G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Chapter 5:
Transient Analysis – Instructor Notes
Chapter 5 was reorganized in the 4th Edition, in response to a number of suggestions forwarded by users.
The chapter is mostly unchanged in this 5th Edition. The chapter begins with a brief description of what is meant by
Transient Analysis in Section 5.1; basic techniques for writing differential equations for dynamic circuits are
described in Section 5.2, while Section 5.3 focuses on DC steady-state solutions and on initial and final conditions.
The analogy between electrical and thermal systems that was introduced in Chapter 3 is now extended to energy
storage elements and transient response (Make The Connection: Thermal Capacitance, p. 218; Make The
Connection: Thermal System Dynamics, p. 219).
Section 5.4 introduces first order transients; a Focus on Methodology box: First Order Transient Response (p. 229)
clearly outlines the methodology that is followed in the analysis of first order circuits; this methodology is then
motivated and explained, and is applied to eight examples, including four examples focusing on engineering
applications (5.8 - Charging a camera flash; 5.9 and 5.11 dc motor transients; and 5.12, transient response of
supercapacitor bank). The box Focus on Measurements: Coaxial Cable Pulse Response (pp. 230-232) illustrates
an important transient analysis computation (this problem was suggested many years ago by a Nuclear Engineering
colleague). The analogy between electrical and thermal systems is carried further in the sidebars Make The
Connection: First-Order Thermal System, p. 232-233;); similarly, the analogy between hydraulic and electrical
circuits, begun in Chapter 2 and continued in Chapter 4, is continued here (Make The Connection: Hydraulic Tank,
pp. 228-229).
Section 5.5 covers second order transients, and, as was done for first order transients, begins with two important
boxes: Focus on Methodology: Roots of Second-Order System (p. 254) and Focus on Methodology: Second Order
Transient Response (pp. 258-259). These boxes clearly outline the methodology that is followed in the analysis of
second order circuits; the motivation and explanations in this section are accompanied by five very detailed
examples in which the methodology is applied step by step. The last of these examples takes a look at an
automotive ignition circuit (with many thanks to my friend John Auzins, formerly of Delco Electronics, for
suggesting a simple but realistic ignition circuit). The analogy between electrical and mechanical systems is
explored in Make The Connection: Automotive Suspension, pp. 253-254 and pp. 259-260.
The homework problems are divided into four sections, and contain a variety of problems ranging from very basic to
the fairly advanced. The focus is on mastering the solution methods illustrated in the chapter text and examples.
The homework problems in this chapter are mostly mathematical exercises aimed at mastery of the techniques. The
5th Edition of this book includes 7 new problems, increasing the end-of-chapter problem count from 74 to 81.
Learning Objectives for Chapter 5
1.
2.
3.
4.
5.
Write differential equations for circuits containing inductors and capacitors.
Determine the DC steady state solution of circuits containing inductors and capacitors.
Write the differential equation of first order circuits in standard form and determine the complete
solution of first order circuits excited by switched DC sources.
Write the differential equation of second order circuits in standard form and determine the
complete solution of second order circuits excited by switched DC sources.
Understand analogies between electrical circuits and hydraulic, thermal and mechanical systems.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Section 5.2: Writing Differential Equations for Circuits Containing
Inductors and Capacitors
Problem 5.1
Write the differential equations for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 in Figure P5.21. How are they related?
Known quantities:
None
Find:
The differential equation for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 in Figure P5.21 and the relationship between 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 .
Analysis:
At 𝑡𝑡 > 0, 𝑉𝑉𝑆𝑆 is not attached to the circuit. Use 𝐾𝐾𝐾𝐾𝐾𝐾 at the bottom of the circuit and note the polarity switch due
to the direction of 𝑖𝑖𝐿𝐿 :
Use Ohm’s law to eliminate 𝑖𝑖1 :
Use Ohm’s law to eliminate 𝑖𝑖3 :
𝑖𝑖𝐿𝐿 = −𝑖𝑖1 − 𝑖𝑖3
𝑖𝑖𝐿𝐿 = −
𝑣𝑣𝐿𝐿
− 𝑖𝑖3
𝑅𝑅1 + 𝑅𝑅2
𝑖𝑖𝐿𝐿 = −
𝑣𝑣𝐿𝐿
𝑣𝑣3
−
𝑅𝑅1 + 𝑅𝑅2 𝑅𝑅3
𝑖𝑖𝐿𝐿 = −
𝑣𝑣𝐿𝐿
𝑣𝑣𝐿𝐿
−
𝑅𝑅1 + 𝑅𝑅2 𝑅𝑅3
Note that 𝑣𝑣𝐿𝐿 and 𝑣𝑣3 are related because they are in parallel:
Substitute the definition of voltage across an inductor:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Simplify:
𝑖𝑖𝐿𝐿 = −
�
𝐿𝐿
𝐿𝐿 𝑑𝑑𝑖𝑖𝐿𝐿
+ �
+ 𝑖𝑖𝐿𝐿 = 0
𝑅𝑅1 + 𝑅𝑅2 𝑅𝑅3 𝑑𝑑𝑑𝑑
Again, the relationship between 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 is:
Thus:
Problem 5.2
𝑑𝑑𝑖𝑖𝐿𝐿
𝐿𝐿
𝐿𝐿 𝑑𝑑𝑖𝑖𝐿𝐿
−
𝑅𝑅1 + 𝑅𝑅2 𝑑𝑑𝑑𝑑 𝑅𝑅3 𝑑𝑑𝑑𝑑
𝑣𝑣3 = 𝑣𝑣𝐿𝐿
𝑣𝑣3 =
𝑑𝑑𝑖𝑖𝐿𝐿
𝑑𝑑𝑑𝑑
The differential equation for t > 0 (switch closed) forVc in
figure P5.23.
Solution:
Known quantities:
Find:
The differential equation for t > 0 (switch closed) for the circuit of P5.23.
Analysis:
Apply KCL at the top node (nodal analysis) to write the circuit equation. Note that the top node voltage is the
capacitor voltage, vC .
v
v − V1
iC + C + C
=0
R2
R1
Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation:
dv
R + R2
V
dvC
R + R2
V
⇒
C C + 1
vC = 1
+ 1
vC = 1
dt
R1R2
R1
dt
C (R1R2 )
CR1
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Problem 5.3
Write the differential equation for 𝑡𝑡 > 0 for 𝑖𝑖𝐶𝐶 in Figure P5.32.
Known quantities:
None.
Find:
The differential equation for 𝑡𝑡 > 0 for the circuit of Figure P5.32.
Analysis:
To determine the differential equation for this circuit, take advantage of nodal analysis and apply KCL to the
circuit.
Let 𝑉𝑉2 be the voltage at the top node. Utilize KCL to determine a relationship between the nodes:
Solve for the current through the capacitor:
𝑉𝑉1 − 𝑉𝑉2 𝑉𝑉2
=
+ 𝑖𝑖𝐶𝐶
𝑅𝑅1
𝑅𝑅3
𝑖𝑖𝐶𝐶 =
𝑉𝑉1 − 𝑉𝑉2 𝑉𝑉2
−
𝑅𝑅1
𝑅𝑅3
𝑉𝑉2 may be eliminated by using Ohm’s law on 𝑅𝑅2 to obtain:
Solving for 𝑉𝑉2 gives the equation:
𝑖𝑖𝐶𝐶 =
𝑉𝑉2 − 𝑉𝑉𝐶𝐶
𝑅𝑅2
𝑉𝑉2 = 𝑅𝑅2 𝑖𝑖𝐶𝐶 + 𝑉𝑉𝐶𝐶
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Substitute the derived equation for 𝑉𝑉2 into the equation for current through the capacitor:
Algebraically simplify:
𝑖𝑖𝐶𝐶 =
𝑉𝑉1 − (𝑅𝑅2 𝑖𝑖𝐶𝐶 + 𝑉𝑉𝐶𝐶 ) 𝑅𝑅2 𝑖𝑖𝐶𝐶 + 𝑉𝑉𝐶𝐶
−
𝑅𝑅1
𝑅𝑅3
𝑖𝑖𝐶𝐶 =
𝑉𝑉1 𝑅𝑅2
𝑉𝑉𝐶𝐶 𝑅𝑅2
𝑉𝑉𝐶𝐶
− 𝑖𝑖𝐶𝐶 − − 𝑖𝑖𝐶𝐶 −
𝑅𝑅1 𝑅𝑅1
𝑅𝑅1 𝑅𝑅3
𝑅𝑅3
𝑖𝑖𝐶𝐶 +
𝑅𝑅2
𝑅𝑅2
𝑉𝑉1 𝑉𝑉𝐶𝐶 𝑉𝑉𝐶𝐶
𝑖𝑖𝐶𝐶 + 𝑖𝑖𝐶𝐶 =
− −
𝑅𝑅1
𝑅𝑅3
𝑅𝑅1 𝑅𝑅1 𝑅𝑅3
Isolate the terms containing 𝑖𝑖𝐶𝐶 :
Distribute 𝑖𝑖𝐶𝐶 out of the left-hand side of the equation:
Solve for 𝑖𝑖𝐶𝐶 and simplify:
𝑖𝑖𝐶𝐶 (1 +
𝑖𝑖𝐶𝐶 =
𝑅𝑅2 𝑅𝑅2
𝑉𝑉1 𝑉𝑉𝐶𝐶 𝑉𝑉𝐶𝐶
+ )=
− −
𝑅𝑅1 𝑅𝑅3
𝑅𝑅1 𝑅𝑅1 𝑅𝑅3
𝑉𝑉1 − 𝑉𝑉𝐶𝐶
𝑉𝑉𝐶𝐶
−
𝑅𝑅2 𝑅𝑅2
𝑅𝑅
𝑅𝑅
𝑅𝑅1 (1 +
+ )
𝑅𝑅3 (1 + 2 + 2 )
𝑅𝑅1 𝑅𝑅3
𝑅𝑅1 𝑅𝑅3
𝑖𝑖𝐶𝐶 =
𝑉𝑉1 − 𝑉𝑉𝐶𝐶
𝑉𝑉𝐶𝐶
−
𝑅𝑅3 𝑅𝑅2
𝑅𝑅2 𝑅𝑅1
𝑅𝑅1 + 𝑅𝑅2 +
𝑅𝑅3 +
+ 𝑅𝑅2
𝑅𝑅3
𝑅𝑅1
Substitute the equation for current through a capacitor:
𝐶𝐶
Simplify:
Problem 5.5
𝑑𝑑𝑉𝑉𝐶𝐶
𝑉𝑉1 − 𝑉𝑉𝐶𝐶
𝑉𝑉𝐶𝐶
=
−
𝑅𝑅
𝑅𝑅
𝑅𝑅
𝑅𝑅
𝑑𝑑𝑑𝑑
𝑅𝑅1 + 𝑅𝑅2 + 2 1
𝑅𝑅3 + 3 2 + 𝑅𝑅2
𝑅𝑅3
𝑅𝑅1
𝑑𝑑𝑉𝑉𝐶𝐶
𝑉𝑉1 − 𝑉𝑉𝐶𝐶
𝑉𝑉𝐶𝐶
=
−
𝑅𝑅 𝑅𝑅
𝑅𝑅 𝑅𝑅
𝑑𝑑𝑑𝑑
𝐶𝐶(𝑅𝑅1 + 𝑅𝑅2 + 2 1 )
𝐶𝐶(𝑅𝑅3 + 3 2 + 𝑅𝑅2 )
𝑅𝑅3
𝑅𝑅1
Write the differential equation for t > 0 for vC in Figure P5.32
Solution:
Known quantities:
I 0 = 17 mA,C = 0.55 µF, R1 = 7 kΩ,R2 = 3.3 kΩ.
Find:
The differential equation for t > 0 for the circuit of P5.32.
Analysis:
Using the definition of capacitor current:
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dvC
dvC I 0
⇒
=
= I0
dt
C
dt
Substituting numerical values, we obtain the following differential equation:
C
dvC
− 30909 = 0
dt
Problem 5.4
Write the differential equation for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 in Figure P5.29.
Known quantities:
None.
Find:
The differential equation for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 of Figure P5.29.
Analysis:
To determine the differential equation for this circuit, take advantage of KVL. Examining the circuit at 𝑡𝑡 > 0
reveals that the current through each element is 𝑖𝑖𝐿𝐿 .
Use KVL starting at 𝑉𝑉𝑆𝑆2 :
𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐿𝐿 𝑅𝑅2 − 𝑉𝑉𝐿𝐿 − 𝑖𝑖𝐿𝐿 𝑅𝑅3 = 0
The goal is to find an equation for 𝑖𝑖𝐿𝐿 , so substitute the equation for voltage across an inductor:
Finally, solve for the inductor voltage:
𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐿𝐿 𝑅𝑅2 −
𝐿𝐿𝐿𝐿𝑖𝑖𝐿𝐿
− 𝑖𝑖𝐿𝐿 𝑅𝑅3 = 0
𝑑𝑑𝑑𝑑
𝑑𝑑𝑖𝑖𝐿𝐿
= 𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐿𝐿 𝑅𝑅2 − 𝑖𝑖𝐿𝐿 𝑅𝑅3
𝑑𝑑𝑑𝑑
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Problem 5.6
Write the differential equations for 𝑡𝑡 > 0 for 𝑖𝑖𝐶𝐶 and 𝑣𝑣3 in Figure P5.34. How are they related?
Known quantities:
None
Find:
The differential equations for 𝑡𝑡 > 0 for 𝑖𝑖𝐶𝐶 and 𝑣𝑣3 .
Analysis:
Note that, at 𝑡𝑡 > 0, only 𝑉𝑉𝑆𝑆2 is connected to the circuit.
Apply nodal analysis at the top node:
For 𝑖𝑖𝐶𝐶 use KCL to write:
𝑖𝑖1 = 𝑖𝑖𝐶𝐶 + 𝑖𝑖3
𝑉𝑉𝑆𝑆2 − 𝑣𝑣1
𝑣𝑣1
= 𝑖𝑖𝐶𝐶 +
𝑅𝑅1
𝑅𝑅3
Note that the voltage at the top node, 𝑣𝑣1 , is equivalent to 𝑣𝑣3 :
Using KVL at the top node gives the equation:
Solve for 𝑣𝑣3 :
Substitute equation for 𝑣𝑣3 into KVL equation:
𝑉𝑉𝑆𝑆2 − 𝑣𝑣3
𝑣𝑣3
= 𝑖𝑖𝐶𝐶 +
𝑅𝑅1
𝑅𝑅3
𝑣𝑣3 − 𝑖𝑖𝐶𝐶 𝑅𝑅2 − 𝑣𝑣𝐶𝐶 = 0
𝑣𝑣3 = 𝑖𝑖𝐶𝐶 𝑅𝑅2 + 𝑣𝑣𝐶𝐶
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Consolidate like terms:
𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐶𝐶 𝑅𝑅2 − 𝑣𝑣𝐶𝐶
𝑖𝑖𝐶𝐶 𝑅𝑅2 + 𝑣𝑣𝐶𝐶
= 𝑖𝑖𝐶𝐶 +
𝑅𝑅1
𝑅𝑅3
𝑉𝑉𝑆𝑆2
𝑅𝑅2 𝑅𝑅2
1
1
− 𝑖𝑖𝐶𝐶 � +
+ 1� − 𝑣𝑣𝐶𝐶 � + � = 0
𝑅𝑅1
𝑅𝑅1 𝑅𝑅3
𝑅𝑅1 𝑅𝑅3
Substitute the definition of current through a capacitor to eliminate 𝑖𝑖𝐶𝐶 :
𝑉𝑉𝑆𝑆2
𝑑𝑑𝑣𝑣𝐶𝐶 𝑅𝑅2 𝑅𝑅2
1
1
− 𝐶𝐶
� +
+ 1� − 𝑣𝑣𝐶𝐶 � + � = 0
𝑅𝑅1
𝑑𝑑𝑑𝑑 𝑅𝑅1 𝑅𝑅3
𝑅𝑅1 𝑅𝑅3
Again, note the relationship between 𝑖𝑖𝐶𝐶 and 𝑣𝑣3 :
𝑣𝑣3 = 𝑖𝑖𝐶𝐶 𝑅𝑅2 + 𝑣𝑣𝐶𝐶
Problem 5.7
Write the differential equation for 𝑡𝑡 > 0 for 𝑣𝑣𝐶𝐶 in Figure P5.41.
Known quantities:
𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, R 3 = 3 Ω, R 4 = 6 Ω, and 𝐶𝐶1 = 𝐶𝐶2 = 4 F
From problem 5.41, assume that switch 𝑆𝑆1 is always held open and that switch 𝑆𝑆2 is closed at 𝑡𝑡 > 0.
Find:
The differential equation for 𝑡𝑡 > 0 for 𝑣𝑣𝐶𝐶 .
Analysis:
Because switch 𝑆𝑆1 is always open and switch 𝑆𝑆2 is closed when 𝑡𝑡 > 0, this circuit can be greatly simplified.
The 20 V voltage source and 𝑅𝑅1 can be ignored since they are not connected to the circuit.
𝐶𝐶1 and 𝐶𝐶2 may be combined into an equivalent capacitor using the equation:
𝐶𝐶1,2,𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶2
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𝐶𝐶1,2,𝑒𝑒𝑒𝑒 = 4 𝐹𝐹 + 4 𝐹𝐹
𝐶𝐶1,2,𝑒𝑒𝑒𝑒 = 8 𝐹𝐹
𝑅𝑅3 and 𝑅𝑅4 may be combined as parallel resistors into an equivalent resistor using the equation:
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 =
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 =
𝑅𝑅3 𝑅𝑅4
𝑅𝑅3 + 𝑅𝑅4
3 Ω∗6 Ω
3 Ω+6 Ω
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 = 2 Ω
Then, 𝑅𝑅3,4,𝑒𝑒𝑒𝑒 and the 4 𝐴𝐴 current source may be viewed as a Norton equivalent circuit and converted to a
Thévenin equivalent circuit.
Solve for the Thévenin equivalent resistance:
𝑅𝑅𝑇𝑇ℎ = 𝑅𝑅3,4,𝑒𝑒𝑒𝑒
𝑅𝑅𝑇𝑇ℎ = 2 Ω
Solve for the Thévenin source:
𝑉𝑉𝑇𝑇ℎ = 𝐼𝐼𝑁𝑁 𝑅𝑅𝑇𝑇ℎ
𝑉𝑉𝑇𝑇ℎ = 4 𝐴𝐴 ∗ 2 Ω
𝑉𝑉𝑇𝑇ℎ = 8 𝑉𝑉
𝑅𝑅2 and 𝑅𝑅𝑇𝑇ℎ may be combined as resistors in series into an equivalent resistor using the equation:
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅𝑇𝑇ℎ
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 = 4 Ω + 2 Ω
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 = 6 Ω
This circuit is now a very simple single loop circuit with a voltage source of 8 𝑉𝑉, a resistor of 6 𝛺𝛺, and a
capacitor of 8 𝐹𝐹, so KVL can be used to write the following equation:
𝑉𝑉𝑇𝑇ℎ − 𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 𝐼𝐼𝐶𝐶 − 𝑣𝑣𝐶𝐶 = 0 𝑉𝑉
Substitute the equation for current through a capacitor:
𝑉𝑉𝑇𝑇ℎ − 𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 𝐶𝐶1,2,𝑒𝑒𝑒𝑒
Write in terms of current through the capacitor:
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 𝐶𝐶1,2,𝑒𝑒𝑒𝑒
5.9
𝑑𝑑𝑣𝑣𝐶𝐶
𝑑𝑑𝑑𝑑
− 𝑣𝑣𝐶𝐶 = 0 𝑉𝑉
𝑑𝑑𝑣𝑣𝐶𝐶
= 𝑉𝑉𝑇𝑇ℎ
𝑑𝑑𝑑𝑑
− 𝑣𝑣𝐶𝐶
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Solve for
𝑑𝑑𝑣𝑣𝐶𝐶
𝑑𝑑𝑑𝑑
:
Substitute known values:
𝑑𝑑𝑣𝑣𝐶𝐶
𝑉𝑉𝑇𝑇ℎ − 𝑣𝑣𝐶𝐶
=
𝑑𝑑𝑑𝑑
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 𝐶𝐶1,2,𝑒𝑒𝑒𝑒
𝑑𝑑𝑣𝑣𝐶𝐶
8 𝑉𝑉 − 𝑣𝑣𝐶𝐶
=
𝑑𝑑𝑑𝑑
6 Ω ∗ 8 𝐹𝐹
𝑑𝑑𝑣𝑣𝐶𝐶 8 𝑉𝑉 − 𝑣𝑣𝐶𝐶
=
𝑑𝑑𝑑𝑑
48 𝑠𝑠
Problem 5.8
Write the differential equation for t > 0 for iC in
Figure P5.47. Assume VS = 9V, RS = 5 kohm,
R1 = 10 kohm, and R2 = R3 = 20 kohm.
Solution:
Known quantities:
C = 1µF , RS =15kΩ , R3 = 30 kΩ .
Find:
The differential equation for t > 0 (switch closed) for the
circuit of P5.47.
Assume:
Assume that VS = 9 V, R1 =10kΩ and R2 = 20kΩ .
Analysis:
Apply KCL at the top node (nodal analysis) to write the circuit equation.
1. Before the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation.
 1
v
v
vC − VS vC
1
1
1 
V
⇒
+
+ iC + C + C = 0
+
+
+ vC + iC − S = 0

R2 R3
RS
R1
RS
 RS R1 R2 R3 
Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation:
 1
1
1
1 
dv
V
+
+
+

vC + C C − S = 0
dt
RS
 RS R1 R2 R3 
Substituting numerical value, we obtain the following differential equation:
dvC
+ 250vC − 600 = 0
dt
2. After the switch opens. Apply KCL at the top node (nodal analysis) to write the circuit equation.
 1
1 
vC − VS vC
V
⇒
+ iC = 0
+
+ vC + iC − S = 0

R1
RS
RS
 RS R1 
Next, use the definition of capacitor current to eliminate the variable iC from the nodal equation:
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 1
V
dv
1 
+ vC + C C − S = 0

