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EGR 2201 Unit 9 First-Order Circuits

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1 EGR 2201 Unit 9 First-Order Circuits
Read Alexander & Sadiku, Chapter 7. Homework #9 and Lab #9 due next week. Quiz next week. -Handouts: Quiz #8, Unit 9 practice sheets -Fire up MATLAB!

2 Review: DC Conditions in a Circuit with Inductors or Capacitors
Recall that when power is first applied to a dc circuit with inductors or capacitors, voltages and currents change briefly as the inductors and capacitors become energized. But once they are fully energized (i.e., “under dc conditions”), all voltages and currents in the circuit have constant values. To analyze a circuit under dc conditions, replace all capacitors with open circuits and replace all inductors with short circuits.

3 What About the Time Before DC Conditions?
We also want to be able to analyze such circuits during the time while the voltages and currents are changing, before dc conditions have been reached. This is sometimes called transient analysis, because the behavior that we’re looking at is short-lived. It’s the focus of Chapters 7 and 8 in the textbook.

4 Four Kinds of First-Order Circuits
The circuits we’ll study in this unit are called first-order circuits because they are described mathematically by first-order differential equations. We’ll study four kinds of first-order circuits: Source-free RC circuits Source-free RL circuits RC circuits with sources RL circuits with sources

5 Natural Response and Step Response
The term natural response refers to the behavior of source-free circuits. And the term step response refers to the behavior of circuits in which a source is applied at some time. So our goal in this unit is to understand the natural response of source-free RC and RL circuits, and to understand the step response of RC and RL circuits with sources.

6 Natural Response of Source-Free RC Circuit (1 of 2)
Consider the circuit shown. Assume that at time t=0, the capacitor is charged and has an initial voltage, V0. As time passes, the initial charge on the capacitor will flow through the resistor, gradually discharging the capacitor. This results in changing voltage v(t) and currents iC(t) and iR(t), which we wish to calculate.

7 Natural Response of Source-Free RC Circuit (2 of 2)
Applying KCL, 𝑖 𝐶 + 𝑖 𝑅 =0 Therefore 𝐶 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅 =0 Therefore 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅𝐶 =0 This equation is an example of a first-order differential equation. How do we solve it for v(t)?

8 Math Detour: Differential Equations
Differential equations arise frequently in science and engineering. Some examples: The equations above are all called linear ordinary differential equations with constant coefficients. A first-order diff. eq. A second-order diff. eq. A fourth-order diff. eq.

9 Solving Differential Equations
The differential equations on the previous slide are quite easy to solve. The ones shown below are more difficult. In a later math course you’ll learn many techniques for solving such equations. Non-constant coefficient Non-linear A partial differential equation

10 A Closer Look at Our Differential Equation
Note that our equation, 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅𝐶 =0, contains two constants, R and C. It also contains two variables, v and t. Also, t is the independent variable, while v is the dependent variable. We sometimes indicate this by writing v(t) instead of just v. Our goal is to write down an equation that expresses v(t) in terms of t, such as: 𝑣(𝑡) =8𝑡 𝑣 𝑡 =𝑅𝐶 sin 𝑡 (But neither of those is right!)

11 Solving Our Differential Equation
To solve our equation, 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅𝐶 =0, use a technique called separation of variables. First, separate the variables v and t: 𝑑𝑣 𝑣 =− 𝑑𝑡 𝑅𝐶 Then integrate both sides: ln 𝑣 =− 𝑡 𝑅𝐶 + ln 𝐴 Then raise e to both sides: 𝑣(𝑡)=𝐴 𝑒 − 𝑡 𝑅𝐶 where A is a constant

12 Apply the Initial Condition
At this point we have: 𝑣(𝑡)=𝐴 𝑒 − 𝑡 𝑅𝐶 The last step is to note that if we set t equal to 0, we get: 𝑣 0 =𝐴 But we assumed earlier that the initial voltage is some value that we called V0. So A must be equal to V0, and therefore: 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶

13 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶 The Bottom Line
I won’t expect you to be able to reproduce the derivation on the previous slides. The important point is to realize that whenever we have a circuit like this the solution for v(t) is: 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶 Do practice questions 1a, 1b.

