Mesh Analysis: Finding $V_o$ In A Circuit

Hey guys! Today, we're diving deep into circuit analysis and tackling a classic problem using the powerful technique of mesh analysis. Specifically, we're going to figure out how to determine the voltage $V_o$ across an inductor in a circuit containing a mix of components – a voltage source, resistors, capacitors, and of course, that inductor we're interested in. So, buckle up and let's get started!

Understanding the Circuit and the Goal

Before we jump into the math, let's paint a clear picture of the circuit we're dealing with. We have an AC voltage source providing a signal of $20 ext{cos}(4t - 15^°)$ volts. This is our power supply, the driving force behind all the action in the circuit. Connected to this source are several key components:

• A resistor with a resistance of $60 ext{Ω}$. Resistors, as you know, impede the flow of current, dissipating electrical energy as heat.
• A capacitor with a capacitance of $10 ext{mF}$ (milliFarads). Capacitors store electrical energy in an electric field, acting like tiny rechargeable batteries.
• An inductor with an inductance of $L$. This is where things get interesting because we're measuring the voltage $V_o$ across this inductor. Inductors store energy in a magnetic field, and their behavior is particularly important in AC circuits.

Our mission, should we choose to accept it (and we do!), is to determine the value of $V_o$, the voltage across the inductor. To do this, we'll be employing mesh analysis, a systematic method for solving circuit problems.

Why Mesh Analysis?

You might be wondering, why mesh analysis? Well, it's a fantastic tool for circuits with multiple loops, like the one we're working with. Mesh analysis allows us to write a set of equations based on Kirchhoff's Voltage Law (KVL), which states that the sum of the voltages around any closed loop in a circuit must equal zero. By solving these equations, we can find the currents flowing in each loop, and from there, we can calculate voltages and other quantities of interest – like our target, $V_o$.

Setting Up for Mesh Analysis

The first step in mesh analysis is to identify the loops in our circuit. A loop is simply a closed path for current to flow. Once we've identified the loops, we assign a mesh current to each loop. Think of these mesh currents as circulating currents within each loop. We'll use these currents as our primary variables in the equations we'll be writing.

Now, let's talk about those equations. For each loop, we'll apply KVL. This means we'll sum the voltage drops across each component in the loop, setting the total equal to zero. Remember, the voltage drop across a component depends on the current flowing through it and the component's impedance (resistance for a resistor, a frequency-dependent value for capacitors and inductors). This is where understanding the relationship between voltage, current, and impedance for each type of component becomes crucial. We will be going through these steps.

Diving into the Math: Applying Mesh Analysis

Alright, let's get our hands dirty with the calculations! This is where the magic happens, so pay close attention.

1. Defining Mesh Currents

The first thing we need to do is define our mesh currents. Let's assume we have two loops in our circuit (you'll need to visualize the specific circuit diagram here, but typically, it would involve two closed loops). We'll call the current in the first loop $I_1$ and the current in the second loop $I_2$. Remember, these are circulating currents, flowing around their respective loops.

2. Applying KVL to Loop 1

Now, we apply Kirchhoff's Voltage Law (KVL) to the first loop. This means we'll sum the voltage drops across each component in Loop 1 and set the result equal to zero. Let's break this down:

• Voltage Source: The voltage source provides a voltage rise of $20 ext{cos}(4t - 15^°)$ V. We'll treat this as a positive voltage in our KVL equation since it's a voltage source driving the current.
• Resistor: The voltage drop across the $60 ext{Ω}$ resistor is given by Ohm's Law: $V = IR$. In this case, the voltage drop is $60I_1$ (since $I_1$ is the current flowing through the resistor). Since this is a voltage drop, we'll treat it as a negative voltage in our KVL equation.
• Common Impedance (if any): If Loop 1 shares a component with Loop 2 (like an inductor or a resistor), the voltage drop across that component will depend on the difference between the two loop currents. For example, if there's a resistor shared between the loops, the voltage drop would be $R(I_1 - I_2)$. We need to carefully consider the direction of the currents when determining the sign of this term.

So, our KVL equation for Loop 1 might look something like this (the exact form depends on the specific circuit configuration):

$20 ext{cos}(4t - 15^°) - 60I_1 - ext{(voltage drop across shared impedance)} = 0$

3. Applying KVL to Loop 2

We repeat the same process for Loop 2. We sum the voltage drops across each component in Loop 2, taking into account the direction of the currents and any shared impedances. The KVL equation for Loop 2 will look similar to the one for Loop 1, but with different components and currents involved.