RS
dt
 RS R1 
Substituting numerical values, we obtain the following differential equation:
dvC 500
+
vC − 600 = 0
3
dt
Problem 5.9
Write the differential equation for t > 0 for iL in
Figure P5.49.
Solution:
Known quantities:
Values of the voltage source, of the inductance and of the
resistors.
Find:
The differential equation for t > 0 (switch open) for the circuit of P5.49.
Analysis:
Apply KCL at the top node (nodal analysis) to write the circuit equation.
v1 − 100 v1
⇒
+ + iL = 0
0.3v1 + i L − 10 = 0
10
5
Note that the node voltage v1 is equal to:
v1 = 2.5i L + v L
Substitute the node voltage v1 in the first equation:
(1.75)i L + (0.3)v L − 10 = 0
Next, use the definition of inductor voltage to eliminate the variable v L from the nodal equation:
di
di L
⇒
(0.3)(0.1) L + (1.75)i L − 10 = 0
+ (58.33)i L − 333 = 0
dt
dt
Problem 5.10
Write the differential equations for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 and 𝑣𝑣1 in Figure P5.52. How are they related?
1 𝐻𝐻 and 𝐿𝐿2 = 5 𝐻𝐻.
Assume 𝐿𝐿1 =
Known quantities:
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𝐿𝐿1 = 1 𝐻𝐻, 𝐿𝐿2 = 5 𝐻𝐻
Find:
The differential equations for 𝑡𝑡 > 0 for 𝑖𝑖𝐿𝐿 and 𝑣𝑣1 .
Analysis:
At 𝑡𝑡 > 0, the switch is closed and the top 10 𝑘𝑘Ω resistor is effectively removed from the circuit.
immediately combine 𝐿𝐿1 and 𝐿𝐿2 in series to write the differential equation for 𝑖𝑖𝐿𝐿 :
𝐿𝐿𝑒𝑒𝑒𝑒 = 𝐿𝐿1 + 𝐿𝐿2
Substitute known values:
Simplify:
𝐿𝐿𝑒𝑒𝑒𝑒 = 1 𝐻𝐻 + 5 𝐻𝐻
𝐿𝐿𝑒𝑒𝑒𝑒 = 6 𝐻𝐻
Use KCL at the top node to get:
Use KVL on the top node to get:
5 𝐴𝐴 =
5 𝐴𝐴 =
Substitute definition of voltage across an inductor:
Simplify and note that 𝑖𝑖𝐿𝐿 𝑒𝑒𝑒𝑒 = 𝑖𝑖𝐿𝐿 :
𝑣𝑣𝑡𝑡𝑡𝑡𝑡𝑡
+ 𝑖𝑖𝐿𝐿
10 𝑘𝑘Ω
𝑣𝑣𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑣𝑣𝐿𝐿
Substitute into KCL equation:
Substitute known values:
Furthermore,
𝑣𝑣𝐿𝐿 𝑒𝑒𝑒𝑒
10 𝑘𝑘Ω
+ 𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
5 𝐴𝐴 =
𝐿𝐿𝑒𝑒𝑒𝑒 𝑑𝑑𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
+ 𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
10 𝑘𝑘Ω 𝑑𝑑𝑑𝑑
5 𝐴𝐴 =
6 𝐻𝐻 𝑑𝑑𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
+ 𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
10 𝑘𝑘Ω 𝑑𝑑𝑑𝑑
5 𝐴𝐴 − 0.0006
𝑑𝑑𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒
𝑑𝑑𝑑𝑑
+ 𝑖𝑖𝐿𝐿𝑒𝑒𝑒𝑒 = 0
For 𝑣𝑣3 , revert to the original circuit and convert it to a Thévenin equivalent:
𝑉𝑉𝑇𝑇ℎ = 5 𝐴𝐴 ∗ 10 𝑘𝑘Ω
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Simplify:
Use KVL to analyze the single-loop circuit:
𝑉𝑉𝑇𝑇ℎ = 50 𝑘𝑘𝑘𝑘
𝑉𝑉𝑇𝑇𝑇𝑇 − 𝑖𝑖𝐿𝐿 𝑅𝑅𝑇𝑇ℎ − 𝑣𝑣1 − 𝑣𝑣2 = 0
Substitute the equation for voltage across an inductor and note that 𝑖𝑖𝐿𝐿 1 = 𝑖𝑖𝐿𝐿 2 :
Combine like terms:
Substitute known values:
𝑉𝑉𝑇𝑇ℎ − 𝑖𝑖𝐿𝐿 𝑅𝑅𝑇𝑇ℎ − 𝐿𝐿1
𝑑𝑑𝑖𝑖𝐿𝐿
𝑑𝑑𝑖𝑖𝐿𝐿
− 𝐿𝐿2
=0
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑉𝑉𝑇𝑇𝑇𝑇 − 𝑖𝑖𝐿𝐿 𝑅𝑅𝑇𝑇ℎ − (𝐿𝐿1 + 𝐿𝐿2 )
𝑑𝑑𝑖𝑖𝐿𝐿
=0
𝑑𝑑𝑑𝑑
50 𝑘𝑘𝑘𝑘 − 10 𝑘𝑘Ω 𝑖𝑖𝐿𝐿 − (𝐿𝐿1 + 𝐿𝐿2 )
These equations are related by the relationship from KVL:
𝑑𝑑𝑖𝑖𝐿𝐿
=0
𝑑𝑑𝑑𝑑
𝑣𝑣𝑡𝑡𝑡𝑡𝑡𝑡 = 𝑣𝑣𝐿𝐿 𝑒𝑒𝑒𝑒 = 𝑣𝑣1 + 𝑣𝑣2
Substitute the equation for voltage across an inductor:
𝑑𝑑𝑖𝑖𝐿𝐿 𝑒𝑒𝑒𝑒
𝑑𝑑𝑑𝑑
= 𝑣𝑣1 + 𝑣𝑣2
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Section 5.3:
DC Steady State Solution of Circuits Containing
Inductors and Capacitors – Initial and Final Conditions
Problem 5.11
Determine the initial and final conditions on 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 in Figure P5.21
Known quantities:
None.
Find:
The initial and final conditions on 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 .
Analysis:
Before the switch is open, the circuit is in a steady state. Inductors behave as a short-circuit in steady state.
Therefore, 𝑅𝑅2 is directly connected to ground, and 𝑖𝑖𝐿𝐿 can be written, using current division, as the equation:
𝑖𝑖𝐿𝐿 (0− ) = 𝑖𝑖𝑆𝑆
Using KCL yields the equation:
Using Ohm’s law, this result can be written as:
𝑅𝑅1
𝑅𝑅1 + 𝑅𝑅2
𝑖𝑖𝑆𝑆 = 𝑖𝑖1 + 𝑖𝑖𝐿𝐿
𝑖𝑖𝑆𝑆 =
𝑉𝑉𝑆𝑆 𝑉𝑉𝑆𝑆
+
𝑅𝑅1 𝑅𝑅2
Substitute this result into the initial result to obtain the equation:
𝑉𝑉𝑆𝑆 𝑉𝑉𝑆𝑆
𝑅𝑅1
+ )
𝑅𝑅1 𝑅𝑅2 𝑅𝑅1 + 𝑅𝑅2
For 𝑣𝑣3 , recognize that the voltage drop across 𝑅𝑅3 is 0, due to the short-circuit.
𝑖𝑖𝐿𝐿 (0− ) = (
𝑣𝑣3 (0− ) = 0 𝑉𝑉
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As time goes to infinity, the source is not connected to the circuit. Therefore, 𝑖𝑖𝐿𝐿 and 𝑣𝑣3 can be written as the
equations:
𝑖𝑖𝐿𝐿 (∞) = 0 𝐴𝐴
Problem 5.12
𝑣𝑣3 (∞) = 0 𝑉𝑉
Determine the initial and final conditions on vC in
Figure P5.23.
Solution:
Known quantities:
Find:
The initial and final conditions for the circuit of P5.23.
Analysis:
Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the
capacitor as an open circuit. When the switch is open, the voltage source is not connected to the circuit. Thus,
vC (0) = 0 V.
After the switch has been closed for a long time, we have again a steady-state condition, and we treat the capacitor
as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R2 :
vC (∞ ) =
R2
V1 V.
R1 + R2
Problem 5.13
Determine the initial and final conditions on 𝑖𝑖𝐶𝐶 in Figure P5.27.
Known quantities:
None.
Find:
The initial and final conditions on 𝑖𝑖𝐶𝐶 .
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Analysis:
Before the switch is closed, there is no source connected to the circuit.
𝑖𝑖𝐶𝐶 (0− ) = 0
Therefore, 𝑖𝑖𝐶𝐶 can be written as follows:
After the switch is closed for a long time, the circuit is in steady-state. In this state, capacitors behave as opencircuits. Therefore, there is no current flowing through the capacitor. Again, 𝑖𝑖𝐶𝐶 can be written as the equation:
𝑖𝑖𝐶𝐶 (∞) = 0
Problem 5.14
Determine the initial and final conditions on 𝑖𝑖𝐿𝐿 in Figure P5.29.
Known quantities:
None.
Find:
The initial and final conditions on 𝑖𝑖𝐿𝐿 .
Analysis:
Initially, the switch is in the left position, and the circuit is in steady-state.
short-circuit. Using KVL at 𝑉𝑉𝑆𝑆1 , the following equation can be derived:
Isolate 𝑖𝑖𝐿𝐿 :
Solve for 𝑖𝑖𝐿𝐿 :
Therefore, the inductor behaves as a
𝑉𝑉𝑆𝑆1 − 𝑖𝑖𝐿𝐿 𝑅𝑅1 − 𝑖𝑖𝐿𝐿 𝑅𝑅2 − 𝑉𝑉𝑆𝑆2 = 0
𝑉𝑉𝑆𝑆1 − 𝑖𝑖𝐿𝐿 (𝑅𝑅1 − 𝑅𝑅2 ) − 𝑉𝑉𝑆𝑆2 = 0
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𝑖𝑖𝐿𝐿 (𝑅𝑅1 − 𝑅𝑅2 ) = 𝑉𝑉𝑆𝑆1 − 𝑉𝑉𝑆𝑆2
𝑖𝑖𝐿𝐿 (0− ) =
𝑉𝑉𝑆𝑆1 − 𝑉𝑉𝑆𝑆2
𝑅𝑅1 − 𝑅𝑅2
Finally, the switch is in the right position, and the circuit in in steady-state.
short-cirtcuit. Using KVL at 𝑉𝑉𝑆𝑆2 , the following equation can be derived:
Isolate 𝑖𝑖𝐿𝐿 :
Solve for 𝑖𝑖𝐿𝐿 :
Therefore, the inductor behaves as a
𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐿𝐿 𝑅𝑅2 − 𝑖𝑖𝐿𝐿 𝑅𝑅3 = 0
𝑉𝑉𝑆𝑆2 − 𝑖𝑖𝐿𝐿 (𝑅𝑅2 − 𝑅𝑅3 ) = 0
𝑖𝑖𝐿𝐿 (𝑅𝑅2 − 𝑅𝑅3 ) = 𝑉𝑉𝑆𝑆2
𝑖𝑖𝐿𝐿 (∞) =
𝑉𝑉𝑆𝑆2
𝑅𝑅2 − 𝑅𝑅3
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Problem 5.15
Determine the initial and final conditions on vC in
Figure P5.32.
Solution:
Known quantities:
Find:
The initial and final conditions for the circuit of P5.32.
Analysis:
Before the switch changes, the capacitor is not connected to the circuit, so we don’t have any information about its
initial voltage.
After the switch has changed, the current source and the capacitor will be in series so the current to the capacitor
will be constant at I 0 . Therefore, the rate at which charge accumulates on the capacitor will also be constant and,
consequently, the voltage across the capacitor will rise at a constant rate, without ever reaching an equilibrium state.
Problem 5.16
Determine the initial and final conditions on iC and
v3 in Figure P5.34.
Solution:
Known quantities:
Find:
The initial and final conditions for the circuit of P5.34.
Analysis:
In a steady-state condition we can treat the capacitor as an open circuit. Before the switch changes, applying the
KVL we have that the voltage across the capacitor is equal to the source voltage VS1:
vC (0) = VS1 = 17 V.
After the switch has changed for a long time, we have again a steady-state condition, and we treat the capacitor as an
open circuit. Since the current flowing through the resistance R2 is equal to zero, the voltage across the capacitor is
equal to the voltage across the resistance R3 :
vC (∞ ) =
R3
VS 2 V.
R1 + R3
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Problem 5.17
Determine the initial and final conditions on vC in
Figure P5.41.
Solution:
Known quantities:
VS = 20 V, R1 = 5 Ω,R2 = 4 Ω,R3 = 3Ω, R4 = 6 Ω,C 1 = 4 F ,C 2 = 4 F, I S = 4 A.
Find:
The initial and final conditions for the circuit of P5.41.
Analysis:
The switch S1 is always open and the switch S2 closes at
t = 0 . Before closing, the switch S2 has been opened for a
long time. Thus we have a steady-state condition, and we treat
the capacitors as open circuits. When the switch is open, the
current source is not connected to the circuit. Thus,
vC1 (0) = 0 V,
vC 2 (0) = 0 V.
After the switch S 2 has been closed for a long time, we have again a steady-state condition, and we treat the
capacitors as open circuits. The voltages across the capacitors are both equal to the voltage across the resistance R3 :
vC1(∞) = vC 2 (∞) = (R3 || R4 )I S =
6⋅3
4 = 8 V.
6+ 3
Problem 5.18
Determine the initial and final conditions on iC in
Figure P5.47. Assume VS = 9V, RS = 5kohm,
R1 = 10 kohm, and R2 = R3 = 20 kohm.
Solution:
Known quantities:
C = 1 µF , RS =15kΩ , R3 = 30 kΩ .
Find:
The initial and final conditions for the circuit of P5.47.
Assume:
Assume that VS = 9 V, R1 =10kΩ and R2 = 20kΩ .
Analysis:
Before opening, the switch has been closed for a long time. Thus, we have a steady-state condition, and we treat
the capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistances R1,
R2 , and R3 . Thus,
vC (0) =
R1 || (R2 || R3)
10k || 12k
VS =
9 = 2.4 V.
RS + R1 || (R2 || R3)
15k + 10k || 12k
After opening, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the
capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R1. Thus,
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vC (∞) =
R1
10000
VS =
9 = 3.6 V.
R1 + RS
10000 + 15000
Problem 5.19
Determine the initial and final conditions on iL in
Figure P5.49. Assume L1 = 1H and L2 = 5H.
Solution:
Known quantities:
Values of the voltage source, of the inductance and of the
resistors.
Find:
The initial and final conditions for the circuit of P5.49.
Analysis:
Before the switch changes, apply KCL at the top node (nodal
analysis) to write the following circuit equation.
v1 − 100 v1 v1
100
⇒
+ +
=0
v1 =
= 0.165 V.
1000
5 2.5
601
v1
40
=
= 66 mA.
2.5 601
After the switch has changed, apply KCL at the top node (nodal analysis) to write the following circuit equation.
v1 − 100 v1 v1
100
⇒
+ +
=0
v1 =
= 14.285 V.
10
5 2.5
7
i L (0) =
i L (∞) =
v1 40
=
= 5.714 A.
2.5 7
Problem 5.20
Determine the initial and final conditions on iL and
v1 in Figure P5.52.
Solution:
Known quantities:
Find:
The initial and final conditions for the circuit of P5.52.
Analysis:
Before closing, the switch has been opened for a long time. Thus we have a steady-state condition, and we treat the
inductors as short circuits. The values of the two resistors are equal so the current flowing through the inductors is:
i L (0) =
IS
2
After the switch has been closed for a long time, we have again a steady-state condition, and we treat the inductors
as short circuits. In this case the resistors are short-circuited and so all the current is flowing through the inductors.
i L (∞) = I S
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Section 5.4:
Transient Response of First Order Circuits
Focus on Methodology
First-order transient response
1.
2.
3.
4.
5.
Solve for the steady-state response of the circuit before the switch changes state (t = 0-),
and after the transient has died out (t → ∞). We shall generally refer to these responses
as x(0-) and x(∞).
Identify the initial condition for the circuit, x(0+), using continuity of capacitor voltages
and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), as illustrated in Section 5.4.
Write the differential equation of the circuit for t = 0+, that is, immediately after the switch
has changed position. The variable x(t) in the differential equation will be either a
capacitor voltage, vC(t), or an inductor current, iL(t). It is helpful at this time to reduce
the circuit to Thévenin or Norton equivalent form, with the energy storage element
(capacitor or inductor) treated as the load for the Thévenin (Norton) equivalent circuit.
Reduce this equation to standard form (Equation 5.8).
Solve for the time constant of the circuit: τ = RTC for capacitive circuits, τ = L/RT for
inductive circuits.
Write the complete solution for the circuit in the form:
x(t) = x(∞) + (x(0) − x(∞))e−t /τ
Problem 5.21
At 𝑡𝑡 = 0− , just before the switch is opened, the current through the inductor in Figure P5.21 is 𝑖𝑖𝐿𝐿 = 140 𝑚𝑚𝑚𝑚. Is
this value the same as that for DC steady-state? Was the circuit in steady-state just before the switch was opened?
Assume 𝑉𝑉𝑆𝑆 = 10 𝑉𝑉, 𝑅𝑅1 = 1 𝑘𝑘Ω, 𝑅𝑅2 = 5 𝑘𝑘Ω, 𝑅𝑅3 = 2 𝑘𝑘Ω, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 = 1 𝑚𝑚𝑚𝑚.
Known quantities:
𝑉𝑉𝑆𝑆 = 10 𝑉𝑉, 𝑅𝑅1 = 1 𝑘𝑘Ω, 𝑅𝑅2 = 5 𝑘𝑘Ω, 𝑅𝑅3 = 2 𝑘𝑘Ω, 𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 = 1 𝑚𝑚𝑚𝑚
Find:
Determine if the DC steady state value is the same as the current in the inductor at 𝑡𝑡 = 0− .
Analysis:
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Assume the circuit is in DC steady-state. The inductor behaves like a short-circuit, connecting 𝑅𝑅2 to ground and
effectively eliminating 𝑅𝑅3 from the circuit. Therefore, 𝑖𝑖𝐿𝐿 is the same as the current through 𝑅𝑅2 . Using Ohm’s
law, the current through 𝑅𝑅2 , 𝑖𝑖2 , can be written as:
𝑖𝑖2 =
Plug in known values:
This step yields 𝑖𝑖2 = 𝑖𝑖𝐿𝐿 = 2 𝑚𝑚𝑚𝑚.
before the switch was opened.
𝑖𝑖2 =
𝑉𝑉𝑆𝑆
𝑅𝑅2
10 𝑉𝑉
5 𝑘𝑘Ω
This value is not 140 𝑚𝑚𝑚𝑚, so the circuit was not in DC steady-state just
Problem 5.22
For t < 0, the circuit shown in Figure P5.22 is at
DC steady-state. The switch is thrown at t = 0.
VS1 = 35V VS2 = 130V
C = 11μF R1 = 17 kohm
R2 = 7k_ R3 = 23 kohm
Determine the initial current through R3 just after the
switch is thrown at t =0+.
Solution:
Known quantities:
Circuit shown in Figure P5.22, VS1 = 35V ,VS 2 = 130V ,C = 11µF , R1 = 17k Ω,R2 = 7kΩ,R3 = 23kΩ.
Find:
At t = 0+ the initial current through R3 just after the
switch is changed.
Assumptions:
None.
Analysis:
To solve this problem, find the steady state voltage across the
capacitor before the switch is thrown. Since the voltage
across a capacitor cannot change instantaneously, this voltage
will also be the capacitor voltage immediately after the switch is thrown. At that instant, the capacitor may be
viewed as a DC voltage source.
At t = 0− :
Determine the voltage across the capacitor. At steady state, the capacitor is modeled as an open circuit:
() ()
iR1 0 − = iR 2 0 − = 0
Apply KVL:
()
VS1 + 0 − VC 0 − + 0 − VS 2 = 0
( )= V
VC 0
+
−
S1 − VS 2
= −95V
At t = 0 :
() ()
(0 )= i (0 )
VC 0 + = VC 0 − = −95V
iR 2 +
R3
Apply KVL:
+
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()
() ()
(0 )= 130 − 95
VS 2 − iR3 0 + R2 + VC 0 + − iR3 0 + R3 = 0
( ) V R+V+ R
iR 3 0 + =
S2
C
2
3
+
7 × 103 + 23 × 103
= 1.167mA
Problem 5.23
Determine the current through the capacitor, 𝑖𝑖𝐶𝐶 , just before and just after the switch is closed in Figure P5.23.
Assume steady-state conditions for 𝑡𝑡 < 0 and 𝑉𝑉1 = 15 𝑉𝑉, 𝑅𝑅1 = 0.5 𝑘𝑘Ω, 𝑅𝑅2 = 2 𝑘𝑘Ω, and 𝐶𝐶 = 0.4 𝜇𝜇𝜇𝜇.
Known quantities:
𝑡𝑡 < 0, 𝑉𝑉1 = 15 𝑉𝑉, 𝑅𝑅1 = 0.5 𝑘𝑘Ω, 𝑅𝑅2 = 2 𝑘𝑘Ω, and 𝐶𝐶 = 0.4 𝜇𝜇𝜇𝜇.
Find:
The current through the capacitor, 𝑖𝑖𝐶𝐶 , just before and just after the switch is closed in Figure P5.23.
Analysis:
At 𝑡𝑡 = 0− , the source is not connected to the circuit.
Therefore, there is no current through the capacitor:
𝑖𝑖𝐶𝐶 (0− ) = 0 𝐴𝐴
At 𝑡𝑡 = 0+ , the source has just been connected to the circuit, so the capacitor may be viewed as a short-circuit.
This property eliminates 𝑅𝑅2 from the circuit, leaving only the voltage source and 𝑅𝑅1 . Therefore, 𝑖𝑖𝐶𝐶 is identical
to the current through 𝑅𝑅1 , 𝑖𝑖1 . Using Ohm’s law provides the equation:
Substitute 𝑖𝑖𝐶𝐶 for 𝑖𝑖1 :
𝑖𝑖1 (0+ ) =
𝑉𝑉1
𝑅𝑅1
𝑖𝑖𝐶𝐶 (0+ ) =
𝑉𝑉1
𝑅𝑅1
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Substitute known values:
This equation yields 𝑖𝑖𝐶𝐶 (0+ ) = 30 𝑚𝑚𝑚𝑚.
𝑖𝑖𝐶𝐶 (0+ ) =
15 𝑉𝑉
0.5 𝑘𝑘Ω
Problem 5.24
Determine the current through the capacitor, 𝑖𝑖𝐶𝐶 , just before and just after the switch is closed in Figure P5.23.
Assume steady-state conditions for 𝑡𝑡 < 0 and 𝑉𝑉1 = 10 𝑉𝑉, 𝑅𝑅1 = 200 𝑚𝑚Ω, 𝑅𝑅2 = 5 𝑘𝑘Ω, and 𝐶𝐶 = 300 𝜇𝜇𝜇𝜇.
Known quantities:
𝑡𝑡 < 0, 𝑉𝑉1 = 10 𝑉𝑉, 𝑅𝑅1 = 200 𝑚𝑚Ω, 𝑅𝑅2 = 5 𝑘𝑘Ω, and 𝐶𝐶 = 300 𝜇𝜇𝜇𝜇.
Find:
The current through the capacitor, 𝑖𝑖𝐶𝐶 , just before and just after the switch is closed in Figure P5.23.
Analysis:
At 𝑡𝑡 = 0− , the source is not connected to the circuit.
Therefore, there is no current through the capacitor:
𝑖𝑖𝐶𝐶 (0− ) = 0 𝐴𝐴
At 𝑡𝑡 = 0+ , the source has just been connected to the circuit, so the capacitor may be viewed as a short-circuit.
This property eliminates 𝑅𝑅2 from the circuit, leaving only the voltage source and 𝑅𝑅1 . Therefore, 𝑖𝑖𝐶𝐶 is identical
to the current through 𝑅𝑅1 , 𝑖𝑖1 . Using Ohm’s law provides the equation:
Substitute 𝑖𝑖𝐶𝐶 for 𝑖𝑖1 :
𝑖𝑖1 (0+ ) =
𝑉𝑉1
𝑅𝑅1
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Substitute known values:
This equation yields 𝑖𝑖𝐶𝐶 (0+ ) = 50 𝐴𝐴.
𝑖𝑖𝐶𝐶 (0+ ) =
𝑖𝑖𝐶𝐶 (0+ ) =
𝑉𝑉1
𝑅𝑅1
10 𝑉𝑉
200 𝑚𝑚Ω
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INSTANTANEOUSLY.
Problem 5.25
Just before the switch is opened at 𝑡𝑡 = 0 in Figure P5.21, the current through the inductor is 𝑖𝑖𝐿𝐿 = 1.5 𝑚𝑚𝑚𝑚.
Determine the voltage 𝑣𝑣3 across 𝑅𝑅3 immediately after the switch is opened. Assume: 𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 𝑅𝑅2 =
6 𝑘𝑘Ω, 𝑅𝑅3 = 3 𝑘𝑘Ω, and 𝐿𝐿 = 0.9 𝑚𝑚𝑚𝑚.
Known quantities:
𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 𝑅𝑅2 = 6 𝑘𝑘Ω, 𝑅𝑅3 = 3 𝑘𝑘Ω, 𝐿𝐿 = 0.9 𝑚𝑚𝑚𝑚
Find:
The voltage across 𝑅𝑅3 , 𝑣𝑣3 .
Analysis:
Immediately after the switch is flipped, the current through 𝐿𝐿 remains the same.
1.5 𝑚𝑚𝑚𝑚 current source in place of L. Combine 𝑅𝑅2 and 𝑅𝑅1 in series:
Substitute known values:
Simplify:
Therefore, view the circuit with a
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅2
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 6 𝑘𝑘Ω + 6 𝑘𝑘Ω
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 12 𝑘𝑘Ω
Now, determine the current through 𝑅𝑅3 , 𝑖𝑖3 as the current division:
Substitute known values:
𝑖𝑖3 = 1.5 𝑚𝑚𝑚𝑚 ∗
𝑖𝑖3 = 1.5 𝑚𝑚𝑚𝑚 ∗
5.26
𝑅𝑅1,2,𝑒𝑒𝑒𝑒
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 + 𝑅𝑅3
12 𝑘𝑘Ω
12 𝑘𝑘Ω + 3 𝑘𝑘Ω
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Simplify:
𝑖𝑖3 = 1.2 𝑚𝑚𝑚𝑚
Use Ohm’s law to realize and note that, according to the problem statement, the polarity across 𝑅𝑅3 is reversed:
Substitute known values:
Simplify:
𝑣𝑣3 = −𝑖𝑖3 𝑅𝑅3
𝑣𝑣3 = −1.2 𝑚𝑚𝑚𝑚 ∗ 3 𝑘𝑘Ω
𝑣𝑣3 = −3.6 𝑉𝑉
Problem 5.26
Assume that steady-state conditions exist in the circuit shown in Figure P5.26 for 𝑡𝑡 < 0. Determine the current
through the inductor, 𝑖𝑖𝐿𝐿 , immediately before and immediately after the switch, that is, at 𝑡𝑡 = 0− and at 𝑡𝑡 = 0+ .
Assume 𝐿𝐿 = 0.5 𝐻𝐻, 𝑅𝑅1 = 100 𝑘𝑘Ω, 𝑅𝑅𝑆𝑆 = 5 Ω, and 𝑉𝑉𝑆𝑆 = 24 𝑉𝑉.
Known quantities:
𝐿𝐿 = 0.5 𝐻𝐻, 𝑅𝑅1 = 100 𝑘𝑘Ω, 𝑅𝑅𝑆𝑆 = 5 Ω, and 𝑉𝑉𝑆𝑆 = 24 𝑉𝑉
Find:
The current through the inductor, 𝑖𝑖𝐿𝐿 , immediately before and immediately after the switch is thrown.
Analysis:
At 𝑡𝑡 = 0− , the circuit is in steady-state, and the inductor behaves like a short-circuit.
to the current through 𝑅𝑅𝑆𝑆 , 𝑖𝑖𝑆𝑆 . Using Ohm’s law provides the equation:
𝑖𝑖𝑆𝑆 (0− ) =
Therefore, 𝑖𝑖𝐿𝐿 is identical
𝑉𝑉𝑆𝑆
𝑅𝑅𝑆𝑆
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Substitute known values:
This equation yields 𝑖𝑖𝑆𝑆 (0− ) = 𝑖𝑖𝐿𝐿 (0− ) = 4.8 𝐴𝐴.
𝑖𝑖𝑆𝑆 (0− ) =
24 𝑉𝑉
5 Ω
At 𝑡𝑡 = 0+ , the switch has just been thrown, and the inductor behaves like a current source. The discharge of the
inductor is represented by the following equation:
𝑅𝑅1
𝑖𝑖𝐿𝐿 (𝑡𝑡) = 𝐼𝐼𝑒𝑒 − 𝐿𝐿 𝑡𝑡
where 𝐼𝐼 is the current through the inductor before discharging began. In this circuit, 𝐼𝐼 = 𝑖𝑖𝐿𝐿 (0− ), and 𝑡𝑡 = 0+ .
Therefore, this equation becomes:
𝑅𝑅1
Substitute known values:
𝑖𝑖𝐿𝐿 (0+ ) = 𝑖𝑖𝐿𝐿 (0− )𝑒𝑒 − 𝐿𝐿 𝑡𝑡
100 𝑘𝑘Ω
0
0.5 H
𝑖𝑖𝐿𝐿 (0+ ) = 4.8 𝐴𝐴 ∗ 𝑒𝑒 −
Evaluating this equation numerically yields 𝑖𝑖𝐿𝐿 (0+ ) = 4.8 𝐴𝐴.
Problem 5.27
Assume that steady-state conditions exist in the circuit shown in Figure P5.27 for 𝑡𝑡 < 0 and that 𝑉𝑉1 = 15𝑉𝑉, 𝑅𝑅1 =
100 Ω, 𝑅𝑅2 = 1.2 𝑘𝑘Ω, 𝑅𝑅3 = 400 Ω, and 𝐶𝐶 = 4.0 𝜇𝜇𝜇𝜇. Determine the current through the capacitor, 𝑖𝑖𝐶𝐶 , at 𝑡𝑡 =
0+ , just after the switch is closed.
Known quantities:
𝑉𝑉1 = 15 𝑉𝑉, 𝑅𝑅1 = 100 Ω, 𝑅𝑅2 = 1.2 𝑘𝑘Ω, 𝑅𝑅3 = 400 Ω, and 𝐶𝐶 = 4.0 𝜇𝜇𝜇𝜇
Find:
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The current through the capacitor, 𝑖𝑖𝐶𝐶 , at 𝑡𝑡 = 0+ .
Analysis:
At 𝑡𝑡 = 0+ , the voltage source has just been connected to the circuit, so the capacitor behaves like a short-circuit.
Therefore, 𝑖𝑖𝐶𝐶 (0+ ) is the same as the current through 𝑅𝑅2 , 𝑖𝑖2 (0+ ). To solve this problem, find the current going
through 𝑅𝑅1 and determine what portion of that current is going through 𝑅𝑅2 . From this point onward, all current
notation assumes 𝑡𝑡 = 0+ .
To find 𝑖𝑖1 , determine an equivalent resistance and find 𝑖𝑖𝑒𝑒𝑒𝑒 .
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 =
Substitute known values:
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 =
Combine 𝑅𝑅2 and 𝑅𝑅3 in parallel:
𝑅𝑅2 𝑅𝑅3
𝑅𝑅2 + 𝑅𝑅3
1.2 𝑘𝑘Ω ∗ 400 Ω
1.2 𝑘𝑘Ω + 400 Ω
Evaluating this equation numerically yields: 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 300 Ω.
Combine 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 and 𝑅𝑅1 in series:
Substitute known values:
𝑅𝑅1,2,3,𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 + 𝑅𝑅1
𝑅𝑅𝑒𝑒𝑒𝑒 = 300 Ω + 100 Ω
Evaluating this equation numerically yields 𝑅𝑅𝑒𝑒𝑒𝑒 = 400 Ω.
Substitute known values:
𝑖𝑖𝑒𝑒𝑒𝑒 =
𝑖𝑖𝑒𝑒𝑒𝑒 =
Use Ohm’s law to determine 𝑖𝑖𝑒𝑒𝑒𝑒 :
𝑉𝑉1
𝑅𝑅𝑒𝑒𝑒𝑒
15 𝑉𝑉
400 Ω
Evaluating this equation numerically provides: 𝑖𝑖𝑒𝑒𝑒𝑒 = 37.5 𝑚𝑚𝑚𝑚.
Using KCL shows the following relationship:
Use current division to determine 𝑖𝑖2 :
Substitute known quantities:
𝑖𝑖𝑒𝑒𝑒𝑒 = 𝑖𝑖1 = 𝑖𝑖2 + 𝑖𝑖3
𝑖𝑖2 = 𝑖𝑖1
𝑅𝑅3
𝑅𝑅2 + 𝑅𝑅3
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𝑖𝑖2 = 37.5 𝑚𝑚𝑚𝑚 ∗
400 Ω
1.2 𝑘𝑘Ω + 400 Ω
Evaluating this expressions provides: 𝑖𝑖2 = 𝑖𝑖𝐶𝐶 = 9.38 𝑚𝑚𝑚𝑚.
Problem 5.28
Assume that steady-state conditions exist in the circuit shown in Figure p5.28 at 𝑡𝑡 < 0. Also, assume: 𝑉𝑉1 =
12 𝑉𝑉, 𝑉𝑉2 = 5 𝑉𝑉, 𝐿𝐿 = 3 𝐻𝐻, 𝑅𝑅1 = 𝑅𝑅2 = 2 Ω, 𝑅𝑅3 = 4 Ω, 𝑅𝑅3 = 29 𝑘𝑘Ω. Find the Norton equivalent network seen
by the inductor. Use it to determine the time constant of the circuit for 𝑡𝑡 > 0.
Known quantities:
𝑉𝑉1 = 12 𝑉𝑉, 𝑉𝑉2 = 5 𝑉𝑉, 𝐿𝐿 = 3 𝐻𝐻, 𝑅𝑅1 = 𝑅𝑅2 = 2 Ω, 𝑅𝑅3 = 4 Ω, 𝑅𝑅3 = 29 𝑘𝑘Ω
Find:
The Norton equivalent network seen by the inductor and the time constant of the circuit for 𝑡𝑡 > 0.
Analysis:
To find the Norton equivalent, find 𝑅𝑅𝑒𝑒𝑒𝑒 from the inductor’s perspective.
Remove the inductor and view each voltage source as a short. Furthermore, 𝑡𝑡 > 0, so 𝑉𝑉2 is not connected to the
inductor. Therefore, the inductor sees 𝑅𝑅1 and 𝑅𝑅2 in series.
Combine 𝑅𝑅1 and 𝑅𝑅2 in series:
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅2
𝑅𝑅𝑒𝑒𝑒𝑒 = 2 Ω + 2 Ω
Calculate the Norton equivalent current source, 𝐼𝐼𝑁𝑁 .
𝑅𝑅𝑒𝑒𝑒𝑒 = 4 Ω
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Substitute known values:
Simplify:
𝐼𝐼𝑁𝑁 =
𝐼𝐼𝑁𝑁 =
𝑉𝑉1
𝑅𝑅𝑒𝑒𝑒𝑒
12 𝑉𝑉
4 Ω
𝐼𝐼𝑁𝑁 = 3 𝐴𝐴
The Norton equivalent circuit has 𝐼𝐼𝑁𝑁 = 3 𝐴𝐴 and 𝑅𝑅𝑁𝑁 = 4 Ω.
Now, the time constant must be calculated. Use 𝑅𝑅𝑒𝑒𝑒𝑒 and 𝐿𝐿 to calculate this value:
Substitute known values:
Simplify:
Problem 5.29
𝜏𝜏 =
𝐿𝐿
𝑅𝑅𝑒𝑒𝑒𝑒
𝜏𝜏 =
3 𝐻𝐻
4 Ω
𝜏𝜏 = 0.75 𝑠𝑠
Assume that steady-state conditions exist in the circuit shown in Figure P5.29 at 𝑡𝑡 < 0. Also, assume: 𝑉𝑉𝑆𝑆1 =
9 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 12 𝑉𝑉, 𝐿𝐿 = 120 𝑚𝑚𝑚𝑚, 𝑅𝑅1 = 2.2 Ω, R 2 = 4.7 Ω, 𝑅𝑅3 = 18 𝑘𝑘Ω. Find the Norton equivalent network
seen by the inductor. Use it to determine the time constant of the circuit for 𝑡𝑡 > 0.
Known quantities:
𝑉𝑉𝑆𝑆1 = 9 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 12 𝑉𝑉, 𝐿𝐿 = 120 𝑚𝑚𝑚𝑚, 𝑅𝑅1 = 2.2 Ω, R 2 = 4.7 Ω, 𝑅𝑅3 = 18 𝑘𝑘Ω.
5.31
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Find:
The Norton equivalent network seen by the inductor and the time constant of the circuit for 𝑡𝑡 > 0.
Analysis:
To find the Norton equivalent, find 𝑅𝑅𝑒𝑒𝑒𝑒 from the inductor’s perspective.
Remove the inductor and view each voltage source as a short. Furthermore, 𝑡𝑡 > 0, so 𝑉𝑉𝑆𝑆1 is not connected to the
inductor. Therefore, the inductor sees 𝑅𝑅2 and 𝑅𝑅3 in series.
Combine 𝑅𝑅2 and 𝑅𝑅3 in series:
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
𝑅𝑅𝑒𝑒𝑒𝑒 = 4.7 Ω + 18 𝑘𝑘Ω
𝑅𝑅𝑒𝑒𝑒𝑒 ≈ 18 𝑘𝑘Ω
Calculate the Norton equivalent current source, 𝐼𝐼𝑁𝑁 .
Substitute known values:
Simplify:
𝐼𝐼𝑁𝑁 =
𝐼𝐼𝑁𝑁 =
𝑉𝑉𝑆𝑆2
𝑅𝑅𝑒𝑒𝑒𝑒
12 𝑉𝑉
18 𝑘𝑘Ω
𝐼𝐼𝑁𝑁 = 0.66 𝑚𝑚𝑚𝑚
The Norton equivalent circuit has 𝐼𝐼𝑁𝑁 = 0.66 𝑚𝑚𝑚𝑚 and 𝑅𝑅𝑁𝑁 = 18 𝑘𝑘Ω.
Now, the time constant must be calculated. Use 𝑅𝑅𝑒𝑒𝑒𝑒 and 𝐿𝐿 to calculate this value:
Substitute known values:
Simplify:
𝜏𝜏 =
𝜏𝜏 =
𝐿𝐿
𝑅𝑅𝑒𝑒𝑒𝑒
120 𝑚𝑚𝑚𝑚
18 𝑘𝑘Ω
𝜏𝜏 = 6.67 𝜇𝜇𝜇𝜇
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Problem 5.30
Find the Thévenin equivalent network seen by the capacitor in Figure P5.30 for 𝑡𝑡 > 0. Use it to determine the
time constant of the circuit for 𝑡𝑡 > 0. Assume: 𝑅𝑅1 = 3 Ω, 𝑅𝑅2 = 1 Ω, 𝑅𝑅3 = 4 Ω, 𝐶𝐶 = 0.2 𝐹𝐹, 𝐼𝐼𝑆𝑆 = 3 𝐴𝐴, 𝑉𝑉𝐶𝐶 (0) =
0.
Known quantities:
𝑅𝑅1 = 3 Ω, 𝑅𝑅2 = 1 Ω, 𝑅𝑅3 = 4 Ω, 𝐶𝐶 = 0.2 𝐹𝐹, 𝐼𝐼𝑆𝑆 = 3 𝐴𝐴, 𝑉𝑉𝐶𝐶 (0) = 0
Find:
The Thévenin equivalent network seen by the capacitor in Figure P5.30 for 𝑡𝑡 > 0 and the time constant of the
circuit for 𝑡𝑡 > 0.
Analysis:
To determine the Thévenin equivalent resistant, 𝑅𝑅𝑇𝑇ℎ , remove 𝐶𝐶 and make 𝐼𝐼𝑆𝑆 an open-circuit. Then, combine 𝑅𝑅1
and 𝑅𝑅2 in series:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅2
Substitute known values:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 3 Ω + 1 Ω
Simplify:
Combine 𝑅𝑅1,2,𝑒𝑒𝑒𝑒 and 𝑅𝑅3 in parallel as 𝑅𝑅𝑇𝑇ℎ :
Substitute known values:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 = 4 Ω
𝑅𝑅1,2,3,𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑇𝑇ℎ =
𝑅𝑅𝑇𝑇ℎ =
5.33
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 𝑅𝑅3
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 + 𝑅𝑅3
4 Ω∗4 Ω
4 Ω+4 Ω
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Simplify:
𝑅𝑅𝑇𝑇ℎ = 2 Ω
Calculate the Thévenin equivalent source, 𝑉𝑉𝑇𝑇ℎ .
𝑉𝑉𝑇𝑇ℎ = 𝐼𝐼𝑆𝑆 𝑅𝑅𝑇𝑇ℎ
Substitute known values:
Simplify:
𝑉𝑉𝑇𝑇ℎ = 3 𝐴𝐴 ∗ 2 Ω
𝑉𝑉𝑇𝑇ℎ = 6 𝑉𝑉
Therefore, the Thévenin equivalent circuit has 𝑉𝑉𝑇𝑇ℎ = 6 𝑉𝑉 and 𝑅𝑅𝑇𝑇ℎ = 2 Ω.
Now, find the time constant. Calculate the time constant, 𝜏𝜏, for 𝑡𝑡 > 0 using 𝑅𝑅𝑇𝑇ℎ :
𝜏𝜏 = 𝑅𝑅𝑇𝑇ℎ ∗ 𝐶𝐶
Substitute known values:
Simplify:
𝜏𝜏 = 2 Ω ∗ 0.2 𝐹𝐹
Problem 5.31
𝜏𝜏 = 0.4 𝑠𝑠
The switch shown in Figure P5.31 is closed at 𝑡𝑡 = 0. Find the Thévenin equivalent network seen by the capacitor
for 𝑡𝑡 > 0, and use it to determine the time constant of the circuit for 𝑡𝑡 > 0. Assume: 𝑅𝑅𝑆𝑆 = 8 𝑘𝑘Ω, 𝑉𝑉𝑆𝑆 =
40 𝑉𝑉, 𝐶𝐶 = 350 𝜇𝜇𝜇𝜇, and 𝑅𝑅 = 24 𝑘𝑘Ω.
Known quantities:
𝑅𝑅𝑆𝑆 = 8 𝑘𝑘Ω, 𝑉𝑉𝑆𝑆 = 40 𝑉𝑉, 𝐶𝐶 = 350 𝜇𝜇𝜇𝜇, 𝑅𝑅 = 24 𝑘𝑘Ω
Find:
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
The Thévenin equivalent network seen by the capacitor for 𝑡𝑡 > 0 and the time constant of the circuit for 𝑡𝑡 > 0.
Analysis:
To find the Thévenin equivalent network, find 𝑅𝑅𝑇𝑇ℎ and 𝑉𝑉𝑇𝑇ℎ . Beginning with 𝑅𝑅𝑇𝑇ℎ , remove the capacitor and turn
𝑉𝑉𝑆𝑆 in a short-circuit. Then, combine 𝑅𝑅𝑆𝑆 and 𝑅𝑅 in parallel, which accounts for all the resistors. Therefore,
𝑅𝑅𝑆𝑆,𝑅𝑅,𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑇𝑇ℎ .
Substitute known values:
Simplify:
𝑅𝑅𝑇𝑇ℎ =
𝑅𝑅𝑇𝑇ℎ =
𝑅𝑅𝑆𝑆 𝑅𝑅
𝑅𝑅𝑆𝑆 + 𝑅𝑅
8 𝑘𝑘Ω ∗ 24 𝑘𝑘Ω
8 𝑘𝑘Ω + 24 𝑘𝑘Ω
𝑅𝑅𝑇𝑇ℎ = 6 𝑘𝑘Ω
In this circuit, there exists a single voltage source, 𝑉𝑉𝑆𝑆 .
Therefore, the Thévenin equivalent source is simply:
𝑉𝑉𝑇𝑇ℎ = 𝑉𝑉𝑆𝑆 = 40 𝑉𝑉
The equivalent Thévenin network can be described by: 𝑉𝑉𝑇𝑇ℎ = 40 𝑉𝑉 and 𝑅𝑅𝑇𝑇ℎ = 6 𝑘𝑘Ω.
Now, use 𝑅𝑅𝑇𝑇ℎ to find the time constant. This circuit is an RC circuit.
written as:
Substitute known values:
Simplify:
Problem 5.32
Therefore, the time constant, 𝜏𝜏,
can be
𝜏𝜏 = 𝑅𝑅𝑇𝑇ℎ 𝐶𝐶
𝜏𝜏 = 6 𝑘𝑘Ω ∗ 350 𝜇𝜇𝜇𝜇
𝜏𝜏 = 2.1 𝑠𝑠
Determine the voltage vC across the capacitor
shown in Figure P5.32 for t > 0. The voltage across
the capacitor just before the switch is thrown is
vC (0−) = −7V. Assume:
Io = 17mA C = 0.55μF
R1 = 7 kohm R2 = 3.3kohm
Solution:
Known quantities:
Circuit shown in Figure P5.32, Vc (0 − ) = −7V ,I 0 = 17mA,C = 0.55µF , R1 = 7k Ω,R2 = 3.3kΩ.
Find:
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The voltage Vc (t ) across the capacitor for t > 0.
Assumptions:
Before the switch is thrown the voltage across the capacitor is –7 V.
Analysis:
The current source and the capacitor will be in series so the current to the capacitor will be constant at I0.
Therefore, the rate at which charge accumulates on the capacitor will also be constant and, consequently, the voltage
across the capacitor will rise at a constant rate. The integral form of the capacitor i-V relationship best expresses
this accumulation process. The continuity of the voltage across the capacitor requires:
( )
( )
VC 0+ = VC 0 − = −7V
iC (t ) = I 0 = 17mA
1 t
1 0
t
∫ iC (t )dt = ( ∫− ∞ iC (t )dt + ∫0 I 0 dt )
C −∞
C
I
I
=VC 0+ + 0 ∫0t dt =VC 0+ + 0 t |t0
C
C
VC (t ) =
( )
( )
= −7 +
17 × 10 − 3
0.55 × 10
−6
t = −7 + 30.91 × 103 t
Problem 5.33
For 𝑡𝑡 < 0, the circuit shown in Figure P5.29 is at steady state. The switch is thrown at 𝑡𝑡 = 0. Determine the
current 𝑖𝑖𝐿𝐿 through the inductor for 𝑡𝑡 > 0. Assume: 𝑉𝑉𝑆𝑆1 = 9 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 12 𝑉𝑉, 𝑅𝑅1 = 2.2 Ω, 𝑅𝑅2 = 4.7 Ω, 𝑅𝑅3 =
18 𝑘𝑘Ω, and 𝐿𝐿 = 120 𝑚𝑚𝑚𝑚.
Known quantities:
𝑉𝑉𝑆𝑆1 = 9 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 12 𝑉𝑉, 𝑅𝑅1 = 2.2 Ω, 𝑅𝑅2 = 4.7 Ω, 𝑅𝑅3 = 18 𝑘𝑘Ω, 𝐿𝐿 = 120 𝑚𝑚𝑚𝑚
Find:
The current through the inductor, 𝑖𝑖𝐿𝐿 , for 𝑡𝑡 > 0.
Analysis:
To determine 𝑖𝑖𝐿𝐿 (𝑡𝑡) for 𝑡𝑡 > 0, use the full equation for the response of current through an inductor:
𝑖𝑖𝐿𝐿 (𝑡𝑡 > 0) = 𝑖𝑖𝐿𝐿 (∞) + [𝑖𝑖𝐿𝐿 (0) − 𝑖𝑖𝐿𝐿 (∞)]𝑒𝑒 −𝑡𝑡/𝜏𝜏
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To utilize this equation, three values are required: 𝑖𝑖𝐿𝐿 (∞), 𝑖𝑖𝐿𝐿 (0), and 𝜏𝜏. To calculate 𝑖𝑖𝐿𝐿 (0), recognize that 𝐿𝐿 is
in steady-state (i.e., short-circuit) when 𝑡𝑡 < 0. Therefore, the 𝑖𝑖𝐿𝐿 (0) is the same as the current through 𝑅𝑅1 and
𝑅𝑅2 . Note that the voltage drop across 𝑅𝑅1 and 𝑅𝑅2 is the difference between 𝑉𝑉𝑆𝑆1 and 𝑉𝑉𝑆𝑆2 :
Substitute known values:
Simplify:
𝑖𝑖𝐿𝐿 (0) =
𝑖𝑖𝐿𝐿 (0) =
𝑉𝑉𝑆𝑆1 − 𝑉𝑉𝑆𝑆2
𝑅𝑅1 + 𝑅𝑅2
9 𝑉𝑉 − 12 𝑉𝑉
2.2 Ω + 4.7 Ω
𝑖𝑖𝐿𝐿 (0) = −0.43 𝐴𝐴
To determine 𝑖𝑖𝐿𝐿 (∞), recognize, again, that 𝐿𝐿 is in steady-state (i.e., short-circuit) as 𝑡𝑡 → ∞.
is 𝑉𝑉𝑆𝑆2 over the addition of 𝑅𝑅2 and 𝑅𝑅3 :
Substitute known values:
𝑖𝑖𝐿𝐿 (∞) =
𝑖𝑖𝐿𝐿 (∞) =
Therefore, 𝑖𝑖𝐿𝐿 (∞)
𝑉𝑉𝑆𝑆2
𝑅𝑅2 + 𝑅𝑅3
12 𝑉𝑉
4.7 Ω + 18 𝑘𝑘Ω
Simplify:
𝑖𝑖𝐿𝐿 (∞) = 0.67 𝑚𝑚𝑚𝑚
For the time constant, calculate the equivalent resistance seen by 𝐿𝐿. To accomplish this goal, remove 𝐿𝐿 from the
circuit and turn 𝑉𝑉𝑆𝑆2 into a short-circuit. Then, 𝐿𝐿 sees 𝑅𝑅2 in series with 𝑅𝑅3 :
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
𝑅𝑅𝑒𝑒𝑒𝑒 = 4.7 Ω + 18 𝑘𝑘Ω
Calculate 𝜏𝜏 as the time constant for an RL circuit:
Substitute known values:
𝑅𝑅𝑒𝑒𝑒𝑒 ≈ 18 𝑘𝑘Ω
𝜏𝜏 =
𝐿𝐿
𝑅𝑅𝑒𝑒𝑒𝑒
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𝜏𝜏 =
Simplify:
120 𝑚𝑚𝑚𝑚
18 𝑘𝑘Ω
𝜏𝜏 = 6.66 𝜇𝜇𝜇𝜇
Plug 𝑖𝑖𝐿𝐿 (0), 𝑖𝑖𝐿𝐿 (∞), and 𝜏𝜏 into the equation for 𝑖𝑖𝐿𝐿 (𝑡𝑡) for 𝑡𝑡 > 0:
Problem 5.34
𝑖𝑖𝐿𝐿 (𝑡𝑡 > 0) = 0.67 𝑚𝑚𝑚𝑚 + [−0.43 𝐴𝐴 − 0.67 𝑚𝑚𝑚𝑚]𝑒𝑒 −𝑡𝑡/6.66
𝜇𝜇𝜇𝜇
For 𝑡𝑡 < 0, the circuit shown in Figure P5.34 is at steady state. The switch is thrown at 𝑡𝑡 = 0. Assume: 𝑉𝑉𝑆𝑆1 =
17 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 11 𝑉𝑉, 𝑅𝑅1 = 14 𝑘𝑘Ω, 𝑅𝑅2 = 13 𝑘𝑘Ω, 𝑅𝑅3 = 14 𝑘𝑘Ω, and 𝐶𝐶 = 70 𝑛𝑛𝑛𝑛. Determine the: (a). Current
through the capacitor, 𝑖𝑖𝐶𝐶 , for 𝑡𝑡 > 0, (b). voltage across 𝑅𝑅3 , 𝑣𝑣3 , for 𝑡𝑡 > 0, and (c). the time require for 𝑖𝑖𝐶𝐶 and 𝑣𝑣3
to change by 98% of their initial values at 𝑡𝑡 = 0+ .
Known quantities:
𝑉𝑉𝑆𝑆1 = 17 𝑉𝑉, 𝑉𝑉𝑆𝑆2 = 11 𝑉𝑉, 𝑅𝑅1 = 14 𝑘𝑘Ω, 𝑅𝑅2 = 13 𝑘𝑘Ω, 𝑅𝑅3 = 14 𝑘𝑘Ω, 𝐶𝐶 = 70 𝑛𝑛𝑛𝑛
Find:
(a). Current through the capacitor, 𝑖𝑖𝐶𝐶 , for 𝑡𝑡 > 0.
(b). Voltage across 𝑅𝑅3 , 𝑣𝑣3 , for 𝑡𝑡 > 0.
(c). The time require for 𝑖𝑖𝐶𝐶 and 𝑣𝑣3 to change by 98% of their initial values at 𝑡𝑡 = 0+ .
Analysis:
(a) Current through the capacitor, 𝒊𝒊𝑪𝑪 , for 𝒕𝒕 > 𝟎𝟎.
To determine 𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) use the equation for current through a capacitor:
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) =
1
[𝑣𝑣 (∞) − 𝑣𝑣𝐶𝐶 (0)]𝑒𝑒 −𝑡𝑡/𝜏𝜏
𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶
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To realize this equation, 𝑣𝑣𝐶𝐶 (0), 𝑅𝑅𝑒𝑒𝑒𝑒 , 𝜏𝜏, and 𝑣𝑣𝐶𝐶 (∞) must be found. For 𝑣𝑣𝐶𝐶 (0), consider the circuit at 𝑡𝑡 < 0. In
steady state, 𝐶𝐶 acts like an open-circuit. Therefore, 𝑣𝑣𝐶𝐶 (0) is the same as 𝑣𝑣3 . Also, note that 𝑅𝑅3 is the only
element in the circuit.
Substitute known values:
Simplify:
𝑣𝑣3 = 𝑣𝑣𝐶𝐶 (0) = 𝑉𝑉𝑆𝑆1
𝑣𝑣𝐶𝐶 (0) = 17 𝑉𝑉 ∗
14 𝑘𝑘Ω
14 𝑘𝑘Ω + 14 𝑘𝑘Ω
𝑣𝑣𝐶𝐶 (0) = 8.5 𝑉𝑉
To find 𝑅𝑅𝑒𝑒𝑒𝑒 , consider the circuit at 𝑡𝑡 > 0. Find 𝑅𝑅𝑒𝑒𝑒𝑒 as seen by 𝐶𝐶. To accomplish this task, remove 𝐶𝐶, turn
𝑉𝑉𝑆𝑆2 into a short-circuit, and combine resistors. Combine 𝑅𝑅1 and 𝑅𝑅3 in parallel:
Substitute known values:
Simplify:
𝑅𝑅1,3,𝑒𝑒𝑒𝑒 =
𝑅𝑅1,3,𝑒𝑒𝑒𝑒 =
𝑅𝑅1 𝑅𝑅3
𝑅𝑅1 + 𝑅𝑅3
14 𝑘𝑘Ω ∗ 14 𝑘𝑘Ω
14 𝑘𝑘Ω + 14 𝑘𝑘Ω
𝑅𝑅1,3,𝑒𝑒𝑒𝑒 = 7 𝑘𝑘Ω
Combine 𝑅𝑅1,3,𝑒𝑒𝑒𝑒 and 𝑅𝑅2 in series and note that this is 𝑅𝑅𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1,3,𝑒𝑒𝑒𝑒 + 𝑅𝑅2
𝑅𝑅𝑒𝑒𝑒𝑒 = 7 kΩ + 13 kΩ
𝑅𝑅𝑒𝑒𝑒𝑒 = 20 𝑘𝑘Ω
Use the time constant equation for an RC circuit to determine 𝜏𝜏:
Substitute known values:
Simplify:
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶
𝜏𝜏 = 20 𝑘𝑘Ω ∗ 70 𝑛𝑛𝑛𝑛
𝜏𝜏 = 1.4 𝑚𝑚𝑚𝑚
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Finally, determine 𝑣𝑣𝐶𝐶 (∞). The strategy to find 𝑣𝑣𝐶𝐶 (∞) is to recognize the elimination of 𝑅𝑅2 , due to the opencircuit caused by the capacitor. Thus, 𝑣𝑣𝐶𝐶 (∞) can be written as a voltage division of 𝑅𝑅1 and 𝑅𝑅3 .
𝑣𝑣𝐶𝐶 (∞) = 𝑉𝑉𝑆𝑆2
Substitute known values:
𝑣𝑣𝐶𝐶 (∞) = 11 𝑉𝑉 ∗
Simplify:
𝑅𝑅3
𝑅𝑅1 + 𝑅𝑅3
14 𝑘𝑘Ω
14 𝑘𝑘Ω + 14 𝑘𝑘Ω
𝑣𝑣𝐶𝐶 (∞) = 5.5 𝑉𝑉
Plug 𝑣𝑣𝐶𝐶 (0), 𝑅𝑅𝑒𝑒𝑒𝑒 , 𝜏𝜏, and 𝑣𝑣𝐶𝐶 (∞) into the equation for current through a capacitor:
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) =
(b). Voltage across 𝑹𝑹𝟑𝟑 , 𝒗𝒗𝟑𝟑 , for 𝒕𝒕 > 𝟎𝟎.
1
[5.5 𝑉𝑉 − 8.5 𝑉𝑉]𝑒𝑒 −𝑡𝑡/(1.3𝑒𝑒−3)
20𝑘𝑘 Ω
To determine 𝑣𝑣3 (𝑡𝑡 > 0), relate 𝑣𝑣𝑐𝑐 (𝑡𝑡 > 0) and 𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) using KVL:
Part (a) determined 𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0).
𝑣𝑣3 (𝑡𝑡 > 0) = 𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0)𝑅𝑅2 + 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0)
To find 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) invoke the equation for voltage across a capacitor:
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) = 𝑣𝑣𝐶𝐶 (∞) + [𝑣𝑣𝐶𝐶 (0) − 𝑣𝑣𝐶𝐶 (∞)]𝑒𝑒 −𝑡𝑡/𝜏𝜏
Substitute previously determine values from (a):
Combine results to get:
Simplify:
𝑣𝑣3 (𝑡𝑡 > 0) =
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) = 5.5 𝑉𝑉 + [8.5 𝑉𝑉 − 5.5 𝑉𝑉]𝑒𝑒 −𝑡𝑡/(1.4𝑒𝑒−3)
𝑅𝑅2
[5.5 𝑉𝑉 − 8.5 𝑉𝑉]𝑒𝑒 −𝑡𝑡/(1.3𝑒𝑒−3) + 5.5 𝑉𝑉 + [8.5 𝑉𝑉 − 5.5 𝑉𝑉]𝑒𝑒 −𝑡𝑡/(1.4𝑒𝑒−3)
𝑅𝑅𝑒𝑒𝑒𝑒
𝑅𝑅2
[5.5 𝑉𝑉 − 8.5 𝑉𝑉] + [8.5 𝑉𝑉 − 5.5 𝑉𝑉]) 𝑒𝑒 −𝑡𝑡/(1.4𝑒𝑒−3)
𝑣𝑣3 (𝑡𝑡 > 0) = 5.5 𝑉𝑉 + (
𝑅𝑅𝑒𝑒𝑒𝑒
(c). The time require for 𝒊𝒊𝑪𝑪 and 𝒗𝒗𝟑𝟑 to change by 98% of their initial values at 𝒕𝒕 = 𝟎𝟎+ .
𝑖𝑖𝐶𝐶 (𝑡𝑡) − 𝑖𝑖𝐶𝐶 (0+ )
= 0.98
𝑖𝑖𝐶𝐶 (∞) − 𝑖𝑖𝐶𝐶 (0+ )
Note that the value at 𝑖𝑖𝐶𝐶 (0+ ) is being determine.
This value can be determined with the equation from part (a):
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𝑖𝑖𝐶𝐶 (0+ ) = −0.15 𝑚𝑚𝑚𝑚
A capacitor is an open-circuit in steady-state, so as 𝑡𝑡 → ∞:
𝑖𝑖𝐶𝐶 (∞) = 0 𝐴𝐴
Substitute these values into the previously determine ratio:
Simplify:
𝑖𝑖𝐶𝐶 (𝑡𝑡) + 0.15 𝑚𝑚𝑚𝑚
= 0.98
0.15 𝑚𝑚𝑚𝑚
𝑖𝑖𝐶𝐶 (𝑡𝑡) = −3 𝜇𝜇𝜇𝜇
Now, substitute the equation from (a) and solve for 𝑡𝑡:
Simplify:
𝑡𝑡
1
[5.5 𝑉𝑉 − 8.5 𝑉𝑉]𝑒𝑒 −1.3𝑒𝑒−3 = −3 𝜇𝜇𝜇𝜇
20𝑘𝑘 Ω
𝑒𝑒 −𝑡𝑡/1.3𝑒𝑒−3 = 0.02
Take the natural log of both sides:
Simplify:
−
𝑡𝑡
= −3.91
1.3 ∗ 10−3
𝑡𝑡 = 5 𝑚𝑚𝑚𝑚
Note that 𝑣𝑣3 will take the same amount of time to change by 98% of its initial conditions.
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Problem 5.35
The circuit in Figure P5.35 is a simple model of an automotive ignition system. The switch models the “points”
that switch electric power to the cylinder when the fuel-air mixture is compressed. 𝑅𝑅 is the resistance across the
gap between the electrodes of the spark plug. Assume: 𝑉𝑉𝐺𝐺 = 12 𝑉𝑉, 𝑅𝑅𝐺𝐺 = 0.37 Ω, and 𝑅𝑅 = 1.7 𝑘𝑘Ω. Determine
the value of 𝐿𝐿 and 𝑅𝑅1 so that the voltage across the spark plug gap just after the switch is changed is 23 𝑘𝑘𝑘𝑘 and
so that this voltage will change exponentially with a time constant, 𝜏𝜏, of 𝜏𝜏 = 13 𝑚𝑚𝑚𝑚.
Known quantities:
𝑉𝑉𝐺𝐺 = 12 𝑉𝑉, 𝑅𝑅𝐺𝐺 = 0.37 Ω, 𝑅𝑅 = 1.7 𝑘𝑘Ω
Find:
The value of 𝐿𝐿 and 𝑅𝑅1 so that the voltage across the spark plug gap just after the switch is changed is 23 𝑘𝑘𝑘𝑘 and
so that this voltage will change exponentially with a time constant of 𝜏𝜏 = 13 𝑚𝑚𝑚𝑚.
Analysis:
First, determine 𝑅𝑅1 .
The current through 𝐿𝐿 when 𝑡𝑡 > 0 must be made such that:
Substitute known values:
Solve for 𝑖𝑖𝐿𝐿 :
23 𝑘𝑘𝑘𝑘 = 𝑖𝑖𝐿𝐿 𝑅𝑅
23 𝑘𝑘𝑘𝑘 = 𝑖𝑖𝐿𝐿 ∗ 1.7 𝑘𝑘Ω
𝑖𝑖𝐿𝐿 = 13.53 𝐴𝐴
Assuming the circuit is in steady-state each time before the switch is flipped, 𝐿𝐿 behaves like a short-circuit.
Therefore, using Ohm’s law and combining 𝑅𝑅𝐺𝐺 and 𝑅𝑅1 in series, 𝑖𝑖𝐿𝐿 can be written as:
Substitute known values:
𝑖𝑖𝐿𝐿 =
𝑉𝑉𝐺𝐺
𝑅𝑅𝐺𝐺 + 𝑅𝑅1
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Solve for 𝑅𝑅1 :
13.53 𝐴𝐴 =
12 𝑉𝑉
0.37 Ω + 𝑅𝑅1
𝑅𝑅1 = 0.52 Ω
To solve for 𝜏𝜏 analyze the circuit at 𝑡𝑡 > 0 from the inductor’s perspective. That is, calculate the equivalent
resistance seen by 𝐿𝐿 at 𝑡𝑡 > 0. Removing the inductor, it sees 𝑅𝑅1 and 𝑅𝑅 in parallel:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1 + 𝑅𝑅
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 0.52 Ω + 1.7 𝑘𝑘Ω
𝑅𝑅𝑒𝑒𝑒𝑒 ≈ 1.7 𝑘𝑘Ω
Use the equation for a time constant to determine 𝐿𝐿:
Substitute known values:
Solve for 𝐿𝐿:
𝜏𝜏 =
𝐿𝐿
𝑅𝑅𝑒𝑒𝑒𝑒
13 𝑚𝑚𝑚𝑚 =
𝐿𝐿
1.7 𝑘𝑘Ω
𝐿𝐿 = 22.1 𝐻𝐻
The final answers are: 𝐿𝐿 = 22.1 𝐻𝐻 and 𝑅𝑅1 = 0.52 Ω.
Problem 5.36
The inductor L in the circuit shown in Figure P5.36
is the coil of a relay. When the current iL through the
coil is equal to or greater than 2mA, the relay is
activated. Assume steady-state conditions at t < 0. If
VS = 12V
L = 10.9mH R1 = 3.1kohm
Determine R2 so that the relay activates 2.3 sec after
the switch is thrown.
Solution:
Known quantities:
Circuit shown in Figure P5.36, when iL ≥ +2 mA , the relay functions.
VS = 12V ,L = 10.9mH ,R1 = 3.1kΩ.
Find:
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R2 so that the relay functions at t = 2.3 s .
Assumptions:
The circuit is in steady-state conditions for t < 0 .
Analysis:
In this problem the current through the inductor is clearly zero before the switch is thrown. The task is determine
the value of the resistance R2 such that the current through the inductor will need 2.3 seconds to rise to 2 mA.
Once again, we must find the complete transient solution, this time for the current through the inductor. Assume a
solution of the form
i L (t) = i∞ + (i 0 − i∞ )e−t /τ
At t = 0− :
The current through the inductor is zero since no source is connected.
i L (0− ) = 0
At t = 0+ :
i 0 = i L (0+ ) = i L (0− ) = 0
For t > 0 :
Determine the Thevenin equivalent resistance as "seen" by the inductor, i.e., with respect to the port or terminals of
the inductor:
RTH = (R1 R2 )=
R1R2
R1 + R2
Hence,
L(R1 + R2 )
L
τ=
=
RTH
R1R2
At
t = infinity:
Steady state is again established. At steady state the inductor is again modeled as a short circuit.
current through R2 is zero and the current through the inductor is given by
Thus, the
V
12
i∞ = iL (∞) = S =
= 3.87mA
R1 3.1 × 103
Plug in the above quantities to the complete solution and set the current through the inductor equal to 2 mA and the
time equal to 2.3 s.
− 2.3
2 