14 Graph of Voltage Versus Time
Here’s a graph of 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶 This curve is called a decaying exponential curve. Note that at first the voltage falls steeply from its initial value (V0). But as time passes, the descent becomes less steep.

15 The Time Constant The values of R and C determine how rapidly the voltage descends. The product RC is given a special name (the time constant) and symbol (): 𝜏=𝑅𝐶 The greater  is, the more slowly the voltage descends.

16 Units of the Time Constant
What are the units of the time constant 𝜏? Since 𝜏=𝑅𝐶 and since R is measured in ohms () and C is measured in farads (F), 𝜏 is measured in ohm-farads (F). Surprisingly, an ohm-farad is equivalent to a second. So 𝜏 is measured in seconds (s).

17 Don’t Confuse t and  𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝜏
Since 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝑅𝐶 and  = RC, we often write 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝜏 t and  are both measured in seconds. But remember that t is a variable (time), and  is a constant (the time constant).

18 Rules of Thumb After one time constant (i.e., when t = ), the voltage has fallen to about 36.8% of its initial value. After five time constants (i.e., when t = 5), the voltage has fallen to about 0.7% of its initial value. For most practical purposes we say that the capacitor is completely discharged and v = 0 after five time constants. -Do practice question 1c and 1d, which is >fplot('15*exp(-455*t)', [0, 0.02]). -In MATLAB, turn on the data cursor (Tools > Data Cursor) to verify specific values.

19 Comparing Different Values of 
The greater  is, the more slowly the voltage descends, as shown below for a few values of .

20 Finding Values of Other Quantities
We’ve seen that 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝜏 From this equation we can use our prior knowledge to find equations for other quantities, such as current, power, and energy. For example, using Ohm’s law we find that 𝑖 𝑅 (𝑡)= 𝑉 0 𝑅 𝑒 − 𝑡 𝜏 Do practice question 1e.

21 The Keys to Working with a Source-Free RC Circuit
Find the initial voltage 𝑣(0)= 𝑉 0 across the capacitor. Find the time constant 𝜏=𝑅𝐶. Once you know these two items, voltage as a function of time is: 𝑣(𝑡)= 𝑉 0 𝑒 − 𝑡 𝜏 Once you know the voltage, solve for any other circuit variables of interest.

22 More Complicated Cases
A circuit that looks more complicated at first might be reducible to a simple source-free RC circuit by combining resistors. Example: Here we can combine the three resistors into a single equivalent resistor, as seen from the capacitor’s terminals. Do practice question 2.

23 Where Did V0 Come From? In previous examples you’ve simply been given the capacitor’s initial voltage, V0. More realistically, you have to find V0 by considering what happened before t = 0. Example: Suppose you’re told that the switch in this circuit has been closed for a long time before it’s opened at t = 0. Can you find the capacitor’s voltage V0 at time t = 0? Do practice question 3.

24 Natural Response of Source-Free RL Circuit (1 of 2)
Consider the circuit shown. Assume that at time t=0, the inductor is energized and has an initial current, I0. As time passes, the inductor’s energy will gradually dissipate as current flows through the resistor. This results in changing current i(t) and voltages vL(t) and vR(t), which we wish to calculate.

25 Natural Response of Source-Free RL Circuit (2 of 2)
Applying KVL, 𝑣 𝐿 + 𝑣 𝑅 =0 Therefore 𝐿 𝑑𝑖 𝑑𝑡 +𝑖𝑅=0 Therefore 𝑑𝑖 𝑑𝑡 + 𝑅 𝐿 𝑖=0 This first-order differential equation is similar to the equation we had for source-free RC circuits.

