INC 122 : RL and RC Circuits
Dr.-Ing. Sudchai Boonto
Assistant Professor
February 15, 2024
Department of Control System and Instrumentation Engineering
King Mongkut’s Unniversity of Technology Thonburi Thailand
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Analysis of RC and RL Circuits: Transient Response, Time Constants, and Applications, Lecture notes of Electrical and Electronics Engineering
A comprehensive analysis of rc and rl circuits, focusing on transient response, time constants, and practical applications. It covers key concepts like initial conditions, differential equations, and step-by-step solution methods. Numerous examples and illustrations to demonstrate the principles and techniques involved in analyzing these circuits. It also explores the use of matlab and simulink for simulating and analyzing rc and rl circuits.
Typology: Lecture notes
2024/2025
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INC 122 : RL and RC Circuits
Dr.-Ing. Sudchai Boonto
Assistant Professor
February 15, 2024
Department of Control System and Instrumentation Engineering King Mongkut’s Unniversity of Technology Thonburi Thailand
Learning Outcomes
Students should be able to: ▶ (^) Calculate the initial values for inductor currents and capacitor voltages in the transient circuits. ▶ (^) Determine the voltages and currents in the first-order transient circuits. ▶ (^) Use Graphical and Symbolic tools to plot and check the calculation results.l
RC Circuits application: Discharge
vs
Rs
vC (t^ +) − C^ R
Discharge period td
KCL for the circuit
C dvC dt
- vC^ (t) R
dvC dt
RC
vC (t) = 0
vC (t) = V 0 e−^ RC^1 t
▶ (^) The solution function is a decaying exponential. ▶ (^) The rate at which it decays is a function of the values of R and C. ▶ (^) The product RC is a very important parameter, called time constant τ.
First-Order Circuit: General Form
A first-order differential equation:
dx dt
- ax(t) = f (t)
There are two solutions for this problem: ▶ (^) x(t) = xp(t) is any solution to the general equation. xp(t) is called the particular integral solution, or forced response. ▶ (^) x(t) = xc(t) is any solution to the homogeneous equation
dx dt
- ax(t) = 0.
xc(t) is called the complementary solution, or natural response. If we consider the situation in which f (t) = A (some constant). The general solution x(t) consists of two parts that are obtained by solving the two equations
dxp dt
axp(t) = A dxc dt
axc(t) = 0 5
First-Order Circuit: General Form
Consider the general solution
x(t) = K 1 + K 2 e−^
(^1) τ t
Each part of the equation has a names that are commonly employed in electrical engineering. ▶ (^) Term K 1 is referred to as the steady-state solution: the value of the variable x(t) as t → ∞, the second term become zero. ▶ (^) The constant τ is called the time constant of the circuit. The second term is a decaying exponential.
K 2 e−^1 τ t^ =
K 2 , τ > 0 and t = 0 0 , τ > 0 and t = ∞
▶ (^) The rate at which the exponential decays is determined by the time constant τ.
First-Order Circuit: General Form
τ (^2) τ 3 τ 4 τ
- 368 K 2
K 2
0
τ
- 632 G (^) G t(s)
xc(t) = K 2 e 1 τ^ t
1 2 3 4 5 6 7 8
1
0
τ = 0. 5 s
τ = 4 s
t(s)
e−^ τ^1 t
▶ (^) The value of xc(t) has fallen from K 2 to a value of 0. 368 K 2 in one time constant, a drop of 63 .2%. ▶ (^) In two time constants the value of xc(t) has fallen to 0. 135 K 2 , a drop of 63 .2% from the value at time t = τ , and the final value of the curve is closed by 63 .2% each time constant. ▶ (^) After five time constants, xc(t) = 0. 0067 K 2 , which is less than 1% ▶ (^) The circuit with a small-time constant has a fast response, and a large time constant circuit has a slow response. 8
Analysis Techniques RC Circuit: Differential Equations
Therefore
vC (t) = Vs + K 2 e−^ RC^1 t
To find the value of K 2 , we need to know the initial condition of vC (0−). Here the capacitor is uncharged at t < 0 , then
0 = Vs + K 2 ⇒ K 2 = −Vs
Hence, the complete solution for the voltage vC (t) is
vC (t) = Vs(1 − e RC^1 t).