= ln 1 −
2 × 10 − 3 = 3.87 × 10 − 3 1 − e − 2.3 / τ or
or τ ≅ 3.16seconds
τ
3
.
87 

Solving for R2:
LR1
L(R1 + R2 )
≅ 3.4mΩ
or R2 =
3.16 =
R1R2
3.16 R1 − L
(
)
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Problem 5.37
Determine the current through the capacitor, 𝑖𝑖𝐶𝐶 , in Figure p5.37 for all time. Assume DC steady-state conditions
for 𝑡𝑡 < 0. Also, assume: 𝑉𝑉1 = 10 𝑉𝑉, 𝐶𝐶 = 200 𝜇𝜇𝜇𝜇, 𝑅𝑅1 = 300 𝑚𝑚Ω, and 𝑅𝑅2 = 𝑅𝑅3 = 1.2 𝑘𝑘Ω.
Known quantities:
𝑉𝑉1 = 10 𝑉𝑉, 𝐶𝐶 = 200 𝜇𝜇𝜇𝜇, 𝑅𝑅1 = 300 𝑚𝑚Ω, 𝑅𝑅2 = 𝑅𝑅3 = 1.2 𝑘𝑘Ω.
Find:
The current through the capacitor, 𝑖𝑖𝐶𝐶 , for all time.
Analysis:
There are two conditions to examine: 𝑡𝑡 < 0 and 𝑡𝑡 > 0. When 𝑡𝑡 < 0, there is no source attached to the circuit.
Therefore, there exists no current through the capacitor when 𝑡𝑡 < 0.
The current through a charging capacitor at any time, 𝑡𝑡, is given by the equation:
𝑖𝑖𝐶𝐶 (𝑡𝑡) =
𝑉𝑉 −𝑡𝑡/𝑅𝑅 𝐶𝐶
𝑒𝑒𝑒𝑒
𝑒𝑒
𝑅𝑅𝑒𝑒𝑒𝑒
To fully realize this equation, the time constant, 𝜏𝜏, and the equivalent resistance, 𝑅𝑅𝑒𝑒𝑒𝑒 , must be determined.
Find 𝑅𝑅𝑒𝑒𝑒𝑒 seen by the capacitor. That is, remove 𝐶𝐶 and make 𝑉𝑉1 a short.
Combine 𝑅𝑅1 and 𝑅𝑅2 in parallel:
Substitute known values:
Simplify:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 =
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 =
Then, combine the resistors.
𝑅𝑅1 𝑅𝑅2
𝑅𝑅1 + 𝑅𝑅2
300 𝑚𝑚Ω ∗ 1.2 kΩ
1.2 𝑘𝑘Ω + 300 𝑚𝑚Ω
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𝑅𝑅1,2,𝑒𝑒𝑒𝑒 ≈ 0.3 Ω
Combine 𝑅𝑅1,2,𝑒𝑒𝑒𝑒 and 𝑅𝑅3 in series. Note that this combination results in the total equivalent resistance, 𝑅𝑅𝑒𝑒𝑒𝑒 .
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅1,2,𝑒𝑒𝑒𝑒 + 𝑅𝑅3
𝑅𝑅𝑒𝑒𝑒𝑒 = 0.3 Ω + 1.2 𝑘𝑘Ω
𝑅𝑅𝑒𝑒𝑒𝑒 ≈ 1.2 𝑘𝑘Ω
Recognize the RC circuit and use 𝑅𝑅𝑒𝑒𝑒𝑒 to calculate the time constant:
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶
Substitute known values:
Simplify:
𝜏𝜏 = 1.2 𝑘𝑘Ω ∗ 200 𝜇𝜇𝜇𝜇
𝜏𝜏 = 0.24 𝑠𝑠
Finally, plug 𝑉𝑉1 , 𝑅𝑅𝑒𝑒𝑒𝑒 , and 𝜏𝜏 into the equation for current through a charging capacitor:
Simplify:
Combine results:
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) =
10 𝑉𝑉 −𝑡𝑡/0.24
𝑒𝑒
1.2 𝑘𝑘Ω
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) = 0.008𝑒𝑒 −𝑡𝑡/0.24
𝑖𝑖𝐶𝐶 (𝑡𝑡) = �
𝑠𝑠
𝑠𝑠
0 𝐴𝐴
𝑡𝑡 ≤ 0
𝑡𝑡 > 0
0.008𝑒𝑒 −𝑡𝑡/0.24 𝑠𝑠 𝐴𝐴
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.38
Determine the voltage across the inductor, 𝑣𝑣𝐿𝐿 , in Figure P5.38 for all time. Assume DC steady-state conditions for
𝑡𝑡 < 0. Also, assume: 𝑉𝑉𝑆𝑆 = 15 𝑉𝑉, 𝐿𝐿 = 100 𝑚𝑚𝑚𝑚, 𝑅𝑅𝑆𝑆 = 1 Ω, and 𝑅𝑅1 = 20 𝑘𝑘Ω.
Known quantities:
𝑉𝑉𝑆𝑆 = 15 𝑉𝑉, 𝐿𝐿 = 100 𝑚𝑚𝑚𝑚, 𝑅𝑅𝑆𝑆 = 1 Ω, 𝑅𝑅1 = 20 𝑘𝑘Ω
Find:
The voltage across the inductor, 𝑣𝑣𝐿𝐿 , for all time.
Analysis:
There are two cases that must be considered: 𝑡𝑡 < 0 and 𝑡𝑡 > 0. For 𝑡𝑡 < 0, the circuit is in a steady-state
condition, as per the problem statement. Furthermore, inductors act as short-circuits under steady-state conditions.
Therefore, the voltage across the inductor is:
𝑣𝑣𝐿𝐿 (𝑡𝑡 < 0) = 0 𝑉𝑉
For 𝑡𝑡 > 0, realize that the inductor is no longer connected to a source and is discharging. The equation for voltage
across a discharging inductor is:
𝑅𝑅𝑒𝑒𝑒𝑒
𝑡𝑡
𝐿𝐿
𝑣𝑣𝐿𝐿 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐼𝐼0 𝑒𝑒 −
where 𝑅𝑅𝑒𝑒𝑒𝑒 is the equivalent resistance of the circuit and 𝐼𝐼0 is the current of the inductor before being discharged.
First, determine 𝐼𝐼0 . Because the circuit was in DC steady-state at 𝑡𝑡 < 0, the current through 𝐿𝐿 is the current
through 𝑅𝑅𝑆𝑆 , 𝑖𝑖. Using Ohm’s law provides:
Substitute known values:
𝑖𝑖𝑆𝑆 = 𝐼𝐼0 =
𝐼𝐼0 =
5.47
𝑉𝑉𝑆𝑆
𝑅𝑅𝑆𝑆
15 𝑉𝑉
1 Ω
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 at 𝑡𝑡 > 0 is simply 𝑅𝑅1 .
𝐼𝐼0 = 15 𝐴𝐴
Recognize the RL circuit and compute the time constant, 𝜏𝜏:
𝜏𝜏 =
Substitute known values:
𝜏𝜏 =
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒
𝐿𝐿
20 𝑘𝑘Ω
100 𝑚𝑚𝑚𝑚
𝜏𝜏 = 0.2 𝑀𝑀𝑀𝑀
Plug 𝐼𝐼0 , 𝑅𝑅𝑒𝑒𝑒𝑒 , and 𝜏𝜏 into the equation for voltage across a discharging inductor:
Simplify:
𝑣𝑣𝐿𝐿 (𝑡𝑡 > 0) = 15 𝐴𝐴 ∗ 20 𝑘𝑘Ω ∗ 𝑒𝑒 −0.2
𝑣𝑣𝐿𝐿 (𝑡𝑡 > 0) = 300 ∗ 𝑒𝑒 −0.2𝑡𝑡
Combine results to determine the complete equation:
𝑣𝑣𝐿𝐿 (𝑡𝑡) = �
0 𝑉𝑉
300𝑒𝑒 −0.2𝑡𝑡
𝑒𝑒6
𝑀𝑀𝑀𝑀
∗ 𝑡𝑡 𝑒𝑒6
𝑉𝑉
𝑡𝑡 < 0
𝑉𝑉
𝑡𝑡 > 0
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.39
For t < 0, the circuit shown in Figure P5.39 is at
DC steady state. The switch is closed at t = 0.
Determine the voltage vC for all time. Assume:
R1 = R3 = 3ohm, R2 = 6ohm V1 = 15V, and C = 0.5F.
Solution:
Known quantities:
Circuit shown in Figure P5.39,
e = 220V ,C = 1F , R1 = R3 = 1k Ω,R2 = 500Ω.
Find:
VC (t )
Assumptions:
The circuit is in steady-state conditions for t < 0 .
Analysis:
CASE 1: t < 0
VC (t ) =
e
⋅ R2 = 73.33V
R1 + R2
CASE 2: t > 0
VC (∞) =
220
e
⋅ ( R2 || R3 ) =
⋅ 333.3 = 54.9V
1333.3
R1 + ( R2 || R3 )
RT = R1 || R2 || R3 = 249.98Ω
VC (t ) = VC (∞) + [VC (0) − VC (∞)]e
−
t
RT C
−3
−3
= 54.9 + (73.33 − 54.9)e −4⋅10 tV = 54.59 + 18.43e −4⋅10 tV
Problem 5.40
For 𝑡𝑡 < 0, the circuit shown in Figure P5.39 is at DC steady steady-state. The switch is opened at 𝑡𝑡 = 0.
Determine the current through the capacitor (?), 𝑖𝑖𝐶𝐶 , for all time. Assume: 𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 𝑅𝑅2 = 400 Ω, 𝑅𝑅3 =
600 Ω, and 𝐶𝐶 = 100 𝑚𝑚𝑚𝑚.
5.49
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Known quantities:
𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 𝑅𝑅2 = 400 Ω, 𝑅𝑅3 = 600 Ω, 𝐶𝐶 = 100 𝑚𝑚𝑚𝑚
Find:
The current through the capacitor, 𝑖𝑖𝐶𝐶 , for all time.
Analysis:
There are two cases to consider: 𝑡𝑡 < 0 and 𝑡𝑡 > 0. For 𝑡𝑡 < 0, the circuit is at DC steady-state, as per the problem
statement. Capacitors act like open circuits in steady state. Therefore, there is no current going through 𝐶𝐶 at
𝑡𝑡 < 0:
𝑖𝑖𝐶𝐶 (𝑡𝑡 < 0) = 0 𝐴𝐴
For 𝑡𝑡 > 0, the capacitor is being charged. The equation for a charging capacitor is given by:
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) =
𝑉𝑉0 −𝑡𝑡/𝑅𝑅 𝐶𝐶
𝑒𝑒𝑒𝑒
𝑒𝑒
𝑅𝑅𝑒𝑒𝑒𝑒
where 𝑉𝑉0 is the initial voltage across the capacitor and 𝑅𝑅𝑒𝑒𝑒𝑒 is the equivalent resistance.
To find 𝑉𝑉0 , determine the voltage across 𝑅𝑅2 , 𝑉𝑉2 , prior to the switch being closed. 𝑉𝑉2 may be written the voltage
division:
Substitute known values:
𝑉𝑉2 = 𝑉𝑉0 = 𝑉𝑉1
𝑉𝑉0 = 12 𝑉𝑉 ∗
𝑅𝑅2
𝑅𝑅1 + 𝑅𝑅2
400 Ω
400 Ω + 400 Ω
Simplify:
𝑉𝑉0 = 4 𝑉𝑉
To determine 𝑅𝑅𝑒𝑒𝑒𝑒 , remove the capacitor from the circuit at 𝑡𝑡 > 0, turn 𝑉𝑉1 into a short circuit, and determine the
resistance it sees. Combine 𝑅𝑅2 and 𝑅𝑅3 in parallel:
Substitute known values:
Simplify:
𝑅𝑅2,𝑒𝑒,𝑒𝑒𝑒𝑒 =
𝑅𝑅2,𝑒𝑒,𝑒𝑒𝑒𝑒 =
𝑅𝑅2 𝑅𝑅3
𝑅𝑅2 + 𝑅𝑅3
400 Ω ∗ 600 Ω
400 Ω + 600 Ω
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𝑅𝑅2,𝑒𝑒,𝑒𝑒𝑒𝑒 = 240 Ω
Combine 𝑅𝑅2,𝑒𝑒,𝑒𝑒𝑒𝑒 and 𝑅𝑅1 in series to realize 𝑅𝑅𝑒𝑒𝑒𝑒 :
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅2,𝑒𝑒,𝑒𝑒𝑒𝑒 + 𝑅𝑅1
Substitute known values:
𝑅𝑅𝑒𝑒𝑒𝑒 = 240 Ω + 400 Ω
Simplify:
The final step is to determine the time constant, 𝜏𝜏.
constant as:
𝑅𝑅𝑒𝑒𝑒𝑒 = 640 Ω
Realize that this circuit is an RC circuit and compute the time
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶
Substitute known values:
𝜏𝜏 = 640 Ω ∗ 100 𝑚𝑚𝑚𝑚
Simplify:
𝜏𝜏 = 64 𝑠𝑠
Plug 𝑉𝑉0 , 𝑅𝑅𝑒𝑒𝑒𝑒 , and 𝜏𝜏 into the equation for current through a charging capacitor:
Simplify:
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) =
4 𝑉𝑉 −𝑡𝑡/64
𝑒𝑒
640 Ω
𝑠𝑠
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) = 0.006𝑒𝑒 −𝑡𝑡/64 𝑠𝑠 𝐴𝐴
Combine results into the equation for 𝑖𝑖𝐶𝐶 (𝑡𝑡):
𝑖𝑖𝐶𝐶 (𝑡𝑡 > 0) = �
Problem 5.41
0 𝐴𝐴
𝑡𝑡 ≤ 0
𝑡𝑡 > 0
0.006𝑒𝑒 −𝑡𝑡/64 𝑠𝑠 𝐴𝐴
For the circuit shown in Figure P5.41, assume that
switch S1 is always held open and that switch S2 is
open until being closed at t = 0. Assume DC
steady-state conditions for t < 0. Also assume
R1 = 5ohm, R2 = 4ohm, R3 = 3ohm, R4 = 6ohm, and
C1 = C2 = 4F.
a. Find the capacitor voltage vC at t = 0+.
b. Find the time constant τ for t > 0.
c. Find vC for all time and sketch the function.
d. Evaluate the ratio vC to vC (∞) at each of the
following times: t = 0, τ, 2τ, 5τ, 10τ
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Solution:
Known quantities:
Circuit shown in Figure P5.41, VS = 20V , R1 = 5Ω,R2 = 4Ω ,R3 = 3Ω,R4 = 6Ω,C1 = 4F ,C2 = 4F ,I S = 4 A.
Find:
+
a) The capacitor voltage VC (t ) at t = 0 .
b) The time constant τ for t ≥ 0 .
c) The expression for VC (t) and sketch the function.
d) Find VC (t) for each of the following values of t :
0,τ ,2τ ,5τ ,10τ .
Assumptions:
Switch S1 is always open and switch S2 closes at t = 0 .
Analysis:
a)
Without any power sources connected the steady state voltages are zero due to relentless dissipation of
energy in the resistors.
VC (0− ) = VC (0+ ) = 0 V
When the initial condition on a transient is zero, the general solution for the transient simplifies to
(
VC (t) = V (∞) 1− e−t /τ
)
b)
The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin
equivalent resistance seen by the 8 F capacitance is found by suppressing the current source (i.e. replacing it with an
open circuit) and computing R2 + R3||R4.
RTH = 4 + (3 6) = 6 Ω
τ = RTH C = (6)(8) = 48 s
c)
The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits
and solving for the voltage across R3. This voltage is found readily by current division
6Ω
VC (∞) =
(4 A)(3Ω) = 8V
3Ω + 6Ω
Plug into the generalized solution given above to find
(
VC (t) = 8 1− e−t / 48
VC (t) = 0
t≥0
, t≤0
d)
VC (0) = 0V ;
VC (2τ ) = 6.9V ;
),
VC (τ ) = 5.06V ;
VC (5τ ) = 7.95V ;
VC (10τ ) = 8.0V
Problem 5.42
For the circuit shown in Figure P5.41, assume that switch 𝑆𝑆1 and 𝑆𝑆2 have been held open and closed, respectively,
for a long time prior to 𝑡𝑡 = 0. Then, simultaneously at 𝑡𝑡 = 0, 𝑆𝑆1 closes and 𝑆𝑆2 opens. Also, assume: 𝑅𝑅1 =
5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, and 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹. Solve the following problems: (a) find the capacitor
voltage, 𝑣𝑣𝐶𝐶 at 𝑡𝑡 = 0+ ; find the time constant, 𝜏𝜏, for 𝑡𝑡 > 0; find 𝑣𝑣𝐶𝐶 for all time and sketch the function; evaluate
the ratio 𝑣𝑣𝐶𝐶 to 𝑣𝑣𝐶𝐶 (∞) at each of the following times: 𝑡𝑡 = 0, 𝜏𝜏, 2𝜏𝜏, 5𝜏𝜏, 10𝜏𝜏.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Known quantities:
𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹
Find:
(a) Find the capacitor voltage, 𝑣𝑣𝐶𝐶 at 𝑡𝑡 = 0+ .
(b) Find the time constant, 𝜏𝜏, for 𝑡𝑡 > 0.
(c) Find 𝑣𝑣𝐶𝐶 for all time and sketch the function.
(d) Evaluate the ratio 𝑣𝑣𝐶𝐶 to 𝑣𝑣𝐶𝐶 (∞) at each of the following times: 𝑡𝑡 = 0, 𝜏𝜏, 2𝜏𝜏, 5𝜏𝜏, 10𝜏𝜏.
Analysis:
(a) Find the capacitor voltage, 𝒗𝒗𝑪𝑪 at 𝒕𝒕 = 𝟎𝟎+ .
To determine 𝑣𝑣𝐶𝐶 (0+ ), realize that the circuit is in steady state at 𝑡𝑡 < 0. It cannot change simultaneously.
Therefore, the conditions at 𝑡𝑡 = 0− and 𝑡𝑡 = 0+ are identical. At steady state, 𝐶𝐶1 and 𝐶𝐶2 act as open-circuits.
Therefore, they have no current going through them. Due to this state, 𝑅𝑅2 has no voltage drop across it, so
𝑣𝑣𝐶𝐶 (0+ ) is the same as the voltage drop across 𝑅𝑅3 or 𝑅𝑅4 . Determine the current through 𝑅𝑅3 , 𝑖𝑖3 , through current
division:
Substitute known values:
Simplify:
𝑖𝑖3 = 4 𝐴𝐴 ∗
𝑖𝑖3 = 4 𝐴𝐴 ∗
𝑅𝑅4
𝑅𝑅3 + 𝑅𝑅4
6 Ω
3 Ω+6 Ω
𝑖𝑖3 = 2.67 𝐴𝐴
Use Ohm’s law to calculate the voltage across 𝑅𝑅3 , 𝑣𝑣3 :
Substitute known values:
𝑣𝑣3 = 𝑖𝑖3 𝑅𝑅3
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Simplify
(b) Find the time constant, 𝝉𝝉, for 𝒕𝒕 > 𝟎𝟎.
𝑣𝑣3 = 2.67 𝐴𝐴 ∗ 3 Ω
𝑣𝑣3 = 8 𝑉𝑉
To find the time constant, consider the circuit at 𝑡𝑡 > 0 with 𝑆𝑆1 closed and 𝑆𝑆2 open. To determine 𝜏𝜏, find 𝑅𝑅𝑒𝑒𝑒𝑒
from 𝐶𝐶𝑒𝑒𝑒𝑒 ’s perspective. Combine 𝐶𝐶1 and 𝐶𝐶2 as capacitors in parallel:
Substitute known values:
Simplify:
𝐶𝐶𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶2
𝐶𝐶𝑒𝑒𝑒𝑒 = 4 𝐹𝐹 + 4 𝐹𝐹
𝐶𝐶𝑒𝑒𝑒𝑒 = 8 𝐹𝐹
Find 𝑅𝑅𝑒𝑒𝑒𝑒 seen by 𝐶𝐶𝑒𝑒𝑒𝑒 . To do this, remove 𝐶𝐶𝑒𝑒𝑒𝑒 from the circuit and turn the voltage source into a short-circuit.
Combine 𝑅𝑅2 and 𝑅𝑅3 in series:
Substitute known values:
Simplify:
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 4 Ω + 3 Ω
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 7 Ω
Combine 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 and 𝑅𝑅1 in parallel and realize that this combination is 𝑅𝑅𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
Calculate 𝜏𝜏 for an RC circuit:
Substitute known values:
𝑅𝑅𝑒𝑒𝑒𝑒 =
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 𝑅𝑅1
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 + 𝑅𝑅1
𝑅𝑅𝑒𝑒𝑒𝑒 =
7 Ω∗5 Ω
7 Ω+5 Ω
𝑅𝑅𝑒𝑒𝑒𝑒 = 2.92 Ω
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶𝑒𝑒𝑒𝑒
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𝜏𝜏 = 2.92 Ω ∗ 8 𝐹𝐹
Simplify:
(c) Find 𝒗𝒗𝑪𝑪 for all time and sketch the function.
For 𝑣𝑣𝐶𝐶 (𝑡𝑡 ≤ 0), use the answer from part (a):
To find 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) use the solution:
𝜏𝜏 = 23.36 𝑠𝑠
𝑣𝑣𝐶𝐶 (𝑡𝑡 ≤ 0) = 8 𝑉𝑉
𝑣𝑣𝐶𝐶 (𝑡𝑡) = 𝑣𝑣𝐶𝐶 (∞) + [𝑣𝑣𝐶𝐶 (0+ ) − 𝑣𝑣𝐶𝐶 (∞)]𝑒𝑒 −𝑡𝑡/𝜏𝜏
From part (a), we have 𝑣𝑣𝐶𝐶 (0+ ). As 𝑡𝑡 → ∞, the capacitors become open-circuits. Therefore, the voltage drop
across the capacitors, 𝑣𝑣𝐶𝐶 (∞), is the same as the voltage drop across 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 . From part (b):
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 7 Ω
The voltage drop across 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 can be written as the voltage division of 𝑅𝑅1 and 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
𝑣𝑣𝐶𝐶 (∞) = 20 𝑉𝑉 ∗
𝑅𝑅2,3,𝑒𝑒𝑒𝑒
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 + 𝑅𝑅1
𝑣𝑣𝐶𝐶(∞) = 20 𝑉𝑉 ∗
7 Ω
7 Ω+5 Ω
𝑣𝑣𝐶𝐶 (∞) = 11.67 𝑉𝑉
Substitute 𝑣𝑣𝐶𝐶 (0+ ), 𝜏𝜏, and 𝑣𝑣𝐶𝐶 (∞) into the equation for 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0):
Combine results:
Sketch 𝑣𝑣𝐶𝐶 (𝑡𝑡):
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) = 11.67 𝑉𝑉 + [8 𝑉𝑉 − 11.67 𝑉𝑉]𝑒𝑒 −𝑡𝑡/23.36
𝑣𝑣𝐶𝐶 (𝑡𝑡) = �
8 𝑉𝑉
𝑡𝑡 ≤ 0
11.67 𝑉𝑉 + [8 𝑉𝑉 − 11.67 𝑉𝑉]𝑒𝑒 −𝑡𝑡/23.36
𝑡𝑡 > 0
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12
11.5
11
C
v (t)
10.5
10
9.5
9
8.5
8
0
20
40
60
80
100
120
time (s)
140
160
180
200
(d) Evaluate the ratio 𝒗𝒗𝑪𝑪 to 𝒗𝒗𝑪𝑪 (∞) at each of the following times: 𝒕𝒕 = 𝟎𝟎, 𝝉𝝉, 𝟐𝟐𝟐𝟐, 𝟓𝟓𝟓𝟓, 𝟏𝟏𝟏𝟏𝟏𝟏.
For brevity’s sake, the numerical simplification will not be shown for each evaluation. Simply plug in the value for
𝑡𝑡 into the 𝑣𝑣𝐶𝐶 (𝑡𝑡) from part (c).
𝑣𝑣𝐶𝐶 (0)
8 𝑉𝑉
=
= 0.69
𝑣𝑣𝐶𝐶 (∞) 11.67 𝑉𝑉
𝑣𝑣𝐶𝐶 (23.36) 10.31 𝑉𝑉
=
= 0.88
𝑣𝑣𝐶𝐶 (∞)
11.67 𝑉𝑉
𝑣𝑣𝐶𝐶 (46.72) 11.17 𝑉𝑉
=
= 0.96
𝑣𝑣𝐶𝐶 (∞)
11.67 𝑉𝑉
𝑣𝑣𝐶𝐶 (116.8) 11.65 𝑉𝑉
=
= 1.00
𝑣𝑣𝐶𝐶 (∞)
11.67 𝑉𝑉
Problem 5.43
𝑣𝑣𝐶𝐶 (233.6) 11.67 𝑉𝑉
=
= 1.00
𝑣𝑣𝐶𝐶 (∞)
11.67 𝑉𝑉
For the circuit shown in Figure P5.41, assume that switch 𝑆𝑆2 is always being held open and that switch 𝑆𝑆1 is
closed until being opened at 𝑡𝑡 = 0. Subsequently, 𝑆𝑆1 closes at 𝑡𝑡 = 3𝜏𝜏 and remains closed. Also, assume DC
steady-state conditions for 𝑡𝑡 < 0 and 𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, and 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹. Find the
following: (a) the capacitor voltage, 𝑣𝑣𝐶𝐶 , at 𝑡𝑡 = 0; (b) 𝑣𝑣𝐶𝐶 for 0 < 𝑡𝑡 < 3𝜏𝜏; (c) 𝑣𝑣𝐶𝐶 at 𝑡𝑡 = 3𝜏𝜏 and 𝑣𝑣𝐶𝐶 at 𝑡𝑡 > 3𝜏𝜏;
(d) compare the two time constants for 0 < 𝑡𝑡 < 3𝜏𝜏 and 𝑡𝑡 > 3𝜏𝜏; (e) sketch 𝑣𝑣𝐶𝐶 for all time.
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Known quantities:
𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹
Find:
(a) Find the capacitor voltage, 𝑣𝑣𝐶𝐶 , at 𝑡𝑡 = 0.
(b) Find 𝑣𝑣𝐶𝐶 for 0 < 𝑡𝑡 < 3𝜏𝜏.
(c) Find 𝑣𝑣𝐶𝐶 at 𝑡𝑡 = 3𝜏𝜏 and 𝑣𝑣𝐶𝐶 at 𝑡𝑡 > 3𝜏𝜏.
(d) Compare the two time constants for 0 < 𝑡𝑡 < 3𝜏𝜏 and 𝑡𝑡 > 3𝜏𝜏.
(e) Sketch 𝑣𝑣𝐶𝐶 for all time.
Analysis:
(a) the capacitor voltage, 𝒗𝒗𝑪𝑪 , at 𝒕𝒕 = 𝟎𝟎
To find 𝑣𝑣𝐶𝐶 (0), consider that the circuit is in steady-state at 𝑡𝑡 < 0. Capacitors behave like open-circuits in steadystate. Therefore, 𝑣𝑣𝐶𝐶 (0) is the same as the voltage drop across 𝑅𝑅2 and 𝑅𝑅3 . Perform a voltage division of 𝑅𝑅1
and 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 and combine 𝑅𝑅2 and 𝑅𝑅3 in series:
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
Substitute known values:
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 4 Ω + 3 Ω
Simplify:
Perform voltage division of 𝑅𝑅1 and 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 :
Substitute known values:
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 7 Ω
𝑣𝑣𝐶𝐶 (0) = 20 𝑉𝑉 ∗
𝑅𝑅2,3,𝑒𝑒𝑒𝑒
𝑅𝑅1 + 𝑅𝑅2,3,𝑒𝑒𝑒𝑒
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Simplify:
𝑣𝑣𝐶𝐶 (0) = 20 𝑉𝑉 ∗
7 Ω
5 Ω+7 Ω
𝑣𝑣𝐶𝐶 (0) = 11.67 𝑉𝑉
(b) 𝒗𝒗𝑪𝑪 for 𝟎𝟎 < 𝒕𝒕 < 𝟑𝟑𝟑𝟑
At 0 < 𝑡𝑡 < 3𝜏𝜏, there are no sources connected to the circuit, so the final voltage across the capacitor before 𝑡𝑡 = 0
is 𝑣𝑣𝐶𝐶 (0). Therefore, to find 𝑣𝑣𝐶𝐶 (0 < 𝑡𝑡 < 3𝜏𝜏), invoke the equation for a discharging capacitor:
𝑣𝑣𝐶𝐶 (0 < 𝑡𝑡 < 3𝜏𝜏) = 𝑣𝑣𝐶𝐶 (0)𝑒𝑒 −𝑡𝑡/𝜏𝜏 𝑉𝑉
To find the time constant, 𝜏𝜏, compute 𝑅𝑅𝑒𝑒𝑒𝑒 and 𝐶𝐶𝑒𝑒𝑒𝑒 of the circuit.
𝐶𝐶𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶2
Substitute known values:
Combine 𝐶𝐶1 and 𝐶𝐶2 in parallel:
𝐶𝐶𝑒𝑒𝑒𝑒 = 4 𝐹𝐹 + 4 𝐹𝐹
Simplify:
𝐶𝐶𝑒𝑒𝑒𝑒 = 8 𝐹𝐹
Combine 𝑅𝑅2 and 𝑅𝑅3 in series and note that this is 𝑅𝑅𝑒𝑒𝑒𝑒 :
𝑅𝑅𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
Substitute known values:
𝑅𝑅𝑒𝑒𝑒𝑒 = 4 Ω + 3 Ω
Simplify:
Compute the time constant for an RC circuit:
𝑅𝑅𝑒𝑒𝑒𝑒 = 7 Ω
Substitute known values:
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶𝑒𝑒𝑒𝑒
Simplify:
𝜏𝜏 = 7 Ω ∗ 8 𝐹𝐹
𝜏𝜏 = 56 𝑠𝑠
Plug 𝑣𝑣𝐶𝐶 (0) and 𝜏𝜏 into the equation for a discharging capacitor:
𝑣𝑣𝐶𝐶 (0 < 𝑡𝑡 < 3𝜏𝜏) = 11.67𝑒𝑒 −𝑡𝑡/56 𝑉𝑉
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(c) 𝒗𝒗𝑪𝑪 at 𝒕𝒕 = 𝟑𝟑𝟑𝟑 and 𝒗𝒗𝑪𝑪 at 𝒕𝒕 > 𝟑𝟑𝟑𝟑
Determine 𝑣𝑣𝐶𝐶 (3𝜏𝜏) by using the result from (b).
𝑣𝑣𝐶𝐶 (3𝜏𝜏) = 0.58 𝑉𝑉
At 𝑡𝑡 = 3𝜏𝜏, switch 𝑆𝑆1 closes, again, and remains closed. To calculate 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 3𝜏𝜏), utilize the full equation for a
capacitor’s response and note the shift of 𝑡𝑡 down by 3𝜏𝜏 to account for the change in time scale:
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 3𝜏𝜏) = 𝑣𝑣𝐶𝐶 (∞) + [𝑣𝑣𝐶𝐶 (3𝜏𝜏) − 𝑣𝑣𝐶𝐶 (∞)]𝑒𝑒 −(𝑡𝑡−3𝜏𝜏)/𝜏𝜏
Now, determine 𝑣𝑣𝐶𝐶 (∞) and 𝜏𝜏.
state response is the same:
The circuit at 𝑡𝑡 > 3𝜏𝜏 is identical to the circuit at 𝑡𝑡 < 0.
𝑣𝑣𝐶𝐶 (∞) = 𝑣𝑣𝐶𝐶 (0) = 11.67 𝑉𝑉
To find 𝜏𝜏, remove the capacitors, turn the voltage source into a short-circuit, and find 𝑅𝑅𝑒𝑒𝑒𝑒 .
in series:
Substitute known values:
Simplify:
Therefore, the steady
Combine 𝑅𝑅2 and 𝑅𝑅3
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 4 Ω + 3 Ω
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 = 7 Ω
Combine 𝑅𝑅1 and 𝑅𝑅2,3,𝑒𝑒𝑒𝑒 in parallel and note that this combination is 𝑅𝑅𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
𝑅𝑅𝑒𝑒𝑒𝑒 =
𝑅𝑅1 𝑅𝑅2,3,𝑒𝑒𝑒𝑒
𝑅𝑅1 + 𝑅𝑅2,3,𝑒𝑒𝑒𝑒
𝑅𝑅𝑒𝑒𝑒𝑒 =
5 Ω∗7 Ω
5 Ω+7 Ω
𝑅𝑅𝑒𝑒𝑒𝑒 = 2.92 Ω
Realize that 𝐶𝐶𝑒𝑒𝑒𝑒 has not changed from part (b) and calculate 𝜏𝜏 in an RC circuit:
Substitute known values:
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶𝑒𝑒𝑒𝑒
𝜏𝜏 = 2.92 Ω ∗ 8 𝐹𝐹
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Simplify:
𝜏𝜏 = 23.3 𝑠𝑠
Plug 𝑣𝑣𝐶𝐶 (∞) and 𝜏𝜏 into the 𝑣𝑣𝐶𝐶 (𝑡𝑡 > 3𝜏𝜏):