26 Solving Our Differential Equation
To solve our equation, 𝑑𝑖 𝑑𝑡 + 𝑅 𝐿 𝑖=0, first separate the variables i and t: 𝑑𝑖 𝑖 =− 𝑅 𝐿 𝑑𝑡 Then integrate both sides: ln 𝑖 =− 𝑅𝑡 𝐿 + ln 𝐴 Then raise e to both sides: 𝑖(𝑡)=𝐴 𝑒 − 𝑅𝑡 𝐿

27 Apply the Initial Condition
At this point we have: 𝑖(𝑡)=𝐴 𝑒 − 𝑅𝑡 𝐿 The last step is to note that if we set t equal to 0, we get: 𝑖 0 =𝐴 But we assumed earlier that the initial current is some value that we called I0. So A must be equal to I0, and therefore: 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑅𝑡 𝐿

28 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑅𝑡 𝐿 The Bottom Line
I won’t expect you to be able to reproduce the derivation on the previous slides. The important point is to realize that whenever we have a circuit like this the solution for i(t) is: 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑅𝑡 𝐿 Do practice questions 4a & b.

29 The Time Constant The values of R and L determine how rapidly the current decreases from its initial value to 0. Recall that for RC circuits we defined the time constant as 𝜏=𝑅𝐶. For RL circuits, we define it as 𝜏= 𝐿 𝑅 The greater  is, the more slowly the current decreases from its initial value.

30 Units of the Time Constant
What are the units of the time constant 𝜏? Since 𝜏= 𝐿 𝑅 and since L is measured in henries (H) and R is measured in ohms (), 𝜏 is measured in henries-per-ohm (H/). Surprisingly, a henry-per-ohm is equivalent to a second. So 𝜏 is measured in seconds (s).

31 Don’t Confuse t and  𝑖(𝑡)= 𝐼 0 𝑒 − 𝑅𝑡 𝐿 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑡 𝜏
Since 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑅𝑡 𝐿 and  = L/R, we often write 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑡 𝜏 t and  are both measured in seconds. But remember that t is a variable (time), and  is a constant (the time constant).

32 Graph of Current Versus Time
Here’s a graph of 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑡 𝜏 It’s a decaying exponential curve, with the current falling steeply from its initial value (I0). But as time passes, the descent becomes less steep.

33 Rules of Thumb After one time constant (i.e., when t = ), the current has fallen to about 36.8% of its initial value. After five time constants (i.e., when t = 5), the current has fallen to about 0.7% of its initial value. For most practical purposes we say that the inductor is completely de-energized and i = 0 after five time constants. Do practice questions 4c & d.

34 Finding Values of Other Quantities
We’ve seen that 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑡 𝜏 From this equation we can use our prior knowledge to find equations for other quantities, such as voltage, power, and energy. For example, using Ohm’s law we find that 𝑣 𝑅 (𝑡)= 𝐼 0 𝑅 𝑒 − 𝑡 𝜏

35 The Keys to Working with a Source-Free RL Circuit
Find the initial current 𝑖(0)= 𝐼 0 through the inductor. Find the time constant 𝜏= 𝐿 𝑅 . Once you know these two items, current as a function of time is: 𝑖(𝑡)= 𝐼 0 𝑒 − 𝑡 𝜏 Once you know the current, solve for any other circuit variables of interest.

36 More Complicated Cases
A circuit that looks more complicated at first might be reducible to a simple source-free RL circuit by combining resistors. Example: Here we can combine the resistors into a single equivalent resistor, as seen from the inductor’s terminals.

37 Where Did I0 Come From? In previous examples you were simply given the inductor’s initial current, I0. More realistically, you have to find I0 by considering what happened before t = 0. Example: Suppose you’re told that the switch in this circuit has been closed for a long time before it’s opened at t = 0. Can you find the inductor’s current I0 at time t = 0? Do practice question 5.

38 Where We Are We’ve looked at: We still need to look at:
Source-free RC circuits Source-free RL circuits We still need to look at: RC circuits with sources RL circuits with sources Before doing this, we’ll look at some mathematical functions called singularity functions (or switching functions), which are widely used to model electrical signals that arise during switching operations.

