flowing through this resistor. Current is the rate of flow of electric charge, so 2 amperes means that a certain amount of charge is passing through the resistor every second. Our mission is to determine the electrical power dissipated by the resistor. Electrical power, measured in watts (W), represents the rate at which electrical energy is converted into another form of energy, in this case, heat. When current flows through a resistor, some of the electrical energy is transformed into heat due to the collisions of electrons with the atoms in the resistor material. This is what we mean by power dissipation. Now that we've clearly defined the problem and the key concepts involved, we can move on to the solution. It's essential to understand the problem statement thoroughly before attempting to solve it. This ensures that we're applying the correct principles and formulas. In summary, we have a resistor resisting the flow of current, and we need to figure out how much power it's using up as heat.

The Power Formula: P = RI²

To figure out the electrical power in this circuit, we're going to use a super important formula in electrical engineering: P = RI². Let's break it down so it makes total sense. The "P" stands for power, which is what we're trying to find – it's the amount of electrical energy the resistor is using up per unit of time, measured in watts (W). Then there's "R", which represents resistance. Remember, this is the measure of how much the resistor opposes the flow of electrical current, and it's measured in ohms (Ω). In our problem, the resistance is given as 25 Ω. Next up is "I", and this stands for current, which is the flow of electrical charge through the circuit. It's measured in amperes (A), and in our problem, the current is 2 A. The little "²" symbol means we need to square the current value, which means multiplying it by itself. So, putting it all together, the formula P = RI² tells us that the power dissipated by a resistor is equal to the resistance multiplied by the square of the current flowing through it. This formula is derived from Ohm's Law (V = IR) and the general power equation (P = VI), where V is voltage. By substituting V = IR into P = VI, we get P = (IR)I = RI². This is a cornerstone equation in circuit analysis, allowing us to calculate power dissipation directly from resistance and current values. It's not just some random equation; it's a direct consequence of the fundamental laws governing electricity. Now that we've thoroughly dissected the formula, we're ready to plug in the values from our problem and solve for the power.

Applying the Formula: Step-by-Step Calculation

Alright, let's put that formula to work and calculate the electrical power! Remember, we have a resistor with a resistance (R) of 25 ohms (Ω) and a current (I) of 2 amperes (A) flowing through it. Our formula is P = RI², so we're going to plug in these values step by step. First, let's substitute the values into the formula. We have P = 25 Ω * (2 A)². See how we've replaced the R with 25 Ω and the I with 2 A? Now, the next step is to take care of that exponent. We need to square the current, which means multiplying it by itself: (2 A)² = 2 A * 2 A = 4 A². So our equation now looks like this: P = 25 Ω * 4 A². We're almost there! Now we just need to multiply the resistance by the squared current. So, we multiply 25 Ω by 4 A², which gives us P = 100 ΩA². And remember, one ohm-ampere squared (ΩA²) is the same as one watt (W), which is the unit of power. So, our final answer is P = 100 W. That means the resistor is dissipating 100 watts of power as heat. This step-by-step approach makes it clear how we arrived at the solution. We started with the formula, plugged in the given values, performed the necessary calculations, and arrived at the final answer in the correct units. It's always a good idea to double-check your calculations and units to ensure accuracy.

Solution and Answer

So, after our calculation, we've found that the electrical power dissipated by the resistor is 100 watts (W). This means that the resistor is converting 100 joules of electrical energy into heat energy every second. Now, let's go back to the multiple-choice options provided in the original problem. We had the following options:

• Opção A 18W
• Opção B 50W
• Opção C 10W
• Opção D 5W
• Opção E 100W

Comparing our calculated answer of 100W with the options, it's clear that the correct answer is Opção E 100W. This confirms that our step-by-step calculation using the power formula P = RI² was accurate. We successfully identified the correct answer among the given choices. This is a great example of how a clear understanding of the principles and a systematic approach to problem-solving can lead to the right solution. It's important not only to arrive at the correct numerical answer but also to understand the meaning of the result in the context of the problem. In this case, 100W represents the rate at which electrical energy is being converted into heat by the resistor. With the correct answer identified, we've effectively solved the problem and can move on to further explorations of electrical circuits.

Key Takeaways and Real-World Applications

Let's wrap things up and highlight the key takeaways from this problem. First, we learned how to calculate electrical power dissipated by a resistor using the formula P = RI². This formula is a cornerstone in circuit analysis and is widely used in electrical engineering. We understood that resistance impedes the flow of current, and the higher the resistance, the more power is dissipated as heat for a given current. We also saw how current, the flow of electrical charge, plays a crucial role in power dissipation. The higher the current, the more power is dissipated. Remember, power is the rate at which electrical energy is converted into other forms of energy, such as heat. Now, you might be wondering, where does this knowledge come in handy in the real world? Well, the principles we've discussed are fundamental to understanding how electrical devices work. For example, the heating element in your toaster uses a resistor to convert electrical energy into heat, which toasts your bread. The filaments in incandescent light bulbs also use resistance to produce light and heat. Understanding power dissipation is also crucial in designing electronic circuits. Engineers need to ensure that components don't overheat and fail due to excessive power dissipation. This involves selecting appropriate resistors and other components that can handle the expected power levels. From designing efficient power supplies to understanding the energy consumption of appliances, the concepts we've covered are essential in many areas of electrical engineering and electronics. So, mastering these principles will definitely come in handy if you're interested in pursuing a career in these fields.

Practice Problems

To really solidify your understanding of electrical power calculations, let's try a couple of practice problems. Working through these will help you internalize the concepts and become more confident in applying the formula P = RI². Here's the first one: Imagine a circuit with a resistor of 50 ohms (Ω) and a current of 3 amperes (A) flowing through it. What is the electrical power dissipated by the resistor? Take a moment to work through the problem using the steps we discussed earlier. Remember to use the formula P = RI², plug in the values, and calculate the result. Pay attention to the units as well. Now, let's try another one: Suppose you have a 100-ohm (Ω) resistor in a circuit, and it's dissipating 25 watts of electrical power. What is the current flowing through the resistor? This problem is a bit different because we're given the power and resistance and need to solve for the current. You'll need to rearrange the formula P = RI² to solve for I. Remember that I = √(P/R). Plug in the values and calculate the current. These practice problems are designed to test your understanding of the concepts and your ability to apply the formula in different scenarios. Don't worry if you don't get them right away. The key is to practice and learn from your mistakes. The more problems you solve, the more comfortable you'll become with electrical power calculations. And remember, if you get stuck, go back and review the explanation and example we worked through earlier. Keep practicing, and you'll become a pro at these calculations in no time!

Conclusion

Alright, guys, we've covered a lot in this article! We tackled a physics problem involving electrical power in a resistor, and we broke it down step by step so it's super clear. We started by understanding the problem statement, defining key concepts like resistance, current, and power. Then, we dove into the power formula P = RI², which is the key to solving these types of problems. We learned what each variable represents and how they relate to each other. Next, we walked through a detailed calculation, plugging in the values from the problem and arriving at the solution. We also made sure to identify the correct answer from the multiple-choice options. But we didn't stop there! We discussed the real-world applications of these concepts, highlighting how they're used in various electrical devices and engineering scenarios. And to make sure you've truly grasped the concepts, we included a couple of practice problems for you to try. The goal was to not just give you the answer but to help you understand the underlying principles and how to apply them. By understanding the relationship between resistance, current, and power, you can analyze and design electrical circuits more effectively. So, keep practicing, keep exploring, and keep learning! Electrical circuits are all around us, and understanding how they work is a valuable skill. You've got this!