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 3𝜏𝜏) = 11.67 𝑉𝑉 + [0.58 𝑉𝑉 − 11.67 𝑉𝑉]𝑒𝑒 −(𝑡𝑡−3𝜏𝜏)/23.3
(d) Compare the two time constants for 𝟎𝟎 < 𝒕𝒕 < 𝟑𝟑𝟑𝟑 and 𝒕𝒕 > 𝟑𝟑𝟑𝟑.
Note that the time constant for 0 < 𝑡𝑡 < 3𝜏𝜏 is greater than that of 𝑡𝑡 > 3𝜏𝜏, due to the addition of 𝑅𝑅1 in parallel with
𝑅𝑅2,3,𝑒𝑒𝑒𝑒 .
(e) Sketch 𝒗𝒗𝑪𝑪 for all time.
12
10
C
v (t)
8
6
4
2
0
0
50
100
150
200
time (s)
250
300
350
400
Problem 5.44
For the circuit shown in Figure P5.41, assume that switches 𝑆𝑆1 and 𝑆𝑆2 have been held open for a long time prior to
𝑡𝑡 = 0 but then close at 𝑡𝑡 = 0. Also, assume: 𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, and 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹.
Do the following parts: (a) find the capacitor voltage, 𝑣𝑣𝐶𝐶 , at 𝑡𝑡 = 0; (b) find the time constant, 𝜏𝜏, for 𝑡𝑡 > 0; (c) find
𝑣𝑣𝐶𝐶 and sketch the function; (d) evaluate the ratio 𝑣𝑣𝐶𝐶 (𝑡𝑡) to 𝑣𝑣𝐶𝐶 (∞) at each of the following times: 𝑡𝑡 =
0, 𝜏𝜏, 2𝜏𝜏, 5𝜏𝜏, 10𝜏𝜏.
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Known quantities:
𝑅𝑅1 = 5 Ω, 𝑅𝑅2 = 4 Ω, 𝑅𝑅3 = 3 Ω, 𝑅𝑅4 = 6 Ω, 𝐶𝐶1 = 𝐶𝐶2 = 4 𝐹𝐹
Find:
(a) Find the capacitor voltage, 𝑣𝑣𝐶𝐶 , at 𝑡𝑡 = 0.
(b) Find the time constant, 𝜏𝜏, for 𝑡𝑡 > 0.
(c) Find 𝑣𝑣𝐶𝐶 and sketch the function.
(d) Evaluate the ratio 𝑣𝑣𝐶𝐶 (𝑡𝑡) to 𝑣𝑣𝐶𝐶 (∞) at each of the following times: 𝑡𝑡 = 0, 𝜏𝜏, 2𝜏𝜏, 5𝜏𝜏, 10𝜏𝜏.
Analysis:
(a) Find the capacitor voltage, 𝒗𝒗𝑪𝑪 , at 𝒕𝒕 = 𝟎𝟎.
At 𝑡𝑡 < 0, there are no sources connected to the circuit.
(b) Find the time constant, 𝝉𝝉, for 𝒕𝒕 > 𝟎𝟎.
𝑣𝑣𝐶𝐶 (0) = 0 𝑉𝑉
To find the time constant, determine 𝑅𝑅𝑒𝑒𝑒𝑒 and 𝐶𝐶𝑒𝑒𝑒𝑒 .
Substitute known values:
Simplify:
Therefore, there is no energy in the circuit:
For 𝐶𝐶𝑒𝑒𝑒𝑒 , combine 𝐶𝐶1 and 𝐶𝐶2 in parallel:
𝐶𝐶𝑒𝑒𝑒𝑒 = 𝐶𝐶1 + 𝐶𝐶2
𝐶𝐶𝑒𝑒𝑒𝑒 = 4 𝐹𝐹 + 4 𝐹𝐹
𝐶𝐶𝑒𝑒𝑒𝑒 = 8 𝐹𝐹
To determine 𝑅𝑅𝑒𝑒𝑒𝑒 , remove the capacitors and turn the voltage source into a short-circuit and the current source into
an open-circuit. Combine 𝑅𝑅3 in parallel with 𝑅𝑅4 :
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Substitute known values:
Simplify:
Combine 𝑅𝑅2 in series with 𝑅𝑅3,4,𝑒𝑒𝑒𝑒 :
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 =
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 =
𝑅𝑅3 𝑅𝑅4
𝑅𝑅3 + 𝑅𝑅4
3 Ω∗6 Ω
3 Ω+6 Ω
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 = 2 Ω
Substitute known values:
𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅3,4,𝑒𝑒𝑒𝑒
Simplify:
𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒 = 4 Ω + 2 Ω
𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒 = 6 Ω
Combine 𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒 in parallel with 𝑅𝑅1 and note that this is 𝑅𝑅𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
Calculate the time constant of an RC circuit:
Substitute known values:
Simplify:
(c) Find 𝒗𝒗𝑪𝑪 and sketch the function.
𝑅𝑅𝑒𝑒𝑒𝑒 =
𝑅𝑅1 𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒
𝑅𝑅1 + 𝑅𝑅2,3,4,𝑒𝑒𝑒𝑒
𝑅𝑅𝑒𝑒𝑒𝑒 =
5 Ω∗6 Ω
5 Ω+6 Ω
𝑅𝑅𝑒𝑒𝑒𝑒 = 2.72 Ω
𝜏𝜏 = 𝑅𝑅𝑒𝑒𝑒𝑒 𝐶𝐶𝑒𝑒𝑒𝑒
𝜏𝜏 = 2.72 Ω ∗ 8 𝐹𝐹
𝜏𝜏 = 21.82 𝑠𝑠
To find 𝑣𝑣𝐶𝐶 invoke the equation for a charging capacitor:
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𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) = 𝑣𝑣𝐶𝐶 (∞)(1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 )
The only term left to compute is 𝑣𝑣𝐶𝐶 (∞). To realize this value, take advantage of Thévenin equivalent networks
and note that 𝐶𝐶1 and 𝐶𝐶2 are open-circuits in steady-state. First, combine 𝑅𝑅3 and 𝑅𝑅4 in parallel (known from
part (c)):
𝑅𝑅3,4,𝑒𝑒𝑒𝑒 = 2 Ω
Now, turn 𝑅𝑅3,4,𝑒𝑒𝑒𝑒 and the current source into a Thévenin equivalent network:
𝑅𝑅𝑇𝑇ℎ = 𝑅𝑅3,4,𝑒𝑒𝑒𝑒
Substitute known values:
𝑉𝑉𝑇𝑇ℎ = 𝑅𝑅𝑇𝑇ℎ ∗ 4 𝐴𝐴
Simplify:
𝑉𝑉𝑇𝑇ℎ = 2 Ω ∗ 4 𝐴𝐴
Next, combine 𝑅𝑅2 and 𝑅𝑅𝑇𝑇ℎ :
Substitute known values:
Simplify:
𝑉𝑉𝑇𝑇ℎ = 8 𝑉𝑉
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 = 𝑅𝑅2 + 𝑅𝑅𝑇𝑇ℎ
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 = 4 Ω + 2 Ω
𝑅𝑅2,𝑇𝑇ℎ,𝐸𝐸𝐸𝐸 = 6 Ω
Now, use the principal of superposition to add the effects of each voltage source as 𝑡𝑡 → ∞.
the contribution can be calculated by a voltage division between 𝑅𝑅1 and 𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 :
Substitute known values:
Simplify:
Next, calculate the contribution from 𝑉𝑉𝑡𝑡ℎ .
𝑣𝑣20𝑉𝑉 = 20 𝑉𝑉 ∗
For the 20 𝑉𝑉 source,
𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒
𝑅𝑅1 + 𝑅𝑅2.𝑇𝑇ℎ,𝑒𝑒𝑒𝑒
𝑣𝑣20𝑉𝑉 = 20 𝑉𝑉 ∗
6 Ω
5 Ω+6 Ω
𝑣𝑣20𝑉𝑉 = 10.90 𝑉𝑉
Calculate the voltage division from 𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒 to 𝑅𝑅1 :
𝑣𝑣𝑡𝑡ℎ = 𝑉𝑉𝑡𝑡ℎ ∗
5.63
𝑅𝑅1
𝑅𝑅1 + 𝑅𝑅2,𝑇𝑇ℎ,𝑒𝑒𝑒𝑒
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Substitute known values:
𝑣𝑣𝑇𝑇ℎ = 8 𝑉𝑉 ∗
Simplify:
5 Ω
5 Ω+6 Ω
𝑣𝑣𝑇𝑇ℎ = 3.64 𝑉𝑉
Combine the contribution from each voltage source:
𝑣𝑣𝐶𝐶 (∞) = 14.54 𝑉𝑉
Plug 𝑣𝑣𝐶𝐶 (∞) into the equation for a charging capacitor:
𝑣𝑣𝐶𝐶 (𝑡𝑡 > 0) = 14.54(1 − 𝑒𝑒 −𝑡𝑡/21.82 )
Sketch the function:
15
C
v (t)
10
5
0
0
20
40
60
80
100
120
time (s)
140
160
180
200
(d) Evaluate the ratio 𝒗𝒗𝑪𝑪 (𝒕𝒕) to 𝒗𝒗𝑪𝑪 (∞) at each of the following times: 𝒕𝒕 = 𝟎𝟎, 𝝉𝝉, 𝟐𝟐𝟐𝟐, 𝟓𝟓𝟓𝟓, 𝟏𝟏𝟏𝟏𝟏𝟏.
Use the result from (c) to evaluate each ratio:
𝑣𝑣𝐶𝐶 (0)
0 𝑉𝑉
=
=0
𝑣𝑣𝐶𝐶 (∞) 14.54 𝑉𝑉
𝑣𝑣𝐶𝐶 (𝜏𝜏)
9.19 𝑉𝑉
=
= 0.63
𝑣𝑣𝐶𝐶 (∞) 14.54 𝑉𝑉
𝑣𝑣𝐶𝐶 (2𝜏𝜏) 12.57 𝑉𝑉
=
= 0.86
𝑣𝑣𝐶𝐶 (∞) 14.54 𝑉𝑉
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𝑣𝑣𝐶𝐶 (5𝜏𝜏) 14.44 𝑉𝑉
=
= 0.99
𝑣𝑣𝐶𝐶 (∞) 14.54 𝑉𝑉
Problem 5.45
𝑣𝑣𝐶𝐶 (10𝜏𝜏) 14.54 𝑉𝑉
=
= 1.00
𝑣𝑣𝐶𝐶 (∞)
14.54 𝑉𝑉
For the circuit shown in Figure P5.41, assume that
switches S1 and S2 have been held closed for a long
time prior to t = 0. S1 then opens at t = 0; however, S2
does not open until t = 48 sec. Also assume R1 = 5ohm,
R2 = 4ohm, R3 = 3ohm, R4 = 6ohm, and C1 = C2 = 4F.
a. Find the capacitor voltage vC at t = 0.
b. Find the time constant τ for 0 < t < 48 sec.
c. Find vC for 0 < t < 48 sec.
d. Find τ for t > 48 sec.
e. Find vC for t > 48 sec.
f. Sketch vC for all time.
Solution:
Known quantities:
Circuit shown in Figure P5.41,
VS = 20 V, R1 = 5 Ω,R2 = 4 Ω,R3 = 3Ω,R4 = 6Ω,C 1 = 4 F ,C 2 = 4 F,I S = 4 A.
Find:
a)
b)
c)
d)
e)
f)
+
The capacitor voltage VC (t) at t = 0 .
The time constant τ for 0 ≤ t ≤ 48s .
The expression for VC (t) valid for 0 ≤ t ≤ 48s .
The time constant τ for t > 48s .
The expression for VC (t) valid for t > 48s .
Plot VC (t) for all time.
Assumptions:
Switch S1 opens at t = 0 ; switch S2 opens at t = 48s .
Analysis:
The approach here is to find the transient solution in the interval 0 < t < 48 seconds and use that solution to
determine the initial condition (the capacitor voltage) for the new transient after the switch S2 opens.
a) S1 and S2 have been closed for a long time and in DC steady state the capacitors can be replaced with open
circuits. Thus, by node analysis and voltage division
V (R + R3 + R4 )
VS − VC (0− )
VC (0− )R2
⇒
=
VC (0− ) = S 2
R1
R2 (R2 + R3 + R4 )
(R1 + R2 + R3 + R4 )
20(4 + 3+ 6) 260
=
V ≅ 14.4V
(5+ 4 + 3+ 6) 18
b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent
resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current
source with an open circuit) and computing (R2 + R3||R4).
RTH = R2 + (R3 || R4 ) = 4 + (3 || 6) = 4 + 2 = 6Ω
τ = RTH (C1 + C 2 ) = 6 ⋅ 8 = 48 s
c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage
when switch S2 opens at t = 48 s. To do so it is first necessary to find the complete transient solution for when only
the switch S1 is open (i.e. as if the switch S2 never opens.). The generalized solution for the transient is
VC (0+ ) = VC (0− ) =
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VC (t) = V (∞) + [V (0+ ) − V (∞)]e−t /τ
The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and
solving for the voltage across R3. This voltage is found readily by current division. Thus,
6Ω
(4 A)(3Ω) = 8V
3Ω + 6Ω
Substitute into the generalized solution given above to find
VC (t) = V (∞) + V (0+ ) − V (∞) e−t /τ = 8 + (14.4 − 8)e−t / 48 = 8 + 6.4e−t / 48 , 0 ≤ t ≤ 48
VC (∞) =
[
]
At t = 48 s, the capacitor voltage is
VC (t = 48− ) = 8 + 6.4e−1 = 10.35V
Continuity of voltage across the capacitors still holds so
VC (48+ ) = VC (48− ) = 10.35V
d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are
opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by
the 8 F capacitance is found by computing (R2 + R3).
RTH = R2 + R3 = 4 + 3 = 7Ω
τ = RTH (C1 + C 2 ) = 7 ⋅ 8 = 56 s
e) The generalized solution for the transient is
VC (t) = V (∞) + [V (t 0+ ) − V (∞)]e−(t−t0 ) /τ
The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources
are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus,
VC (∞) = 0V
Substitute into the generalized solution given above to find
VC (t) = V (48+ )e−t /τ = 10.35e−(t−48) / 56 , t > 48
f) The plot of VC (t) for all time is shown in the following figure.
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Problem 5.46
For the circuit shown in Figure P5.41, assume that
switches S1 and S2 have been held closed for a long
time prior to t = 0. S2 then opens at t = 0; however, S1
does not open until t = 96 sec. Also assume R1 = 5ohm,
R2 = 4ohm, R3 = 3ohm, R4 = 6ohm, and C1 = C2 = 4F.
a. Find the capacitor voltage vC at t = 0.
b. Find the time constant for 0 < t < 96 sec.
c. Find vC for 0 < t < 96 sec.
d. Find the time constant for t > 96 sec.
e. Use part (c) to find the capacitor voltage vC at
t = 96 sec and use it to find vC for t > 96 sec.
Solution:
Known quantities:
Circuit shown in Figure P5.41,
VS = 20 V, R1 = 5 Ω,R2 = 4 Ω,R3 = 3Ω, R4 = 6 Ω,C 1 = 4 F ,C 2 = 4 F, I S = 4 A.
Find:
+
a) The capacitor voltage VC (t) at t = 0 .
b) The time constant τ for 0 ≤ t ≤ 96s .
c) The expression for VC (t) valid for 0 ≤ t ≤ 96s .
d) The time constant τ for t > 96 s .
e) The expression for VC (t ) valid for t > 96s .
f) Plot VC (t) for all time.
Assumptions:
Switch S1 opens at t = 96s ; switch S2 opens at t = 0 .
Analysis:
The approach here is to find the transient solution in the interval 0 < t < 96 seconds and use that solution to
determine the initial condition (the capacitor voltage) for the new transient after the switch S1 opens.
a) S1 and S2 have been closed for a long time and in DC steady state the capacitors can be replaced with open
circuits. Thus, by node analysis and voltage division
VS − VC (0 − )
VC (0 − ) R2
V (R + R3 + R4 )
⇒
VC (0 − ) = S 2
=
(R1 + R2 + R3 + R4 )
R1
R2 (R2 + R3 + R4 )
(
)
260
6
4
3
20
+
+
V ≅ 14.4V
VC (0 + ) = VC (0 − ) =
=
(5 + 4 + 3 + 6) 18
b) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. The Thevenin equivalent
resistance seen by the 8 F capacitance is found by suppressing the independent sources (i.e. by replacing the current
source with an open circuit) and computing R1 (R2 + R3 ).
RTH = R1 (R2 + R3) = 5 (4 + 3) = 2.92Ω
τ = RTH (C1 + C 2 ) = 2.92 ⋅8 = 23.3s
c) As mentioned above, to find the complete transient solution for t > 0 it is necessary to find the capacitor voltage
when switch S1 opens at t = 96 s. To do so it is first necessary to find the complete transient solution for when only
the switch S2 is open (i.e. as if the switch S1 never opens.). The generalized solution for the transient is
VC (t) = V (∞) + [V (0+ ) − V (∞)]e−t /τ
The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and
solving for the voltage across R2 and R3 . This voltage is found readily by voltage division. Thus,
4Ω + 3Ω
VC (∞) =
(20V ) = 11.67V
5Ω + 4Ω + 3Ω
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Substitute into the generalized solution given above to find
[
]
VC (t ) = V (∞) + V (0 + ) − V (∞) e −t / τ = 11.67 + (14.4 − 11.67 )e −t / 23.3
− t / 23.3
,0 ≤ t ≤ 96
VC (t ) = 11.67 + 2.73e
At t = 96 s, the capacitor voltage is
VC (t = 96− ) = 11.67 + 2.73e−96 / 23.3 = 11.71V
Continuity of voltage across the capacitors still holds so
VC (96 + ) = VC (96 − ) = 11.71V
d) The two capacitors in parallel can be combined into one 8 F equivalent capacitor. When both the switches are
opened, there are no independent sources connected to the circuit. Thus, the Thevenin equivalent resistance seen by
the 8 F capacitance is found by computing (R2 + R3).
RTH = R2 + R3 = 4 + 3 = 7Ω
τ = RTH (C1 + C 2 ) = 7 ⋅ 8 = 56 s
e) The generalized solution for the transient is
VC (t) = V (∞) + [V (t 0+ ) − V (∞)]e−(t−t0 ) /τ
The long-term steady state capacitor voltage after the switch has been opened is zero since no independent sources
are connected and all the initial energy in the circuit is eventually dissipated by the resistors. Thus,
VC (∞) = 0V
Plug in to the generalized solution given above to find
VC (t) = V (96+ )e−t /τ = 11.71e−(t−96) / 56 , t > 96
f) The plot of VC (t) for all time is shown in the following figure.
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Problem 5.47
For the circuit in Figure P5.47, determine the value
of resistors R1 and R2, knowing that the time constant
before the switch opens is 1.5ms, and it is 10ms after
the switch opens. Assume: RS = 15 kohm, R3 = 30 kohm,
and C = 1μF.
Solution:
Known quantities:
RS =15kΩ , τ =1.5 ms, τ ' =10 ms, R3 = 30 kΩ , C = 1 µF .
Find:
The value of resistors R1 and R2 .
Assumptions:
None.
Analysis:
Before the switch opens:
Req = RS // R1 // R2 // R3
τ = Req C = 1.5ms
After the switch opens:
'
Req
= RS // R1
'
τ ' = Req
C = 10 ms
Solving the system of equations we have,
RSτ '
15000 ⋅ 0.01
=
= 30kΩ
R1 =
RS C − τ ' 15000 ⋅10−6 − 0.01
−1
−1 
C 1
1
1 
10−6
1
1
1 
 = 1875Ω
R2 =  −
−
−  = 
−
−
−
 τ RS R1 R3 
 0.0015 15000 30000 30000 
Problem 5.48
For the circuit in Figure P5.47, assume
VS = 100V, RS = 4 kohm, R1 = 2 kohm, R2 = R3 = 6kohm,
C = 1μF, and the circuit is in a steady-state condition
before the switch opens. Find the value of vC 2.666 ms
after the switch opens.
Solution:
Known quantities:
Circuit shown in Figure P5.47, VS = 100V , RS = 4 kΩ, R1 = 2 kΩ,R2 = R3 = 6 kΩ,C = 1 µF.
Find:
The value of the voltage across the capacitor after t = 2.666 ms.
Assumptions:
None.
Analysis:
Before opening, the switch has been closed for a long time. Thus we have a steady-state condition, and we treat the
capacitor as an open circuit. The voltage across the capacitor is equal to the voltage across the resistance R1. Thus,
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RS || R1 || R2 || R3
300
VS =
V ≅ 23.077V
RS
13
After the switch opens, the time constant of the circuit is
4000
1
Ω
τ = Req C =
Req = RS // R1 =
ms ≅ 1.3ms
3
750
+
−(t−t0 ) / τ
the generalized solution for the transient is VC (t) = V (∞) + [V (t 0 ) − V (∞)]e
The long-term steady state voltage across the capacitors is found by replacing them with DC open circuits and
solving for the voltage across R1. This voltage is found readily by voltage division. Thus,
R1
2000Ω
20
VC (∞) =
VS =
V ≅ 6.67V
20V =
RS + R1
4000Ω + 2000Ω
3
Substitute into the generalized solution given above to find
20  300 20  −750(t−t0 ) 20 640 −750(t−t0 )
e
+
=
− e
+
VC (t) = V (∞) + [V (t 0+ ) − V (∞)]e−(t−t0 ) /τ =
39
3
3
3  13
Finally,
20 640 −750(0.002666)
VC (t 0 + 2.666ms) =
+
e
= 8.888V
3
39
VC (t −0 ) = VC (t 0+ ) =
Problem 5.49
In the circuit in Figure P5.49, how long after the
switch is thrown at t = 0 will iL = 5A? Plot iL (t).
Solution:
Known quantities:
As described in Figure P5.49.
Find:
The time at which the current through the inductor is equal to
5 A, and the expression for iL (t ) for t ≥ 0 .
Assumptions:
None.
Analysis:
At t < 0 :
Using the current divider rule:
100
5
iL 0 − = (
)(
) = 66.5mA
1000 + 5 // 2.5 5 + 2.5
At t > 0 :
Using the current divider rule:
100
5
)(
) = 5.71A
iL (∞ )= (
10 + 5 // 2.5 5 + 2.5
()
To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor:
L
0.1
Req = 10 // 5 + 2.5 = 5.83Ω
τ=
=
= 17.1 ms
Req 5.83
Finally, we can write the solution:
−
iL (t ) = iL (∞)− (iL (∞ )− iL (0 ))e τ = 5.71 − (5.71 − 0.0665)e
Solving the equation we have,
t
()
− t
17.1×10 −3
= 5.71 − 5.64e
− t
17.1×10 −3
A
− tö
−3
iL tö = 5.71 − 5.64e 17.1×10 = 5
t? = 35.437ms
⇒
The plot of i L (t) for all time is shown in the figure.
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Problem 5.50
Refer to Figure P5.49 and assume that the switch
takes 5 ms to move from one contact to the other. Also
assume that during this time neither switch position
has electrical contact. Find:
a. iL (t) for 0 < t < 5ms;
b. The maximum voltage between the contacts during
the 5-ms duration of the switching.
Hint: This problem requires solving both a turn-off and a
turn-on transient problem.
Solution:
Known quantities:
As described in Figure P5.49.
Find:
The expression for i L (t) for 0 ≤ t ≤ 5ms . The maximum voltage between the contacts during the 5-ms duration of
the switch.
Assumptions:
The mechanical switching action requires 5 ms.
Analysis:
a)
For t < 0 :
Using the current divider rule:
100
5
iL 0 − = (
)(
) = 66.5mA
1000 + 5 // 2.5 5 + 2.5
For 0 ≤ t ≤ 5ms :
The long term steady state inductor current after the switch has been opened is zero since no independent source is
connected to the circuit and all the initial energy in the circuit is eventually dissipated by the resistors. Thus,
()
iL (∞ ) = 0 A
To find the time constant for the circuit we must find the Thevenin resistance seen by the inductor:
Req = 5+ 2.5 = 7.5Ω
L
0.1
= 13.33ms
=
Req 7.5
Finally, we can write the solution:
τ=
−t
− t
iL (t ) = iL (0 )e τ = (0.0665)e 13.33×10 A
(0 ≤ t ≤ 5ms)
b)
The voltage between the contacts during the 5-ms duration of the switching is equal to:
Vcont (t) = VS − V5Ω = 100 − (VL + 2.5i L )
where,
− t
di (t)
(0.1)(0.0665) − t 13.33×10 −3
−3
VL (t) = L L
=−
e
= (−0.5)e 13.33×10 V
−3
dt
13.33 × 10
Therefore,
−3
− t
− t
Vcont (t) = 100 − [(2.5)(0.0665) − 0.5]e 13.33×10 = 100 + (0.33)e 13.33×10 V
Thus, the maximum voltage between the contacts during the 5-ms duration of the switch is:
MAX
= Vcont (t = 0) = 100.33V
Vcont
−3
−3
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Problem 5.51
The circuit in Figure P5.51 includes a
voltage-controlled switch. The switch closes or opens
when the voltage across the capacitor reaches the value
VMC or V0G respectively. V0M 1V and the period of
the capacitor voltage waveform is 200 ms, find VMC
Solution:
Known quantities:
As described in Figure P5.51.
The switch closes when
C
;
the voltage across the capacitor voltage reaches vM
voltage reaches
vO
M
The switch opens when the voltage across the capacitor
= 1V . The period of the capacitor voltage waveform is 200 ms.
Find:
The voltage v C
M.
Assumptions:
The initial capacitor voltage is 1V and the switch has just opened.
Analysis:
With the switch open:
VC (∞ ) = 10 V
τ = RC = 0.15 s
−
−
VC (t ) = VC (∞)− [VC (∞)− VC (0 )]e τ = 10 − (10 − 1)e
t
Now we must determine the time when VC (t) =
Using the expression for the capacitor voltage:
t
0.15 = 10 − 9e
− t 0.15
V
vC
M.
 10 − v C 
M 
t 0 = −0.15ln