39 Three Singularity Functions
The three singularity functions that we’ll study are: The unit step function The unit impulse function The unit ramp function

40 The Unit Step Function 𝑢 𝑡 = 0, 𝑡<0 1, 𝑡>0
The unit step function u(t) is equal to 0 for negative values of t and equal to 1 for positive values of t: 𝑢 𝑡 = 0, 𝑡<0 1, 𝑡>0

41 Shifting and Scaling the Unit Step Function
We can obtain other step functions by shifting the unit step function to the left or right… …or by multiplying the unit step function by a scaling constant: 𝑢 𝑡−2 = 0, 𝑡<2 s 1, 𝑡>2 s 3𝑢 𝑡 = 0, 𝑡<0 3, 𝑡>0

42 Flipping the Unit Step Function Horizontally
As with any mathematical function, we can “flip” the unit step function horizontally by replacing t with t: 𝑢 −𝑡 = 1, 𝑡<0 0, 𝑡>0

43 Adding Step Functions By adding two or more step functions we can obtain more complex “step-like” functions, such as the one shown below from the book’s Practice Problem 7.6. Do practice questions 6 & 7.

44 Using Step Functions to Model Switched Sources
Step functions are useful for modeling sources that are switched on (or off) at some time:

45 The Unit Impulse Function
The unit step function’s derivative is the unit impulse function (t), also called the delta function. The unit impulse function is 0 everywhere except at t =0, where it is undefined: It’s useful for modeling “spikes” that can occur during switching operations. 𝛿 𝑡 = 0, 𝑡<0 Undefined, 𝑡=0 0, 𝑡>0

46 Shifting and Scaling the Unit Impulse Function
We can obtain other impulse functions by shifting the unit impulse function to the left or right, or by multiplying the unit impulse function by a scaling constant: We won’t often use impulse functions in this course, but they’re used in more advanced courses.

47 The Unit Ramp Function 𝑟 𝑡 = 0, 𝑡<0 𝑡, 𝑡≥0
The unit step function’s integral is the unit ramp function r(t). The unit ramp function is 0 for negative values of t and has a slope of 1 for positive values of t: 𝑟 𝑡 = 0, 𝑡<0 𝑡, 𝑡≥0

48 Shifting and Scaling the Unit Ramp Function
We can obtain other ramp functions by shifting the unit ramp function to the left or right… …or by multiplying the unit ramp function by a scaling constant: 𝑟 𝑡−0.2 = 0, 𝑡<0.2 s 𝑡, 𝑡≥0.2 s 4𝑟 𝑡 = 0, 𝑡<0 4𝑡, 𝑡≥0

49 Adding Step and Ramp Functions
By adding two or more step functions or ramp functions we can obtain more complex functions, such as the one shown below from the book’s Example 7.7. Do practice questions 8 & 9.

50 Step Response of a Circuit
A circuit’s step response is the circuit’s behavior due to a sudden application of a dc voltage source or current source. We can use a step function to model this sudden application.

51 t = 0 versus t = 0+ We distinguish the following two times:
t = 0 (the instant just before the switch closes) t = 0+ (the instant just after the switch closes) Since a capacitor’s voltage cannot change abruptly, we know that v(0) = v(0+) in this circuit. But on the other hand, i(0)  i(0+) in this circuit.

52 Step Response of RC Circuit (1 of 2)
Assume that in the circuit shown, the capacitor’s initial voltage is V0 (which may equal 0 V). As time passes after the switch closes, the capacitor’s voltage will gradually approach the source voltage VS. This results in changing voltage v(t) and current i(t), which we wish to calculate.