Since τ = RC we can change the time constant by changing the value of RC.
Analysis Techniques RC Circuit: Example
2 Ω
4 Ω
10 μF −vC (t)
t = 0
− 9 V^ +
At t < 0 , we have vC (0−) = 9 V.
2 Ω
4 Ω
10 μF −vC (t)
We have
C
dvC dt
vC (t) R
vC (t) = vC (0−)e−^ RC^1 t
= 9e−^
1 60 × 10 −^6 t^ V
Note: The differential equation of this question is
dvC dt
(6)(10 × 10 −^6 )
vC (t) = 0
We can solve this problem by using Symbolic computational program. 11
Analysis Techniques RC Circuit: Example (Simscape)
,--------<14----------1 L
Simulink-PS Converter (^) Step
20hm
f(x) = O ConfigurationSolver 1
40hm
10 uF
PS-SimulinkConverter
Vs
Consider vC (t) after the fully charge period from t > 0. 3 ms.
13
Analysis Techniques RL Circuit: Differential Equation
Determine vR(t) of the circuit below at time t > 0.
−
−
Vs
t = 0 (^) R
+vR(t)−
L
Using KVL for t > 0 , we have
L diL dt
- RiL(t) = Vs diL dt
+ R
L
iL(t) = Vs L
From the standard from, we have
iL(t) = K 1 + K 2 e−^
(^1) τ t
Substituting the solution into the differential equation yields
− K^2
τ
e−^ τ^1 t^ + R L
K 1 + R
L
K 2 e−^1 τ^ t^ = Vs L
Equating the constant and exponential terms, we obtain
K 1 = Vs R
and τ = L R (^14)
Analysis Techniques RL Circuit: Example
−
−
VS 1
12 V
R 1
2 Ω L 2 H
t = 0
R 2 2 Ω
− 4 V (^) + VS 2
R 3 2 Ω
−
vo(t)
The switch in the network opens at t = 0. Let us find the output voltage vo(t) for t > 0.
The circuit at t < 0
−
−
VS 1
12 V
R 1
2 Ω (^) iL(0−)
R 2 2 Ω
− 4 V (^) + VS 2
R 3 2 Ω
A
There are several ways to find iL(0−). Here we use KCL.
vA − 12 2
vA^ + 4 2
vA 2
vA = 4 ⇒ vA =
V
iL(0−) =^4 3
A
Analysis Techniques RL Circuit: Example
The circuit at t > 0
−
−
VS 1
12 V
R 1
2 Ω (^) iL(t) L 2 H
R 3 2 Ω
diL dt
- 2iL(t) = 6
iL(t) = K 1 + K 2 e−^
RL t
= K 1 + K 2 e−^2 t K 1 = 3 and τ = 0. 5 s
Since iL(0−) = 43 A, we have
4 3
= 3 + K 2 ⇒ K 2 = −
Thus,
iL(t) = 3 −
e−^2 t^ A ⇒ vo(t) = 6 −
e−^2 t^ V
Analysis Techniques RL Circuit: Example (Simscape)
Consider vo(t) from t > 3 s.
Analysis Techniques RC and RL Circuits
▶ (^) We will not consider the step-by-step method. It is no benefit. ▶ (^) To use the step-by-step method, we need to store more formulas, which are not necessary. ▶ (^) Simple using KVL and KCL analysis are more than enough. ▶ (^) Just keep in your mind that
vC (0−) = vC (0+) = vC (0) iL(0−) = iL(0+) = iL(0)
This phenomenon is from the physical behavior of inductors and capacitors.