9


With the switch closed, the capacitor sees the Thevenin equivalent defined by:
10
Veq = VC (∞) =
× 10 ≈ 1× 10 − 2 V (voltage division) Req = 10kΩ 10Ω ≅10Ω τ = Req C = 0.15ms
10 + 10000
C
The initial value of this part of the transient is v M at t = t0. With these values we can write the expression for the
capacitor voltage:
vC
M = 10 − 9e
− t0 0.15
⇒
e
− t0 0.15
=
10 − v C
M
9
VC (t ) = VC (∞)− [VC (∞)− VC (t 0 )]e
−
⇒
(
(t−t0 )
)
τ = 0.01+ v C − 0.01 e
M
−
(t−t 0 )
0.15×10
−3
V
The end of one full cycle of the waveform across the 10-Ω resistor occurs when the second transient reaches
vO
M = 1V . If we call the time at which this event occurs t1,
at t = t1 = 200ms
then: VC (t ) = 1V
and so
(
)
1 = 0.01+ v C
M − 0.01 e
−(t1
−t0 )
0.15×10
−3
Graphically, the solution is the intersection between the
following two functions:
vC
M = 10 − 9e
− t0 0.15
(blue line)
(1 − 0.01)
vC
M = 0.01+
e
−(0.2−t0 )
0.15×10
(red line)
−3
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
C
which correspond to v M = 7.627 V and t 0 ≅ 0.1997 ms.
Problem 5.52
At t = 0 the switch in the circuit in Figure P5.52
closes. Assume that L1 = 1H, L2 = 5H, and that the
circuit is in DC steady-state for t < 0. Find iL (t) for all
time.
Solution:
Known quantities:
As describes in Figure P5.52. At t = 0 , the switch closes.
Find:
a)
i L (t) for t ≥ 0 .
Assumptions:
i L (0) = 0 A .
Analysis:
a)
In the long-term DC steady state after the switch is closed the inductors may be modeled as short circuits and
so all of the current from the source will travel through the inductors.
iL (∞) = 5 A
With the current source suppressed (treated as an open circuit) the Thevenin equivalent resistance seen by the
inductors in series and the associated time constant are
Req = 10 kΩ
Leq = L1 + L2 = 6 H
τ=
Leq
Req
= 0.6 ms