53 Step Response of RC Circuit (2 of 2)
Applying KCL, for t>0, 𝐶 𝑑𝑣 𝑑𝑡 + 𝑣− 𝑉 𝑆 𝑅 =0 Therefore 𝑑𝑣 𝑑𝑡 =− 𝑣− 𝑉 𝑆 𝑅𝐶 Separating variables, 𝑑𝑣 𝑣− 𝑉 𝑆 =− 𝑑𝑡 𝑅𝐶 Integrating both sides, ln (𝑣− 𝑉 𝑆 ) =− 𝑡 𝑅𝐶 + ln (𝐴)

54 Solution of Our Differential Equation
Raising e to both sides and rearranging: 𝑣 𝑡 = 𝑉 𝑆 +𝐴 𝑒 − 𝑡 𝑅𝐶 Applying the initial condition and letting  = RC: 𝑣 𝑡 = 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 Finally, 𝑣 𝑡 = 𝑉 0 , 𝑡<0 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 , 𝑡≥0

55 𝑣 𝑡 = 𝑉 0 , 𝑡<0 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 , 𝑡≥0
The Bottom Line I won’t expect you to be able to reproduce the derivation on the previous slides. The important point is to realize that for a circuit like this: the solution for v(t) is: Do practice questions 10a, b. 𝑣 𝑡 = 𝑉 0 , 𝑡<0 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 , 𝑡≥0

56 Graph of Voltage Versus Time
Here’s a graph of (assuming V0< VS). This is a saturating exponential curve. 𝑣 𝑡 = 𝑉 0 , 𝑡<0 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 , 𝑡≥0 Note that the voltage at first rises steeply from its initial value (V0), and then gradually approaches its final value (VS).

57 Rules of Thumb, etc. We can repeat many of the same remarks as for source-free circuits, such as: The greater  is, the more slowly v(t) approaches its final value. For most practical purposes, v(t) reaches its final value after 5. Knowing v(t), we can use Ohm’s law to find current, and we can use other familiar formulas to find power, energy, etc. Do practice questions 10c, d.

58 Another Way of Looking At It
We can rewrite the complete response 𝑣 𝑡 = 𝑉 𝑆 +( 𝑉 0 − 𝑉 𝑆 ) 𝑒 − 𝑡 𝜏 as: 𝑣 𝑡 =𝑣(∞)+(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 Here v(0) is the initial value and v() is the final, or steady-state, value. This same equation works for source-free RC circuits too, since setting v() to 0 gives 𝑣 𝑡 =𝑣(0) 𝑒 − 𝑡 𝜏 .

59 The Keys to Finding an RC Circuit’s Step Response
Find the capacitor’s initial voltage 𝑣(0). Find the capacitor’s final voltage 𝑣(∞). Find the time constant 𝜏=𝑅𝐶. Once you know these items, voltage is: 𝑣 𝑡 =𝑣(∞)+(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 Once you know the voltage, solve for any other circuit variables of interest.

60 More Complicated Cases
A circuit that looks more complicated at first might be reducible to a simple RC circuit by combining resistors. Example: Here, for t > 0 we can combine the resistors into a single equivalent resistor, as seen from the capacitor’s terminals. Do practice problems 11 & 12.

61 Two Ways of Breaking It Down
There are two useful ways of looking at the response 𝑣 𝑡 =𝑣(∞)+(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 As the sum of a natural response and a forced response. As the sum of a transient response and a steady-state response. This way is of more interest to us.

62 Natural Response Versus Forced Response
We can think of the complete response 𝑣 𝑡 =𝑣(∞)+(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 as being the sum of: A natural response 𝑣 𝑛 𝑡 =𝑣(0) 𝑒 − 𝑡 𝜏 that depends the capacitor’s initial charge. Plus a forced response 𝑣 𝑓 𝑡 =𝑣(∞)−𝑣(∞) 𝑒 − 𝑡 𝜏 that depends on the voltage source.

63 Transient Response Versus Steady-State Response
We can think of the complete response 𝑣 𝑡 =𝑣(∞)+(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 as being the sum of: A transient response 𝑣 𝑡 𝑡 =(𝑣(0)−𝑣(∞)) 𝑒 − 𝑡 𝜏 that dies away as time passes. Plus a steady-state response 𝑣 𝑠𝑠 𝑡 =𝑣(∞) that remains after the transient response has died away.

