In the circuit shown in Figure 4, a generator supplies a voltage of E(t)=40 sin60 t volts, the i

step 1 For this problem, we have a circuit. The details of the circuit don't really matter themselves,

In the circuit shown in Figure 4, a generator supplies a...

Key Concepts

Time Constant in RL Circuits

The time constant in an RL circuit is given by the ratio L/R, and it characterizes the rate at which the transient response decays. A smaller time constant means the circuit reaches its steady state more quickly, whereas a larger time constant indicates a slower decay of the transient effects.

Graphical Interpretation of Dynamics

Graphing the current as a function of time provides visual insight into both the transient and steady-state behaviors of the circuit. The graph typically shows the initial deviation due to the transient response and the eventual periodic behavior corresponding to the steady-state response induced by the sinusoidal voltage source.

Transient and Steady-State Response

The solution to the differential equation in an RL circuit includes a transient component, which decays over time, and a steady-state component, which persists as long as the forcing function (such as a sinusoidal input) is applied. The transient response is influenced by the circuit's time constant, while the steady-state response reflects the behavior of the circuit once the transient effects have diminished.

Integrating Factor Method

The integrating factor method is a standard technique for solving first-order linear differential equations. It involves multiplying the entire differential equation by an integrating factor to rewrite the left-hand side as the derivative of a product. This transformation simplifies the equation, making it easier to integrate and solve for the function, such as the current in the circuit.

First-Order Linear Differential Equations

The governing equation for an RL circuit is a first-order linear differential equation. Its general solution comprises a homogeneous solution that represents the transient response and a particular solution that represents the steady-state response. The process of solving this equation involves determining both parts and then applying initial conditions to finalize the overall solution.

Kirchhoff’s Voltage Law (KVL)

Kirchhoff’s Voltage Law is fundamental in deriving the dynamic equations for electrical circuits. It states that the algebraic sum of all voltages around any closed loop in a circuit must equal zero. In an RL circuit, this principle leads to an equation that relates the applied voltage to the voltage drops across the resistor (using Ohm's law) and across the inductor (using L dI/dt).

RL Circuit Analysis

RL circuit analysis involves studying circuits that include a resistor and an inductor. In such circuits, the behavior of the current is governed by a first?order differential equation derived from applying Kirchhoff's voltage law, which equates the sum of the voltage drops across the resistor (proportional to the current) and the inductor (proportional to the rate of change of the current) to the applied voltage.

Transcript

00:01 For this problem, we have a circuit.

00:05 The details of the circuit don't really matter themselves, or the figure doesn't really matter.

00:10 But we have that the generator supplies a voltage of e of t equals 40 sine 60t volts.

00:16 The inductance is 1 henry.

00:18 The resistance is 20 oms, and the i of 0 is equal to 1 amp.

00:22 We're asked to find i of t.

00:23 So we need to keep in mind that our differential equation for i here is ldi by dt plus, i believe it was...

00:36 Sorry.

00:38 So it's plus the resistance times the current equals the voltage e of t.

00:46 So in our case, we have the inductance l is 1, so we just have a d i by dt out front, plus now r times i, the resistance is 20, so we have plus 20 i, and then that is going to equal 40 sine 60t.

01:06 Now this is a linear equation so we can use our integrating factor.

01:11 We specifically want to have our integrating factor.

01:15 I have learned to label it as mu, but you can call it basically whatever you want.

01:21 E to the power of the integral of our coefficient on the i term, so we'd have e to the power of 20t here.

01:27 So the left -hand side, after multiplying through everything by the integrating factor, the left -hand side should become e to the power of 20 -t -i, and the right -hand side, side should become 40e to the power of 20t times sign of 60t.

01:44 Now the next step, excuse me, that should be all prime on the left hand side there.

01:50 The next step is to integrate both sides.

01:53 So we'd have e to the power of 20t times i is it equal to 40 times the integral indefinite of e to the power of 20t, sign of 60t, dt.

02:06 One moment here.

02:08 So we can find the integral there just by doing integration by parts.

02:12 It's pretty quick, so i won't go into the details there.

02:17 But the result should be e to the power of 2t over 200 times sign of 60t minus 3 cos of 60t.

02:32 So out front we have the 40 over 200.

02:37 So that will become just 1 over 5 out front.

02:46 So we have each the power of 2t over 5 times sign of 60t.

02:53 Ooh, you need to be careful.

02:55 There is a plus c there.