−
−t 
i L (t) = i L (∞)1 − e τ  = 51 − e

 
t

0.6×10 −3  A

Problem 5.53
At t = 0 the switch in the circuit in Figure P5.52
closes. Assume that L1 = 1H, L2 = 5H, and that the
circuit is in DC steady-state for t < 0. Find iL (t) for all
time.
Solution:
Known quantities:
As describes in Figure P5.52. At t = 0 , the switch closes.
Find:
a)
VL1 (t) for t ≥ 0 .
Assumptions:
i L (0) = 0 A .
Analysis:
a)
The voltage across either of the inductors is derived directly from the differential relationship between current
and voltage for an inductor.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
VL1 (t) = L1
−
di L (t)
d
= (1)(5) (1− e
dt
dt
t
0.6×10 −3 )

−
1
= 5
e
−3
 0.6 × 10
t

0.6×10 −3 

= 8.333e
−t
0.6×10 −3
kV
Problem 5.54
The analogy between electrical and thermal
systems can be used to analyze the behavior of a pot
heating on an electric stove. The heating element is
modeled as shown in Figure P5.54. Find the “heat
capacity” of the burner, CS , if the burner reaches 90%
of the desired temperature in 10 seconds. Assume
RS = 1.5ohm.
Solution:
Known quantities:
As describes in Figure P5.54.
Find:
The heat capacity of the burner, if the burner reaches 90 percent of the desired temperature in 10 seconds.
Analysis:
It is a first order dynamic system. We just need to calculate the time constant and related with the value of
capacitor.
τ Burner = RS CS = 1.5CS VCSS
So,
0.9 = 1-e-t/τ , −
So,
CS = 2.9 F
=
(
RS I S , VC (0 ) = 0 VC (t ) = RS I S 1 - e-t/τ
)
10
t
= 2.3
= −2.3 ,
1.5CS
τ
Problem 5.55
The burner and pot of Problem 5.54 can be
modeled as shown in Figure P5.55. R0 models the
thermal loss between the burner and the pot. The pot is
modeled by a thermal capacitance CP in parallel with a
thermal resistance RP .
a. Find the final temperature of the water in the pot—
that is, find V0 as t →∞– if: IS = 75 A; CP = 80
F; R0 = 0.8ohm; RP = 2.5ohm, and the burner is the
same as in Problem 5.54.
b. How long will it take for the water to reach 80% of
its final temperature?
[Hint: Neglect CS since CS _ CP .]
Solution:
Known quantities:
As describes in Figure P5.55.
Find:
a) Final temperature of the water in the pot.
b) Time takes for the water to reach 80 percent of its final
temperature.
Assumptions:
CS << C P
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Analysis:
a) As t→∞, the capacitors become open circuits, and we can
+
compute an equivalent circuit, as shown below.
R
+
-
In the circuit above VOC = I S RS = 75 ×1.5 = 112.5 V
RL
S
Voc
Rp
Vpss
Therefore,
VPSS =
VOC
112.5
RP =
2.5 = 58.6 V
RS + RL + RP
1.5 + 0.8 + 2.5
-
(b) The Thèvenin equivalent seen by the capacitance, Cp, is
shown:
RT
In the circuit above RT = (RS + RL )||Rp = 1.2 Ω
and the time constant for this circuit is: τ ′ = RT C p = 96 s
To find the 80% time we set 0.8 = 1-e
t = 155 s
-t/τ
VT
+
-
Cp
and solve for t:
Problem 5.56
The circuit in Figure P5.56 is used as a variable
delay in a burglar alarm. The alarm is a siren with
internal resistance of 1 k_. The alarm will not sound
until the current i0 exceeds 100 μA. Use a graphical or
numerical solution to find the range of the variable
resistor, R, for which the delay is between 1 and 2 s.
Assume the capacitor is initially uncharged.
Solution:
Known quantities:
As described in Figure P5.56.
Find:
The range of variable resistor.
Analysis:
When iL = 100 “A ,
1000 + R
1000 + R
vC =
× 100 × 10-6 × 1000 =
10000
1000
A general expression for the capacitor current is: VC (t ) =
At the time the alarm sounds, we have
1000 + R
R + 1000
vC =
=
×10 × 1 - e-t/™ ,
10000
21000 + R
(
)
(
)
R + 1000
× 10 × 1 - e -t/™
20000 + 1000 + R
 21000 + R 
-
- 1 = e-t/™
100000


The time constant is given by the expression τ = 100 “F [(1000 + R )||20000]
Substituting the expression for the time constant into the above equation we have one equation in one unknown, R.
This is a transcendental equation, and can be solved by iteration or by graphical analysis.
the solution is that R
must be between 33,580 Ω and 53,510 Ω or 33,580Ω ≤ R ≤ 53,510Ω
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.57
For t > 0, find the voltage v1 across C1 shown in
Figure P5.57. Let C1 = 5μF; C2 = 10μF. Assume the
capacitors are initially uncharged.
Solution:
Known quantities:
As describes in Figure P5.57.
Find:
The voltage across the C1 .
Analysis:
Since both two resistance and two capacitors are series connected, we can first calculate the equivalent capacity and
equivalent resistance, and then calculate the time constant.
CC
50
10
10
Ceq = 1 2 =
“F = “F ™= ReqCeq = (19 + 1) = 66.7 “s
C1 + C2 15
3
3
i(0 ) =
VC1(t ) =
[
]
t
1 t
66.7“S
− ‘/ 66.7“S
(0.5)e−λ/ 66.7“S
d‘ =
∫ 0.5e
0
5“F o
5“F
10V
= 0.5A
(1 + 19)ô
VC (t ) =
i (t )= i (0 )e - t/™= 0.5e - t/ 66.7“S
1
∫ i (λ )dλ
C
= 6.67- 6.67e−t/ 66.7“S V
Problem 5.58
The switch shown in Figure P5.58 opens at t = 0.
It closes at t = 10 seconds.
a. What is the time constant for 0 < t < 10 s?
b. What is the time constant for t > 10 s?
Solution:
Known quantities:
As describes in Figure P5.58
Find:
a)
b)
time constant for 0<t<10 s
time constant for t>10 s
Analysis:
a) With the switch open we must consider the following circuit.
5Ω
+
S -
1kΩ
4kΩ
v
5Ω
20 Ω
5.77
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1Ω
2.5 Ω
G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
To find the time constant for this circuit we must find the Thèvenin resistance seen by the capacitor:
RT = 1000 + 4000 + 2.5 + 5||(5||20 + 1) = 5005 ô
™= RT C = (5005 Ω )(1 “F ) = 5.005 ms
b) With the switch closed, the 1-kΩ and 4-kΩ resistors are not included since there is a short circuit across them
with the switch closed.
The resulting circuit is shown
5Ω
below.
vS +-
Now the Thèvenin resistance seen by the
capacitor is:
5Ω
RT = 2.5 + 5||(5||20 + 1) = 5 ô
20Ω
1µF
2.5Ω
1Ω
and the time constant is,
™= RT C = (5 Ω )(1 “F ) = 5 “s
Problem 5.59
The circuit in Figure P5.59 models the charging
circuit of an electronic camera flash. The flash should
be charged to VC ≥ 7.425V for each use. Assume
C = 1.5mF, R1 = 1kohm, and R2 = 1ohm.
a. How long does it take the flash to recharge after
taking a picture?
b. The shutter button stays closed for 1/30 s. How
much energy is delivered to the flash bulb R2 in that
interval? Assume the capacitor is fully charged.
c. If the shutter button is pressed 3 s after a flash, how
much energy is delivered to the bulb R2?
Solution:
Known quantities:
As describes in Figure P5.59
Find:
a) Time to wait for 99% recharge
b) Energy delivered to flush during 1/30 seconds.
c) Energy delivered to flush when not fully recharged.
Analysis:
a)
VC (t )= VBatt − VBatt e −t/™a
™
a = RC = (1000ô )(1500“F ) = 1.5 s
(
)
VC tready = 0.99 × 7.5 = 7.5 − 7.5e
−t ready /™
a
tready = -1.5 log e (0.01) = 6.9 s
b) We must find an expression for the voltage across the flash bulb (R2) after the shutter button has closed. This
is the circuit:
™= RT C = (1||1000)C ≈ 0.0015 s = 1.5 ms
i flash(t)
1K
5.78
+
7.5V
+
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only) to teachers
7.5=VC(t
µF
1Ω and
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-
G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
(
(
))
V falsh (t ) = VC (t ) = VC (∞ )− VC (∞ )− VC t flash e
−
t − t flash
™
t − t flash
V (t )
i flash (t )= C
R2
−
1
(7.5)−  1 (7.5)− 7.5 e 1.5ms
=
1000 + 1

 1000 + 1
−
t − t flash
= 7.5 + 7.4925e 1.5ms mV
1
30
∫ Vc λ i flash
t flash +
W=
()
(λ )dλ =
t flash
c)
1
30 
∫  7.5 + 7.4925e
0 

−
λ
1.5ms
λ 

−

.
1
5
ms dλ = 42.27 mJ
 7.5 + 7.4925e



In this case the capacitor has not fully charged but has achieved the value of
−
3s
VC (t = 3s ) = 7.5 − 7.5e 1.5s = 6.48 V
after the shutter switch closes at t = 3 s.
The voltage V flash becomes:
V falsh (t )= 7.5 + 6.4775e
−
t −3
1.5ms
mV
1
30 
λ 
λ 
−
−


W = ∫  7.5 + 6.4775e 1.5ms  7.5 + 6.4775e 1.5ms dλ = 31.6 mJ


0 



Problem 5.60
The ideal current source is (t) in Figure P5.60
switches levels as shown. Determine and sketch the
voltage vo(t) across the inductor for 0 < t < 2s.
Assume the inductor current is zero before t = 0,
R = 500_, and L = 50 H.
Known quantities:
As describes in Figure P5.60
Find:
Voltage-time curve
Analysis:
Applying KCL :
vo 1
+ ∫ v o dt = i S (t )
R L
Differentiating, we have
R
dvo
= − vo
dt
L
Solve it, we can have
vo (t )= vo (t0 )e
region I: t < 0
−
R
(t −t 0 )
L
= vo t0
( )e −10(t −t
0
)
v0 (t )= 0
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region II:
0< t < 0.8 s
Since the current through an inductor cannot change instantaneously
iL (t = 0 ) = 0 , iR (t = 0 ) = 10mA, vo (t = 0 ) = 5 V
voII (t )= 5 e −10t V
iRII (t )=
voII (t )
= 10 e −10t mA, iLII (t )= iS (t )− iRII (t )= 10 − 10 e −10t mA
R
region III:
(
0.8 < t < 0.9 sec
)
(
) ( )
( )
(t = 0.8 )= −(i (t = 0.8 )+ i (t = 0.8 ))= −12mA, v (t = 0.8 )= −6V
iL t = 0.8− = 10mA = iL t = 0.8+ , iR t = 0.8− = 0 , vo t = 0.8− = 0 V
iRIII
+
voIII (t ) = −6 e
region IV:
(
+
L
−10(t − 0.8 )
+
oIII
V
0.9 < t < 1.3 sec
)
+
SIII
(
)
(
)
( )
(t = 0.9 )= −0.2V
vo t = 0.9− = −2.2 V, iR t = 0.9− = −4.4mA,iL t = 0.9− = 2.4mA = iL t = 0.9+
(
iRIV t = 0.9
+
)= −(i (t = 0.9 )+ i (t = 0.9 ))= −0.4mA, v
voIV (t ) = −0.2 e
region V:
+
L
−10(t − 0.9 )
+
SIV
oIV
+
V
1.3 < t < 1.4 sec
( )
( ) ( )
( )
(t = 1.3 )= −(i (t = 1.3 )+ i (t = 1.3 ))= −2mA, v (t = 1.3 )= −1V
vo t = 1.3− = 0 V, iR t = 1.3− = 0 ,iL t = 1.3− = 2mA = iL t = 1.3−
+
iRV
voV (t )= − e
region VI:
(
+
L
−10(t −1.3)
+
oV
+
V
t > 1.4 sec
)
SV
(
)
( )
(
)− i (t = 1.4 )= 9.36mA, v (t = 1.4 )= 4.68V
vo t = 1.4− = −0.37 V, iR t = 1.4− = −0.74mA,iL t = 1.4− = 0.74mA = iL t = 1.4−
(
iRVI t = 1.4
+
)= i (t = 1.4
voV (t )= 4.68 e
Section 5.5:
SVI
−10(t −1.4 )
+
L
+
oVI
)
+
V
Transient response of Second-Order Circuits
Focus on Methodology – roots of second order systems
Case 1: Real and distinct roots. This case occurs when ζ>1, since the term under the square root is
positive in this case, and the roots are: s1,2 = −ζω n ± ω n ζ 2 − 1 .
overdamped response.
Case 2:
This leads to an
Real and repeated roots. This case holds when ζ=1, since the term under the square root is
zero in this case, and s1,2 = −ζω n = −ω n . This leads to a critically damped response.
Case 3: Complex conjugate roots. This case holds when ζ<1, since the term under the square root is
negative in this case, and s1,2 = −ζω n ± jω n 1 − ζ 2 . This leads to an underdamped response.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Focus on Methodology
Second-order transient response
1.
2.
3.
4.
5.
Solve for the steady-state response of the circuit before the switch changes state (t = 0-), and after
the transient has died out (t → ∞). We shall generally refer to these responses as x(0-) and x(∞).
Identify the initial conditions for the circuit, x(0+ ), and x?(0+ ) using continuity of capacitor
voltages and inductor currents (vC(0+) = vC(0-), iL(0+) = iL(0-)), and circuit analysis. This will be
illustrated by examples.
Write the differential equation of the circuit for t = 0+, that is, immediately after the switch has
changed position. The variable x(t) in the differential equation will be either a capacitor voltage,
vC(t), or an inductor current, iL(t). Reduce this equation to standard form (Equation 5.9, or 5.48).
Solve for the parameters of the second-order circuit: ωn and ζ .
Write the complete solution for the circuit in one of the three forms given below, as appropriate:
Overdamped case (ζ > 1):
 −ζω n +ω n ζ 2 −1 t

x(t) = x N (t) + x F (t) = α1e
Critically damped case (ζ = 1):
x(t) = x N (t) + x F (t) = α1e(
Underdamped case (ζ = 1):
−ζω n )t


−ζω n + jω n 1−ζ 2  t

x(t) = x N (t) + x F (t) = α1e
6.


−ζω n −ω n ζ 2 −1 t

+ α 2e 
−ζω n )t
+ α 2 te(
+ x(∞) t ≥ 0
+ x(∞) t ≥ 0


−ζω n − jω n 1−ζ 2 t

+ α 2e
+ x(∞) t ≥ 0
Apply the initial conditions to solve for the constants α1 and α2.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.61
In the circuit shown in Figure P5.61:
VS1 = 15V VS2 = 9V
RS1 = 130ohm RS2 = 290ohm
R1 = 1.1kohm R2 = 700ohm
L = 17mH C = 0.35μF
Assume that DC steady-state conditions exist for
t < 0. Determine the voltage vC across the capacitor
and the current iL through the inductor as t →∞.
Solution:
Known quantities:
Circuit shown in Figure P5.61,
VS 1 = 15V ,VS 2 = 9V , RS1 = 130 Ω,RS 2 = 290 Ω,R1 = 1.1kΩ,R2 = 700Ω,L = 17mH ,C = 0.35µF .
Find:
The voltage across the capacitor and the current through the
inductor and RS2 as t approaches infinity.
Assumptions:
The circuit is in DC steady-state conditions for t < 0 .
Analysis:
The conditions that exist at t < 0 have no effect on the long-term DC steady state conditions at t → ∞ . In the
long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit. In
this case, the inductor short circuits the R1 branch and the R2 C branch. Thus, the voltage across these branches
and the current through them are zero. In other words all of the current produced by the 9V source travels through
the inductor in this case.
V
9
i L (∞) = S2 =
= 31.03mA
RS2 290
Of course, this current is also the current traveling through the 290 Ω resistor.
i RS2 (∞) = i L (∞) = 31.03mA
And since the voltage across the inductor in the long-term DC steady state is zero (short circuit)
0 + VC (∞) + i R2 (∞)R2 = 0
VC (∞) = 0
Problem 5.62
For t > 0, determine the current iL through the
inductor and the voltage vC across the capacitor in
Figure P5.62. Assume VS = −1V for t < 0 but is
reversed to VS = 1V for t > 0. Also assume
R = 10ohm, L = 5mH, C = 100 μF, and that the circuit
was in DC steady-state prior to when the source was
reversed.
Solution:
Known quantities:
Circuit shown in Figure P5.62,
R = 10 Ω,L = 5mH ,C = 100µF , e(t ).
Find:
The current
iL (t )
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Assumptions:
The circuit is in DC steady-state conditions for t < 0 .
Analysis:
The conditions that exist at t < 0 have no effect on the long-term DC steady state conditions at t → ∞ . In the
long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit.
t<0
iL (t ) = e / R = −1 / 10 = −0.1A
VC (t ) = −e = 1V
t>0
iL (∞) = e∞ / R = 1 / 10 = 0.1A
Req = R || R
The circuit equation is:
1 diL
1
diL2
+
⋅
+
iL = 0
2
dt
Req C dt LC
iL (0) − 0.1A
With initial conditions diL
dt
=
t =0
v L (0)
= 400
L
The natural frequencies of the circuit are:
λ1, 2 = −1000 ± j1000 = 6 ± jω
Let
i (t ) = e 6t ( A cos ωt + B sin ωt ) + i∞
Imposing the initial conditions:
 A = I 0 − i∞ = 0.2
 A = −0.2


