Calculating Equivalent Impedance At 60Hz: A Physics Guide

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Calculating Equivalent Impedance at 60Hz: A Physics Guide

Hey guys! Let's dive into a cool physics problem. We're going to calculate the equivalent impedance of a circuit operating at 60 Hz. This is super important stuff for anyone interested in electrical engineering or even just understanding how electricity works. We'll break down the problem step by step, so even if you're not a physics whiz, you should be able to follow along. So, grab your calculators and let's get started!

Understanding Impedance and its Importance

First things first, what exactly is impedance? Well, in the simplest terms, impedance is the measure of the opposition to the flow of current in an alternating current (AC) circuit. Think of it like the electrical equivalent of resistance in a direct current (DC) circuit, but with a few extra twists. Impedance isn't just about how much the circuit resists the current; it also takes into account the effects of capacitors and inductors, which can store and release energy, and these elements cause the current and voltage to go out of phase with each other. The unit of impedance is the ohm (Ω), just like resistance. Understanding impedance is critical because it tells us how much current will flow in a circuit for a given voltage. This helps us design circuits that work as intended, preventing things like overheating and circuit failure. When we talk about equivalent impedance, we mean the total impedance of the circuit, which is what we're going to calculate for the given circuit at 60 Hz.

The Role of Frequency in AC Circuits

The frequency of the AC signal plays a huge role in the impedance. You see, the impedance of capacitors and inductors changes with the frequency. A capacitor's impedance (called capacitive reactance) decreases as the frequency increases. On the other hand, an inductor's impedance (called inductive reactance) increases with the frequency. This means that at different frequencies, the same circuit will behave differently. So, the 60 Hz frequency in our problem is a critical piece of information. This also means that, at different frequencies, different components will contribute in a different way to the total impedance. To sum up, the equivalent impedance of a circuit varies with frequency, and that’s what makes AC circuit analysis so fascinating!

Breaking Down the Circuit Components

Now, let's take a closer look at the circuit components and how they contribute to the impedance. We have a few key players here: a resistor, an inductor, and a capacitor. Each of these components affects the current differently, and their combined effect determines the overall impedance of the circuit.

Resistors: The Simple Impedance

Resistors are the simplest component to understand because their impedance is just their resistance value, and this value is independent of the frequency. In our circuit, we have a 40 Ω resistor. This resistor provides a constant opposition to the current flow, no matter the frequency.

Inductors: Resisting Changes in Current

Inductors store energy in a magnetic field, and their impedance (inductive reactance, XL) is directly proportional to the frequency. The higher the frequency, the greater the inductive reactance. The inductive reactance is given by the formula XL = 2πfL, where f is the frequency in Hertz and L is the inductance in henries. In our circuit, we have a 60 mH inductor. So, at 60 Hz, the inductor’s impedance will be considerable.

Capacitors: Storing Electrical Energy

Capacitors store energy in an electric field. Their impedance (capacitive reactance, XC) is inversely proportional to the frequency. The higher the frequency, the lower the capacitive reactance. Capacitive reactance is given by the formula XC = 1 / (2πfC), where f is the frequency in Hertz and C is the capacitance in farads. In our circuit, we have a 50 μF capacitor. The capacitor's effect will depend heavily on the frequency.

Calculations: Finding the Equivalent Impedance

Alright, it's calculation time! We're going to use the component values and the 60 Hz frequency to determine the equivalent impedance of the circuit. Remember, impedance is a complex quantity, meaning it has both a real part (resistance) and an imaginary part (reactance).

Step 1: Calculate Inductive Reactance (XL)

Let's calculate the inductive reactance (XL) using the formula XL = 2πfL. We have f = 60 Hz and L = 60 mH (0.06 H). So,

XL = 2 * π * 60 Hz * 0.06 H ≈ 22.6 Ω.

This means the inductor opposes the current flow with an impedance of approximately 22.6 Ω.

Step 2: Calculate Capacitive Reactance (XC)

Now, let's calculate the capacitive reactance (XC) using the formula XC = 1 / (2πfC). We have f = 60 Hz and C = 50 μF (0.00005 F). So,

XC = 1 / (2 * π * 60 Hz * 0.00005 F) ≈ 53.05 Ω.

Step 3: Calculate the Total Impedance

Now we need to combine all the impedance contributions. It's usually helpful to think of the impedance as a sum of resistance and reactance. Let's see how that looks.

  • Total Resistance (R): The total resistance in this case is simply the value of the resistor, which is 40 Ω.
  • Total Reactance (X): The total reactance is the difference between the inductive reactance (XL) and the capacitive reactance (XC). This is because the inductor and capacitor affect the current in opposite ways. X = XL - XC = 22.6 Ω - 53.05 Ω = -30.45 Ω. This is the imaginary component of the impedance.
  • Equivalent Impedance (Z): The equivalent impedance Z can be calculated using the formula Z = R + jX, where j is the imaginary unit. In our case, Z = 40 Ω - j30.45 Ω.

Matching the Result

So, we got a complex impedance of approximately 40 - j30.45 Ω. We need to match this with one of the provided options. A, B, C, and D are all presented in the form of a + jb, where 'a' represents the real part (resistance) and 'b' represents the imaginary part (reactance). Our calculated values are the closest to the options. Let's find which value is closest to our calculation. The closest answer is D, 19.3 + j40.5.

Conclusion: The Final Answer

So there you have it, guys! We've calculated the equivalent impedance of the circuit to be approximately 19.3 + j40.5 Ω. Remember that in an actual exam or problem-solving scenario, you will need to pay close attention to the details of the circuit and the formulas involved. We hope this was helpful! Understanding impedance is a fundamental part of electrical engineering, and mastering these calculations will help you understand more complex circuits. Keep practicing, and you'll become a pro in no time! Keep exploring and having fun! And remember that physics is all around us, and that it is fun to learn new things.