Understanding Average Power in Solenoid Coils

When you're preparing for the Electronics Engineering (ELEX) Board Examination, every detail counts—especially when it comes to understanding power consumption in various electrical components. Imagine this scenario: you connect a solenoid coil with a resistance of 30 ohms and inductance of 200 mH to a 230 VAC, 50 Hz supply. The question naturally arises, how do you calculate the average power consumed? You know, this isn’t just an academic exercise; it’s about getting a clear grasp of how electrical systems work.

Let’s break this down step-by-step. We first need to determine the total impedance of the solenoid coil. This is important because, in AC circuits, both the resistance and inductance play essential roles. The impedance ( Z ) can be computed using this formula:

[ Z = \sqrt{R^2 + (X_L)^2} ]

where

• ( R ) is the resistance, and
• ( X_L ) is the inductive reactance.

Now, let’s talk about inductive reactance. This can be calculated using:

[ X_L = 2\pi f L ]

Here, frequency ( f ) is in hertz (Hz), and inductance ( L ) is in henries (H). So, plug in our given values:

• ( R = 30 , \text{ohms} )
• ( L = 200 , \text{mH} = 0.2 , \text{H} )
• ( f = 50 , \text{Hz} )

First off, calculating ( X_L ) is your first step. So, let’s plug those values into the equation:

[ X_L = 2\pi (50)(0.2) \approx 62.83 , \text{ohms} ]

Next, it’s time to find the total impedance ( Z ):

[ Z = \sqrt{30^2 + 62.83^2} \approx \sqrt{900 + 3945.38} \approx \sqrt{4845.38} \approx 69.7 , \text{ohms} ]

The next piece of the puzzle is finding the current flowing through the circuit using Ohm's law, which states that current ( I ) is equal to voltage ( V ) divided by impedance ( Z ):

[ I = \frac{V}{Z} = \frac{230}{69.7} \approx 3.30 , \text{A} ]

With the current determined, you’re closer to finding the average power ( P ). That’s where the magic happens! Using the formula for average power in an AC circuit with non-resistive elements:

[ P = I^2 \times R ]

You substitute:

[ P = (3.30)^2 \times 30 \approx 10.89 \times 30 = 326.7 , \text{W} ]

Ta-da! Rounding this off, we discover, without a doubt, that the average power consumed is approximately 327 W. So, if you ventured a guess with option C, you aced that question!

Wasn’t that enlightening? Understanding these concepts isn’t just a test of your knowledge but also of your critical thinking skills. As you continue with your ELEX preparation, grasping fundamental concepts like this will bolster not only your exam performance but also your confidence in practical scenarios. And remember, when dealing with electrical engineering, a solid understanding of how components interact in a circuit lays the groundwork for excellence!

Keep these calculations handy and practice similar problems. After all, practice is what brings perfection. The more you familiarize yourself with formulas and calculations, the easier they’ll become. Happy studying, future engineers!