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RL Series Circuit Analysis
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Today, we're going to solve an example involving an RL series circuit. We have a resistor of 5 Ξ© and an inductor of 2 H with a step input voltage of 10u(t). Can anyone tell me the first step in using the Laplace Transform for this circuit?
We need to transform the circuit into the s-domain.
That's right! We start by finding the Laplace transform of the voltage. So, how do we express the voltage in the s-domain?
The voltage V(s) would be 10/s.
Excellent! Now, can anyone recall what the total impedance Z(s) will look like for this RL circuit?
It will be R + sL, which is 5 + 2s.
Great job! Now using Ohm's law, how can we find the current I(s) in the s-domain?
I(s) = V(s) / Z(s). So I(s) becomes 10/s divided by (5 + 2s).
Exactly! Can you simplify that?
I(s) = 10 / [s(5 + 2s)].
Good! Next up, letβs use partial fractions to find A and B. What will I(s) look like when we break it down?
It would be A/s + B/(5 + 2s).
Well done! After solving, we found that A = 2 and B = -2. Now, how would we find the inverse Laplace Transform to express the current i(t)?
We'd take the inverse Laplace of the simplified I(s).
Exactly! So, whatβs the final result for i(t)?
i(t) = 2 - 2e^(-2.5t) A.
Great summary! This process illustrates the application of the Laplace Transform step by step.
RC Series Circuit Analysis
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Now, letβs move on to our second example involving an RC series circuit where R = 10 Ξ© and C = 0.1 F, with a step input voltage of 20u(t). Can anyone outline the first step?
We start by calculating the Laplace Transform of the voltage.
Correct! So, what would V(s) look like?
V(s) = 20/s.
Exactly! What about the impedance of the capacitor?
Z(s) = 1/(sC), which becomes 10/s.
Perfect! Now to find the total impedance of the circuit, what do we need to do?
Add the resistor impedance and capacitor impedance together.
Correct! It becomes 10 + 10/s. Now, how do we find I(s)?
I(s) = V(s)/Z(s). So, it simplifies to 20/s(10 + 10/s).
Good job! Now can someone explain how we find V_C(s) using I(s)?
V_C(s) equals I(s) multiplied by the impedance of the capacitor.
Thatβs right! Now letβs apply the inverse Laplace Transform. Whatβs V_C(t)?
It will become 200(1 - e^(-t)) V.
Excellent recap! These examples help solidify your understanding of Laplace Transforms in circuit analysis.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
Quick Overview
Standard
The section provides two key example problemsβone focusing on an RL series circuit and another on an RC series circuit. Each example illustrates the process of applying Laplace Transforms to find current and voltage across circuit elements, emphasizing the steps of transforming equations and applying the inverse Laplace Transform.
Detailed
In this section, we delve into practical applications of the Laplace Transform in electrical circuit analysis, focusing on two example problems to illustrate the process in detail. The first example explores a series RL circuit with a step input voltage, detailing the steps to find the time-domain expression for the current. After determining the Laplace transform of the voltage, we apply the principles of partial fractions to find the current in the s-domain and subsequently use the inverse Laplace Transform to return to the time domain.
The second example examines a series RC circuit, where we similarly analyze the voltage across a capacitor with a step input. By calculating the impedance and current using the Laplace Transform, we highlight how the inverse transform is applied to find the voltage across the capacitor. These examples reinforce the utility of Laplace Transforms in simplifying circuit analysis involving differential equations and emphasize the significance of finding both transient and steady-state responses.
Audio Book
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Example 1: RL Series Circuit
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Given: A series RL circuit with π = 5 Ξ©, πΏ = 2 π», and step input π(π‘) = 10π’(π‘). Find the current π(π‘).
Step 1: Laplace Transform of the circuit
π(π ) = 10/s, π(π ) = π + π πΏ = 5 + 2π
πΌ(π ) = π(π ) / π(π ) = (10/π ) / (5 + 2π ) = 10/(π (5 + 2π )).
Step 2: Partial Fractions
10/(π (5 + 2π )) = A/π + B/(5 + 2π ).
Solving gives π΄ = 2, π΅ = β2. Therefore, πΌ(π ) = 2/π - 2/(5 + 2π ).
Step 3: Inverse Laplace
π(π‘) = 2 β 2π^(β2.5π‘) (A)
Detailed Explanation
In this example, we analyze a series RL circuit with a resistor (R) and inductor (L) using the Laplace Transform.
- Step 1: Laplace Transform of the circuit involves converting the time-domain functions into the s-domain. We calculate the total voltage in the s-domain as π(π ) = 10/s, and find the impedance π(π ) = 5 + 2π . The current πΌ(π ) is then derived by dividing the voltage by the impedance: πΌ(π ) = (10/π )/(5 + 2π ).