64 “Under DC Conditions” = Steady-State
Recall that earlier we used the term “under dc conditions” to refer to the time after an RC or RL circuit’s currents and voltages have “settled down” to their final values. This is just another way of referring to what we’re now calling steady-state values. So way can say that in the steady state, capacitors look like open circuits and inductors look like short circuits.

65 Step Response of RL Circuit (1 of 2)
Assume that in the circuit shown, the inductor’s initial current is I0 (which may equal 0 A). As time passes, the inductor’s current will gradually approach a steady-state value. This results in changing current i(t) and voltage v(t), which we wish to calculate.

66 Step Response of RL Circuit (2 of 2)
Using the same sort of math we used previously for RC circuits, we find where  = L/R, just as for source-free RL circuits. 𝑖 𝑡 = 𝐼 0 , 𝑡<0 𝑉 𝑆 𝑅 + 𝐼 0 − 𝑉 𝑆 𝑅 𝑒 − 𝑡 𝜏 , 𝑡>0

67 Another Way of Looking At It
We can rewrite 𝑖 𝑡 = 𝑉 𝑆 𝑅 + 𝐼 0 − 𝑉 𝑆 𝑅 𝑒 − 𝑡 𝜏 as: 𝑖 𝑡 =𝑖(∞)+(𝑖(0)−𝑖(∞)) 𝑒 − 𝑡 𝜏 Here i(0) is the initial value and i() is the final, or steady-state, value. This same equation works for source-free RL circuits too, since setting i() to 0 gives 𝑖 𝑡 =𝑖(0) 𝑒 − 𝑡 𝜏 .

68 Graph of Voltage Versus Time
Here’s a graph of 𝑖 𝑡 =𝑖(∞)+(𝑖(0)−𝑖(∞)) 𝑒 − 𝑡 𝜏 (assuming i(0) > i(∞)).

69 Rules of Thumb, etc. We can repeat many of the same remarks as for previous circuits, such as: The greater  is, the more slowly i(t) approaches its final value. For most practical purposes, i(t) reaches its final value after 5. Knowing i(t), we can use Ohm’s law to find voltage, and we can use other familiar formulas to find power, energy, etc.

70 The Keys to Finding an RL Circuit’s Step Response
Find the inductor’s initial current 𝑖(0). Find the capacitor’s final current 𝑖(∞). Find the time constant 𝜏=𝐿/𝑅. Once you know these items, current is: 𝑖 𝑡 =𝑖(∞)+(𝑖(0)−𝑖(∞)) 𝑒 − 𝑡 𝜏 Once you know the current, solve for any other circuit variables of interest.

71 More Complicated Cases
A circuit that looks more complicated at first might be reducible to a simple RL circuit by combining resistors. Example: Here we can combine the resistors into a single equivalent resistor, as seen from the inductor’s terminals. Do practice problems 13, 14.

72 A General Approach for First-Order Circuits (1 of 3)
As noted on page 276 of the textbook: Once we know 𝑥(0), 𝑥(∞), and , almost all the circuit problems in this chapter can be solved using the formula 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏 What is x here? It could be any current or voltage in a first-order circuit.

73 A General Approach for First-Order Circuits (2 of 3)
So to find x(t) in a first-order circuit, where x could be any current or voltage: Find the quantity’s initial value 𝑥(0). Find the quantity’s final value 𝑥(∞). Find the time constant: 𝜏=𝑅𝐶 for an RC circuit. 𝜏=𝐿/𝑅 for an RL circuit. Once you know these items, solution is: 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏

74 A General Approach for First-Order Circuits (3 of 3)
The equation from the previous slide, 𝑥 𝑡 =𝑥(∞)+(𝑥(0)−𝑥(∞)) 𝑒 − 𝑡 𝜏 always graphs as either: A decaying exponential curve if the initial value x(0) is greater than the final value x(). Or a saturating exponential curve if the initial value x(0) is less than the final value x().


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