4. Dealing with Capacitors and Inductors: Impedance

Here's where things get a little more interesting. When we have capacitors and inductors in our AC circuit, we need to use the concept of impedance. Impedance is the AC equivalent of resistance; it opposes the flow of alternating current. However, unlike resistance, impedance depends on the frequency of the AC signal.

• Capacitive Impedance: The impedance of a capacitor is given by Z_C = rac{1}{j ext{ω}C}, where:

  • $j$ is the imaginary unit ($ ext{√-1}$)
  • $ ext{ω}$ is the angular frequency of the AC signal (in radians per second)
  • $C$ is the capacitance (in Farads)

  Notice that the capacitive impedance is inversely proportional to the frequency. This means that capacitors offer low impedance (easy current flow) at high frequencies and high impedance (difficult current flow) at low frequencies.

• Inductive Impedance: The impedance of an inductor is given by $Z_L = j ext{ω}L$, where:

  • $j$ is the imaginary unit ($ ext{√-1}$)
  • $ ext{ω}$ is the angular frequency of the AC signal (in radians per second)
  • $L$ is the inductance (in Henries)

  Inductive impedance is directly proportional to the frequency. This means inductors offer high impedance at high frequencies and low impedance at low frequencies.

So, when we calculate the voltage drop across a capacitor or inductor, we'll use the impedance instead of the resistance. For example, the voltage drop across an inductor with impedance $Z_L$ carrying current $I$ is $V = IZ_L$.

5. Solving the System of Equations

After applying KVL to each loop, we'll have a system of equations (usually two equations for a two-loop circuit, three equations for a three-loop circuit, and so on). These equations will be in terms of our mesh currents ($I_1$, $I_2$, etc.). The next step is to solve this system of equations to find the values of the mesh currents.

There are several methods we can use to solve these equations:

• Substitution: Solve one equation for one current and substitute that expression into the other equation(s).
• Elimination: Multiply the equations by constants so that the coefficients of one of the currents are equal (but opposite in sign), then add the equations together to eliminate that current.
• Matrix Methods: For larger circuits with many loops, matrix methods (like Cramer's Rule or matrix inversion) are often the most efficient way to solve the system of equations.

6. Finding $V_o$: The Voltage Across the Inductor

Once we've found the mesh currents, we're in the home stretch! Our goal is to find $V_o$, the voltage across the inductor. We can calculate this using Ohm's Law (in its generalized form, using impedance): $V_o = I imes Z_L$, where:

• $I$ is the current flowing through the inductor. This might be a single mesh current (if the inductor is only in one loop) or a combination of mesh currents (if the inductor is shared between loops).
• $Z_L$ is the inductive impedance, which we calculated earlier.

By plugging in the values we've found for the current and the impedance, we can finally determine $V_o$! Huzzah!

Example Scenario

Let's say after performing mesh analysis, we found that the current flowing through the inductor is $I = 0.2 ext{∠}30^°$ amps (this is a phasor representation, indicating both magnitude and phase). And let's say the inductive impedance at the given frequency is $Z_L = 50j ext{Ω}$. Then, the voltage across the inductor would be:

$V_o = I imes Z_L = (0.2 ext{∠}30^°) imes (50j) = 10 ext{∠}120^° ext{ volts}$

This tells us that the voltage across the inductor has a magnitude of 10 volts and a phase angle of 120 degrees relative to the reference (usually the voltage source).

Key Takeaways and Pro Tips

• Master KVL: Kirchhoff's Voltage Law is the foundation of mesh analysis. Make sure you understand it thoroughly.
• Impedance is Key: Remember to use impedance when dealing with capacitors and inductors in AC circuits.
• Sign Conventions Matter: Pay close attention to the sign conventions when applying KVL. A voltage drop is negative, a voltage rise is positive.
• Solve Systematically: Use a systematic approach to solve the system of equations. Substitution, elimination, or matrix methods can all be effective.
• Practice Makes Perfect: The best way to master mesh analysis is to practice, practice, practice! Work through plenty of example problems.

Conclusion

So there you have it, guys! We've walked through the process of determining $V_o$ using mesh analysis. It might seem like a lot of steps, but with a solid understanding of the fundamentals and a little practice, you'll be solving circuit problems like a pro in no time. Remember to break down the problem into smaller steps, stay organized, and don't be afraid to ask for help when you need it. Now go forth and conquer those circuits!