- Step 2: Partial Fractions is a technique used to simplify the expression of πΌ(π ) further. We express it as a sum of fractions, allowing us to easily apply the inverse Laplace Transform later. Here, we solve for constants A and B and rewrite πΌ(π ) accordingly.
- Step 3: Inverse Laplace transforms our current back to the time-domain, indicating how the current behaves over time after the step input is applied. The resulting time-domain function is π(π‘) = 2 - 2π^(β2.5π‘) (A), showing the current starts at 2 Amps and decays over time.
Examples & Analogies
Think of this RL circuit as a water pipeline where we control the water flow (current) using a valve (resistor) and a tank (inductor). When the valve opens suddenly (like the step input), the water (current) quickly fills the tank but then starts to settle at a steady amount. The Laplace Transform helps us calculate how quickly the water fills and settles down, giving us a clear view of the system's behavior over time.
Example 2: RC Series Circuit
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Given: π = 10 Ξ©, πΆ = 0.1 πΉ, input π£(π‘) = 20π’(π‘). Find capacitor voltage π£ (π‘).
Step 1: π(π ) = 20/s, impedance of capacitor π = 1/(sC) = 1/(0.1s) = 10/s.
Total impedance: π = π + π_C = 10 + 10/s (total).
Current: πΌ(π ) = π(π )/π = (20/s) / (10 + 10/s) = 20/(s(s + 1)).
Voltage across capacitor:
π_C(s) = πΌ(s) β 1/(sC) = πΌ(s)*10/s = 200/s(s + 1).
Inverse Laplace:
π£_C(t) = 200(1 β π^(βπ‘)) (V)
Detailed Explanation
In this example, we work with a series RC circuit, which has a resistor (R) and capacitor (C). We will find how the voltage across the capacitor changes over time when a step input is applied.
- Step 1: Laplace Transform of the circuit involves converting the circuit's time-domain voltage into the s-domain. We express the input voltage as π(π ) = 20/s and the capacitor impedance as π_C = 10/s. The total impedance then combines the resistor and capacitor together, giving us a total impedance of π = 10 + 10/s.
- Step 2: Current calculation is done by applying Ohmβs law, allowing us to find πΌ(π ) = π(π )/π. This involves solving for πΌ(π ) which will be helpful in determining the voltage across the capacitor.
- Step 3: Voltage across Capacitor (π_C) is calculated by multiplying the current πΌ(π ) by the capacitor's impedance. Once we have π_C(s), we can easily transform it back to the time-domain using the inverse Laplace transform.
- The final time-domain function is π£_C(t) = 200(1 β π^(βπ‘)), indicating how the voltage across the capacitor builds up over time, reaching a maximum value as time progresses.
Examples & Analogies
Imagine the RC circuit as a sponge (capacitor) in a bucket (circuit) being filled with water (voltage). When the tap (input voltage) is turned on, water starts to pour into the bucket. The water fills the sponge up gradually instead of instantly. The Laplace Transform allows us to predict how quickly the sponge absorbs the water over time, giving us insights into the electrical behavior of our circuit similar to the filling process of the sponge.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
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Circuit Analysis: The process of determining the voltages and currents in a circuit.
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Laplace Transform: A method for converting differential equations into algebraic equations.
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Initial and Final Value Theorems: Theorems that provide insights into circuit behavior at initial and final time states.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
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Example 1 demonstrates finding the current in an RL series circuit by applying Laplace Transforms.
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Example 2 illustrates how to find the voltage across a capacitor in an RC series circuit using the Laplace Transform.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
π΅ Rhymes Time
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To find I, remember the flow, V over Z is the way to go!
π Fascinating Stories
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Imagine a circuit as a river, with currents flowing easily through pathsβvoltages create pressure, and impedance is like rocks that make the flow slow. Use the Laplace to smooth the waters, making analysis easy!
π§ Other Memory Gems
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Remember the order: Transform, Solve, Inverse! TSIV - Transform to solve, then Inverse!
π― Super Acronyms
RLC
- Resistor
- Inductor
- Capacitor; Remember they show circuit's behavior!
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
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Term: Laplace Transform
Definition:
A mathematical operation that transforms a time-domain function into a complex frequency-domain function.
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Term: sdomain
Definition:
The domain in which Laplace Transforms operate, represented by the complex frequency variable s.
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Term: Initial Value Theorem
Definition:
A theorem stating that the initial value of a time-domain function can be found by evaluating the limit of s times the Laplace transformed function as s approaches infinity.
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Term: Final Value Theorem
Definition:
A theorem that helps determine the steady-state value of a function as time approaches infinity by evaluating the limit of the Laplace transformed function as s approaches zero.
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Term: Impedance
Definition:
A measure of the opposition that a circuit presents to a current when a voltage is applied, combining resistance and reactance.