VL ⇒ 
6 A + ωB =
 B = 0.2

L

Problem 5.63
If the switch shown in Figure P5.63 is closed at
t = 0 and
VS = 170V RS = 7kohm
R1 = 2.3kohm R2 = 7 kohm
L = 30mH C = 130 μF
determine the current iL through the inductor and the
voltage vC across the capacitor as t →∞.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Solution:
Known quantities:
Circuit shown in Figure P5.63,
VS =170 V,RS = 7 kΩ, R1 = 2.3kΩ,R2 = 7 KΩ,L = 30 mH ,C = 130 µF.
Find:
The current through the inductor and the voltage across the
capacitor and R1 at steady state.
Assumptions:
None.
Analysis:
As t → ∞ , the circuit will return to DC steady state conditions. In the long-term DC steady state the inductor may
be modeled as a short circuit and the capacitor as an open circuit. Therefore, in this case, the current through R2 is
zero and thus the voltage across the capacitor must be equal to the voltage across R1. Furthermore, the current
through the inductor and R1 is simply VS/(RS + R1).
VS
170
iC (∞) = 0
i L (∞) = i R1 (∞) =
=
≅ 18.3 mA
RS + R1 7000 + 2300
VR1 (∞) = i L (∞) R1 = 18.28 ×10−3 × 2.3×10 3 = 42.04 V
and
VC (∞) = VR1 (∞) = 42.04 V
Problem 5.64
If the switch in the circuit shown in Figure P5.64 is
closed at t = 0 and
VS = 12V C = 130 μF
R1 = 2.3kohm R2 = 7 kohm
L = 30mH
determine the current iL through the inductor and the
voltage vC across the capacitor as t →∞.
Solution:
Known quantities:
Circuit shown in Figure P5.64,
VS =12 V,C = 130 µF, R1 = 2.3kΩ,R2 = 7 KΩ,L = 30 mH .
Find:
The current through the inductor and the voltage across the
capacitor and R1 at steady state.
Assumptions:
None.
Analysis:
As t → ∞ , the circuit will return to DC steady state conditions (practically after about 5 time constants.) In the
long-term DC steady state the inductor may be modeled as a short circuit and the capacitor as an open circuit.
Therefore, in this case, the voltage across the capacitor must be equal to the voltage across R2. Furthermore, the
current through the inductor and R2 is simply VS/(R1 + R2).
iC (∞) = 0
VL (∞) = 0
VS
12
i L (∞) = i R 2 (∞) =
=
= 1.29 mA
R1 + R2 9.3 × 10 3
VR1 (∞) = iS (∞) R1 = 1.29 × 10−3 × 2.3× 10 3 = 2.968V
And by observation
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VC (∞) = VR 2 (∞) = (1.29 × 10−3 )(7 × 10 3 ) = 9.03V
All answers are positive indicating that the directions of the currents and polarities of the voltages assumed initially
are correct. (You did do that, didn't you?)
Problem 5.65
If the switch shown in Figure P5.65 is thrown at 𝑡𝑡 = 0 and 𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 31 𝑘𝑘Ω, 𝑅𝑅2 = 22 𝑘𝑘Ω, 𝑅𝑅𝑆𝑆 =
100 Ω, 𝐿𝐿 = 0.9 𝑚𝑚𝑚𝑚, and 𝐶𝐶 = 0.5 𝜇𝜇𝜇𝜇, determine the current through 𝑅𝑅1 , 𝑖𝑖1 , and the voltage across 𝑅𝑅2 , 𝑣𝑣2 , as
𝑡𝑡 → ∞.
Known quantities:
𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 31 𝑘𝑘Ω, 𝑅𝑅2 = 22 𝑘𝑘Ω, 𝑅𝑅𝑆𝑆 = 100 Ω, 𝐿𝐿 = 0.9 𝑚𝑚𝑚𝑚, 𝐶𝐶 = 0.5 𝜇𝜇𝜇𝜇
Find:
The current through 𝑅𝑅1 , 𝑖𝑖1 , and the voltage across 𝑅𝑅2 , 𝑣𝑣2 , as 𝑡𝑡 → ∞.
Analysis:
At 𝑡𝑡 > 0, the source is not connected to the circuit.
Therefore, 𝑖𝑖1 and 𝑣𝑣2 are:
Problem 5.66
Therefore, as 𝑡𝑡 → ∞, the state of the circuit will decay to 0.
𝑖𝑖1 = 0 𝐴𝐴, 𝑣𝑣2 = 0 𝑉𝑉
For 𝑡𝑡 < 0, the circuit shown in Figure P5.66 is at DC steady-state, and the voltage across the capacitor, 𝑣𝑣𝐶𝐶 = 7 𝑉𝑉.
The switch is thrown at 𝑡𝑡 = 0. Assume: 𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 9.1 𝑘𝑘Ω, 𝑅𝑅2 = 𝑅𝑅3 = 4.3 𝑘𝑘Ω, and 𝐿𝐿 = 16 𝑚𝑚𝑚𝑚.
Determine the initial current through the inductor, 𝑖𝑖𝐿𝐿 , and the current through 𝑅𝑅2 , 𝑖𝑖2 , at 𝑡𝑡 = 0+ .
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Known quantities:
𝑉𝑉𝑆𝑆 = 12 𝑉𝑉, 𝑅𝑅1 = 9.1 𝑘𝑘Ω, 𝑅𝑅2 = 𝑅𝑅3 = 4.3 𝑘𝑘Ω, 𝐿𝐿 = 16 𝑚𝑚𝑚𝑚
Find:
The initial current through the inductor, 𝑖𝑖𝐿𝐿 , and the initial current through 𝑅𝑅2 , 𝑖𝑖2 , at 𝑡𝑡 = 0+ .
Analysis:
The current through an inductor cannot change instantaneously. Therefore, 𝑖𝑖𝐿𝐿 (0+ ) will be the same as its DC
steady-state value. Inductors behave like short-circuits in steady-state. Treat 𝐿𝐿 as such and use Ohm’s law to
determine 𝑖𝑖𝐿𝐿 (0+ ):
𝑖𝑖𝐿𝐿 (0+ ) =
Substitute known values:
Simplify:
𝑖𝑖𝐿𝐿 (0+ ) =
𝑉𝑉𝑆𝑆
𝑅𝑅3
12 𝑉𝑉
4.3 𝑘𝑘Ω
𝑖𝑖𝐿𝐿 (0+ ) = 2.79 𝑚𝑚𝑚𝑚
For 𝑖𝑖2 , recognize that current through a resistor can change instantaneously. At 𝑡𝑡 = 0+ , the 𝐶𝐶 and 𝐿𝐿 may be
treated like ideal voltage and current sources, respectively. Use the principal of superposition to solve for 𝑖𝑖2 . For
the ideal capacitor voltage source, 𝑉𝑉𝐶𝐶 , turn the ideal current source, 𝐼𝐼𝐿𝐿 , into an open-circuit. This action eliminates
𝑅𝑅3 from the circuit. Use Ohm’s law:
Substitute known values:
Simplify:
𝑖𝑖𝑉𝑉𝐶𝐶 =
𝑖𝑖𝑉𝑉𝐶𝐶 =
𝑉𝑉𝐶𝐶
𝑅𝑅1 + 𝑅𝑅2
7 𝑉𝑉
9.1 𝑘𝑘Ω + 4.3 𝑘𝑘Ω
𝑖𝑖𝑉𝑉𝐶𝐶 = 0.52 𝑚𝑚𝑚𝑚
Now, use superposition on 𝐼𝐼𝐿𝐿 . That is, turn 𝑉𝑉𝐶𝐶 into a short-circuit. Simplify circuit elements to realize 𝑖𝑖𝐼𝐼𝐿𝐿 .
First, combine 𝑅𝑅1 and 𝑅𝑅2 in parallel:
Substitute known values:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 =
𝑅𝑅1 𝑅𝑅2
𝑅𝑅1 + 𝑅𝑅2
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Simplify:
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 =
9.1 𝑘𝑘Ω ∗ 4.3 𝑘𝑘Ω
9.1 𝑘𝑘Ω + 4.3 𝑘𝑘Ω
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 ≈ 2.9 𝑘𝑘Ω
𝑅𝑅3 is in series with 𝑅𝑅1,2,𝑒𝑒𝑒𝑒 , so they share the current from 𝐼𝐼𝐿𝐿 .
𝑅𝑅1,2,𝑒𝑒𝑒𝑒 using Ohm’s law:
𝑣𝑣1,2,𝑒𝑒𝑒𝑒 = 𝑅𝑅1,2,𝑒𝑒𝑒𝑒 𝐼𝐼𝐿𝐿
Substitute known values:
Simplify:
Use this fact to calculate the voltage drop across
𝑣𝑣1,2,𝑒𝑒𝑒𝑒 = 2.9 𝑘𝑘Ω ∗ 2.79 𝑚𝑚𝑚𝑚
Use this voltage and Ohm’s law to calculate 𝑖𝑖2 :
Substitute known values:
𝑣𝑣1,2,𝑒𝑒𝑒𝑒 = 8.1 𝑉𝑉
𝑖𝑖𝐼𝐼𝐿𝐿 =
𝑖𝑖𝑉𝑉𝐿𝐿 =
Simplify:
𝑣𝑣1,2,𝑒𝑒𝑒𝑒
𝑅𝑅2
8.1 𝑉𝑉
4.3 𝑘𝑘Ω
𝑖𝑖𝐼𝐼𝐿𝐿 = 1.88 𝑚𝑚𝑚𝑚
Note that, according to P5.66, 𝑖𝑖𝐼𝐼𝐿𝐿 is flowing opposite 𝑖𝑖𝑉𝑉𝐶𝐶 .
answer:
𝑖𝑖2 = 𝑖𝑖𝑉𝑉𝐶𝐶 − 𝑖𝑖𝐼𝐼𝐿𝐿
Substitute known values:
Simplify:
Therefore, subtract these two values to obtain the final
𝑖𝑖2 = 0.52 𝑚𝑚𝑚𝑚 − 1.88 𝑚𝑚𝑚𝑚
𝑖𝑖2 = −1.36 𝑚𝑚𝑚𝑚
Problem 5.67
For t < 0, the circuit shown in Figure P5.67 is in
DC steady-state. Determine at t = 0+, immediately
after the switch is opened, the current iL through the
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inductor and the voltage vC across the capacitor.
VS1 = 15V VS2 = 9V
RS1 = 130ohm RS2 = 290ohm
R1 = 1.1 kohm R2 = 700ohm
L = 17mH C = 0.35μF
Solution:
Known quantities:
Circuit shown in Figure P5.67,
VS 1 =15 V, VS2 = 9 V, RS1 =130 Ω, RS2 = 290 Ω,R1 = 1.1 kΩ, R2 = 700 Ω, L = 17 mH ,C = 0.35 µF.
Find:
The current through and the voltage across the inductor
and the capacitor and the current through RS2 at t = 0+ .
Assumptions:
The circuit is in DC steady-state conditions for t < 0 .
Analysis:
Since this was not done in the specifications above, you
must note on the circuit the assumed polarities of voltages
and directions of currents.
At t = 0− :
Assume that steady state conditions exist. At steady state the inductor is modeled as a short circuit and the
capacitor as an open circuit. Choose a ground. Note that because the inductor is modeled as a short circuit, there
is no voltage drop from the top node to the bottom node and so before the switch there is no current through R1.
()
V L 0− = 0
Apply KCL:
()
iC 0− = 0
0 − VS1
0
0 − VS2
+ i L (0− ) +
+ 0+
=0
RS1
R1
RS2
i L (0− ) =
VS1 VS2
15
9
+
=
+
= 146.4 mA
RS1 RS2 130 290
() ()
(0 )= 0
VC 0− + iC 0− R2 = 0
Apply KVL:
VC
−
+
At t = 0 :
() ()
VC (0 )= VC (0− )= 0
iL 0+ = iL 0− = 146.4 mA
+
−VL (0+ ) + 0 + iC (0+ )R2 = 0
Apply KVL:
iC (0+ ) =
VL (0+ )
R2
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i L (0+ ) +
Apply KCL:
VL (0+ ) VL (0+ ) VL (0+ ) − VS2
+
+
=0
R1
R2
RS2
VS2
− i L (0+ )
R
V − i (0+ ) RS 2
S2
VL (0+ ) =
= S2 L
1
1
1
RS 2 RS2
+
+
+
+1
R1 R2 RS 2
R1
R2
=
9 − 146.4 × 10−3 × 0.29 × 10 3
= −19.94 V
0.29 0.29
+
+1
1.1
0.7
VL (0+ )
−19.94
=
= −28.49 mA
R2
0.7 × 10 3
−VL (0+ ) + iRS2 (0+ )RS 2 + VS2 = 0
iC (0+ ) =
Apply KVL again:
iRS2 (0+ ) =
VL (0+ ) − VS2 −19.94 − 9
= −99.79 mA
=
RS2
0.29 × 10 3
Problem 5.68
For 𝑡𝑡 < 0, the circuit shown in Figure P5.68 is in DC steady-state. The switch is closed at 𝑡𝑡 = 0. Determine the
current through the inductor, 𝑖𝑖𝐿𝐿 , for 𝑡𝑡 > 0. Assume: 𝑅𝑅 = 3 𝑘𝑘Ω, 𝑅𝑅2 = 600 Ω, 𝑉𝑉2 = 2 𝑉𝑉, 𝐶𝐶 = 2 𝑚𝑚𝑚𝑚, and 𝐿𝐿 =
1𝑚𝑚𝑚𝑚.
Known quantities:
𝑅𝑅 = 3 𝑘𝑘Ω, 𝑅𝑅𝑆𝑆 = 600 Ω, 𝑉𝑉𝑆𝑆 = 2 𝑉𝑉, 𝐶𝐶 = 2 𝑚𝑚𝑚𝑚, 𝐿𝐿 = 1𝑚𝑚𝑚𝑚
Find:
The current through the inductor, 𝑖𝑖𝐿𝐿 , for 𝑡𝑡 > 0.
Analysis:
This circuit is not in any usually recognizable format, so write and solve a differential equation that models it. First,
recognize that, at 𝑡𝑡 > 0, the switch is closed, eliminating 𝑅𝑅 from the circuit. Use KCL to write the equation:
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𝑖𝑖𝑆𝑆 = 𝑖𝑖𝐶𝐶 + 𝑖𝑖𝐿𝐿
where 𝑖𝑖𝑆𝑆 is the current through 𝑅𝑅𝑆𝑆 . Now, use nodal analysis to remove 𝑖𝑖𝑆𝑆 and the definition of current through
an inductor to remove 𝑖𝑖𝐿𝐿 :
𝑉𝑉𝑆𝑆 − 𝑣𝑣𝐶𝐶
= 𝑖𝑖𝐶𝐶 + 𝑖𝑖𝐿𝐿
𝑅𝑅𝑆𝑆
Simplify and substitute the definition of current through a capacitor:
𝑉𝑉𝑆𝑆
1
−
∫ 𝑖𝑖 𝑑𝑑𝑑𝑑 = 𝑖𝑖𝐶𝐶 + 𝑖𝑖𝐿𝐿
𝑅𝑅𝑆𝑆 𝐶𝐶𝑅𝑅𝑆𝑆 𝐶𝐶
Now, find a relationship between 𝑖𝑖𝐶𝐶 and 𝑖𝑖𝐿𝐿 . Realize that 𝐿𝐿 and 𝐶𝐶 share voltages:
𝑣𝑣𝐶𝐶 = 𝑣𝑣𝐿𝐿
Substitute the definition for voltage across a capacitor:
1
𝑑𝑑𝑖𝑖𝐿𝐿
∫ 𝑖𝑖𝐶𝐶 𝑑𝑑𝑑𝑑 = 𝐿𝐿
𝑑𝑑𝑑𝑑
𝐶𝐶
Simplify:
∫ 𝑖𝑖𝐶𝐶 𝑑𝑑𝑑𝑑 = 𝐶𝐶𝐶𝐶
Take the derivative:
𝑖𝑖𝐶𝐶 = 𝐶𝐶𝐶𝐶
𝑑𝑑𝑖𝑖𝐿𝐿
𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝑖𝑖𝐿𝐿
𝑑𝑑𝑡𝑡 2
Substitute these values into the equation derived using KCL:
𝑉𝑉𝑆𝑆
1
𝑑𝑑𝑖𝑖𝐿𝐿
𝑑𝑑 2 𝑖𝑖𝐿𝐿
−
𝐶𝐶𝐶𝐶
= 𝐶𝐶𝐶𝐶 2 + 𝑖𝑖𝐿𝐿
𝑅𝑅𝑆𝑆 𝐶𝐶𝑅𝑅𝑆𝑆
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡
Simplify:
𝐶𝐶𝐶𝐶
Simplify:
𝐿𝐿
𝑉𝑉𝑆𝑆
𝑑𝑑 2 𝑖𝑖𝐿𝐿 𝐿𝐿 𝑑𝑑𝑖𝑖𝐿𝐿
+
+ 𝑖𝑖𝐿𝐿 −
=0
2
𝑑𝑑𝑡𝑡
𝑅𝑅𝑆𝑆
𝑅𝑅𝑆𝑆 𝑑𝑑𝑑𝑑
𝐿𝐿 𝑑𝑑𝑖𝑖𝐿𝐿 1
𝑉𝑉𝑆𝑆
𝑑𝑑 2 𝑖𝑖𝐿𝐿
+
+ 𝑖𝑖𝐿𝐿 =
𝑑𝑑𝑑𝑑
𝑅𝑅𝑆𝑆
𝐶𝐶𝑅𝑅𝑆𝑆 𝑑𝑑𝑑𝑑 𝐶𝐶
Now, determine whether this circuit is critically-damped, over-damped, or under-damped.
this circuit is
𝐿𝐿
𝐶𝐶𝑅𝑅𝑆𝑆
Also, note that “R” for
:
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𝑅𝑅2 = �
Substitute known values:
𝑅𝑅2 = �
Simplify:
𝐿𝐿 2
�
𝐶𝐶𝑅𝑅𝑆𝑆
2
1 𝑚𝑚𝑚𝑚
�
2 𝑚𝑚𝑚𝑚 ∗ 600 Ω
𝑅𝑅2 = 6.94 ∗ 10−7
Note that 𝐿𝐿 and 𝐶𝐶 are identical to that of the typical RLC circuit:
4𝐿𝐿 4 ∗ 1 𝑚𝑚𝑚𝑚
=
𝐶𝐶
2 𝑚𝑚𝑚𝑚
Simplify:
𝑅𝑅2 <
4𝐿𝐿
𝐶𝐶
4𝐿𝐿
=2
𝐶𝐶
, so this circuit is under-damped, with solution:
𝑖𝑖𝐿𝐿 (𝑡𝑡) = 𝑒𝑒 −𝜁𝜁𝜁𝜁 (𝐴𝐴 cos(𝜔𝜔𝑛𝑛 𝑡𝑡) + 𝐵𝐵 sin(𝜔𝜔𝑛𝑛 𝑡𝑡))
For this circuit, the damping coefficient is:
𝐿𝐿
𝐶𝐶𝑅𝑅𝑆𝑆
𝜁𝜁 =
2𝐿𝐿
Simplify:
𝜁𝜁 =
Substitute known values:
Simplify:
And 𝜔𝜔𝑛𝑛 is:
Simplify:
𝜁𝜁 =
1
2𝐶𝐶𝑅𝑅𝑆𝑆
1
2 ∗ 2 𝑚𝑚𝑚𝑚 ∗ 600 Ω
𝜁𝜁 = 0.6
𝐿𝐿2
�1
𝐶𝐶 2 𝑅𝑅𝑆𝑆2
𝜔𝜔𝑛𝑛 =
−
𝐿𝐿𝐿𝐿
4𝐿𝐿2
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𝜔𝜔𝑛𝑛 = �
Substitute known values:
𝜔𝜔𝑛𝑛 = �
Simplify and take the positive answer:
1
1
− 2 2
𝐿𝐿𝐿𝐿 4𝐶𝐶 𝑅𝑅𝑆𝑆
1
1
−
1 𝑚𝑚𝑚𝑚 ∗ 2 𝑚𝑚𝑚𝑚 4 ∗ (2 𝑚𝑚𝑚𝑚)2 ∗ (600 Ω)2
𝜔𝜔𝑛𝑛 = 707 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
The solution to the differential equation becomes:
𝑖𝑖𝐿𝐿 (𝑡𝑡) = 𝑒𝑒 −0.6𝑡𝑡 (𝐴𝐴 cos(707𝑡𝑡) + 𝐵𝐵 sin(707𝑡𝑡))
Now, determine initial conditions on the circuit to solve. For 𝑖𝑖𝐿𝐿 (0), consider the circuit at steady-state.
capacitor is an open-circuit, and the inductor is a short-circuit. Therefore, 𝑖𝑖𝐿𝐿 (0) is:
Substitute known values:
𝑖𝑖𝐿𝐿 (0) =
𝑖𝑖𝐿𝐿 (0) =
Simplify:
The
𝑉𝑉𝑆𝑆
𝑅𝑅 + 𝑅𝑅𝑆𝑆
2 𝑉𝑉
3 𝑘𝑘Ω + 600 Ω
𝑖𝑖𝐿𝐿 (0) = 0.56 𝑚𝑚𝑚𝑚
Since the current through a capacitor cannot change instantaneously:
𝑑𝑑𝑖𝑖𝐿𝐿 (0)
= 0 𝐴𝐴
𝑑𝑑𝑑𝑑
Use these initial conditions to solve the differential equation:
𝑖𝑖𝐿𝐿 (0) = 𝑒𝑒 − 0 (𝐴𝐴 cos(0) + 𝐵𝐵 sin(0)) = 0.56 𝑚𝑚𝑚𝑚
Simplify:
Determine the derivative:
𝐴𝐴 = 0.56 𝑚𝑚𝑚𝑚
𝑑𝑑𝑖𝑖𝐿𝐿
= −0.6𝑒𝑒 −0.6𝑡𝑡 (𝐴𝐴 cos(707𝑡𝑡) + 𝐵𝐵 sin(707𝑡𝑡)) + (−707𝐴𝐴 sin(707𝑡𝑡) + 707𝐵𝐵 cos(707𝑡𝑡))𝑒𝑒 −0.6𝑡𝑡
𝑑𝑑𝑑𝑑
𝑑𝑑𝑖𝑖𝐿𝐿 (0)
= −𝐴𝐴 + 707 𝐵𝐵 = 0
𝑑𝑑𝑑𝑑
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Substitute known values:
−0.56 𝑚𝑚𝑚𝑚 + 707 𝐵𝐵 = 0
Simplify:
𝐵𝐵 = 0.79 𝜇𝜇𝜇𝜇
Plug A and B back into the solution to the differential equation:
Problem 5.69
𝑖𝑖𝐿𝐿 (𝑡𝑡) = 𝑒𝑒 −0.6𝑡𝑡 (0.56 𝑚𝑚𝑚𝑚 cos(707𝑡𝑡) + 0.79𝜇𝜇𝜇𝜇 sin(707𝑡𝑡))
Assume the switch in the circuit in Figure P5.69
has been closed for a very long time. It is suddenly
opened at t = 0 and then reclosed at t = 5 s.
Determine the inductor current iL and the voltage v
across the 2Ω resistor for t ≥ 0.
Solution:
Known quantities:
As described in Figure P5.69.
Find:
The inductor current iL and the voltage v
across the 2Ω resistor for t ≥ 0.
Assumptions:
The switch has been closed for a long time. It is suddenly opened at
. Variables observe the passive sign convention.
then reclosed at
and
Analysis:
For t < 0, the switch has been closed for a very long time so the circuit is in a DC steady-state, where the capacitor
acts as an open circuit and the inductor acts as a short-circuit. Thus, the voltage across the capacitor is 6V and the
current through the inductor is 2A.
:
For
The switch is now open. KCL at the upper right node yields:
or
(1)
KVL around the outer most loop yields:
or
(2)
Use (2) to substitute for vc in (1) to obtain a 2nd-order ODE in iL. The result is
Combining terms and multiplying both sides by -1:
Plug in for L and C to obtain:
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Finally, multiply both sides of the equation by 3/5 to obtain a 2nd-order ODE in standard form in iL.
(3)
Compare this 2nd-order ODE to the generalized 2nd-order ODE to write
Solving for ω0 and ζ:
Since ζ < 1 the response is underdamped. The roots are:
Therefore, the inductor current is of the form:
From the initial conditions:
Solving the above equations:
The solution for capacitor voltage will have the same form.
From the initial conditions:
Solving the above equations:
From the above results:
These are the initial conditions for the solution after the switch recloses. The voltage across the 2Ω resistor for 0 <
t < 5 sec is simply given by Ohm’s law
:
For
The switch is now closed. KCL at the upper right node yields:
or
(4)
KVL around the outer most loop yields:
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or
(5)
Use (5) to substitute for vc in (4) to obtain a 2nd-order ODE in iL. The result is
Plug in for L and C to obtain:
Compare this 2nd-order ODE to the generalized 2nd-order ODE to write
Solving for ω0 and ζ:
Since ζ < 1 the response is underdamped. The roots are:
Therefore, the inductor current is of the form:
From the initial conditions:
Solving the above equations:
The voltage across the 2Ω resistor for t > 5 sec is simply 6V.
Problem 5.70
Determine if the circuit in Figure P5.70 is overdamped or underdamped for t < 0 and t > 0. Also find the capacitance that results
in critical damping in both intervals. Assume VS = 15V, R = 200Ω, L = 20 mH, and C = 0.1μF.
Solution:
Known quantities:
As described in Figure P5.70.
Find:
Determine if the circuit is underdamped or overdamped. The capacitor value that results in critical damping.
Assumptions:
The circuit initially stores no energy.
The switch is closed at
t = 0.
Analysis:
a)
For t ≥ 0 :
The characteristic polynomial is:
1
Ls 2 + Rs + = 0
C
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The damping ratio is:
ζ =
1
200 ⋅ 10 −7
=
LC
2
RC
2
1010
<1
20
The system is underdamped, in fact we have the following complex conjugate roots:
s1, 2 = −5 × 10 3 ± j (2.1794 × 10 4 )
b)
The capacitor value that results in critical damping is:
RC
2
1
=1
LC
10 4 C 2
100
=1
2C
ζ =
⇒
⇒
100
=1
2C
2
C = 6 F = 2 µF
10
100C
Problem 5.71
For t < 0, assume the circuit in Figure P5.70 is in DC steady-state. If the switch is thrown at
a. Initial capacitor voltage v at t = 0+.
t = 0, find the:
C
b.
c.
d.
Capacitor voltage vC at t = 20 μs.
Capacitor voltage vC as t → ∞.
Maximum capacitor voltage.
Solution:
Known quantities: As described in Figure P5.70.
Find: The quantities listed in a.-d. listed above.
Assumptions: The circuit initially stores no energy. The switch is thrown at t = 0.
Analysis:
a.
b.
In DC steady-state the capacitor acts as an open-circuit and the inductor acts as a short-circuit. Both of
these state variables are continuous across the event of the switch being thrown. Thus, vC = 15V and iL=0 at
t = 0 +.
For a series LC circuit
Since ζ<1 the transient response is underdamped. The long-term DC steady-state response is a completely
discharged circuit because there is no source present for t > 0. Thus:
and
At t = 0+:
The latter expression can be solved for the constant B = 3.44 V. Finally, at t = 20µs:
c.
d.
The capacitor voltage as t → ∞ is zero because no source is present.
The maximum capacitor voltage is found when the derivative of vC is zero. That is, when the capacitor
current is zero. For the circuit shown in Figure 5.70 the capacitor current equals the inductor current.
Thus, set:
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But the inductor current is zero at t = 0 so that is when the capacitor voltage is at its maximum value. In
other words, the maximum capacitor voltage is
Problem 5.72
Assume the switch in the circuit in Figure P5.69 has been open for a very long time. It is suddenly closed at t = 0
and then reopened at t = 5 s. Determine the inductor current iL, the capacitor voltage vC, and the voltage v across the
2Ω� resistor for t ≥ 0.
Solution:
Known quantities: As described in Figure P5.69.
Find: The inductor current, the capacitor voltage, and the voltage across the 2Ω� resistor for t ≥ 0.
Assumptions: The circuit is in its DC steady-state for t < 0. The switch is closed at t = 0.
Variables observe the passive sign convention.
Analysis:
For 0 < t < 5 sec, KCL at the upper right node yields:
or
(1)
KVL around the outer most loop yields:
or
(2)
Use (2) to substitute for vc in (1) to obtain a 2nd-order ODE in iL. The result is
Plug in for L and C to obtain:
Compare this 2nd-order ODE to the generalized 2nd-order ODE to write
Solving for ω0 and ζ:
Since ζ < 1 the response is underdamped. The roots are
Therefore, the inductor current, the capacitor voltage, and the voltage v are
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where 2A and 6V are the long-term DC steady-state values for the inductor current and the capacitor voltage,
respectively. For t < 0 the voltage source is disconnected so that the initial conditions are trivial.
Plug in to find:
Or:
Also, from (1) and (2) above:
Using the above expressions for the inductor current and the capacitor voltage:
and
Solve for B1 and B2:
At t = 5 seconds, the inductor current and the capacitor voltage are
These values form the initial conditions on the t > 5 solution.
For t > 5, the switch is open such that KCL at the upper right node yields:
or
(3)
KVL around the outer most loop yields:
or
(4)
Use (4) to substitute for vc in (3) to obtain a 2nd-order ODE in iL. The result is
Combining terms and multiplying both sides by -1:
Plug in for L and C to obtain:
Finally, multiply both sides of the equation by 3/5 to obtain a 2nd-order ODE in standard form in iL.
(3)
Compare this 2nd-order ODE to the generalized 2nd-order ODE to write
Solving for ω0 and ζ:
Since ζ < 1 the response is underdamped. The roots are:
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Therefore, the inductor current and the capacitor voltage are of the form:
The initial conditions on these functions and their first derivatives are
and
and
Applying these conditions on the solutions for the inductor current and capacitor voltage yields:
Solve these linear algebraic equations for the unknown constants to find:
The voltage across the 2Ω resistor for t > 5 is easily determined from Ohm’s law since the current through the 2Ω
resistor is the inductor current but oppositely directed with respect to the passive sign convention. Thus:
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Problem 5.73
Assume that the circuit shown in Figure P5.70 is underdamped, and for t < 0, the circuit is in DC steady-state with
vC = VS . After the switch is closed at t = 0, the first two zero crossings of the capacitor voltage vC occur at t = 5π/3μs
and t = 5π μs. At t = 20π/3μs, the capacitor voltage vC peaks at 0.6VS . If C = 1.6μF, what are the values of R and L?
Solution:
Known quantities:
Circuit shown in Figure P5.70 where C = 1.6 µF. Assume DC steady-state
for t < 0. For t > 0, the capacitor voltage is zero at 5π/3 µs and 5π µs. At
20π/3 µs the capacitor voltage reaches a local maximum of 0.6VS.
Find:
The resistance R and the inductance L.
Assumptions:
The circuit is underdamped. Variables observe the passive sign convention.
Analysis:
The general solution for an underdamped response is
where
and
For t < 0, the capacitor acts as an open-circuit such that the initial conditions on the state variables are
KCL applied to the upper right node yields
or
Therefore, the initial condition on the derivative of the capacitor voltage is
These initial conditions on vC and its first derivative yield
and
or
At t1 = 5π/3 µs and t2 = 5π µs, the capacitor voltage is zero. Thus:
and
Since the exponential term is non-zero the result is
or
The peaks of the decaying sinusoid occur when:
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Collect coefficients of the sinusoidal functions to find:
Or
Plug in for the constants C1 and C2 to express the right side as
The result is that peaks of the decaying sinusoid occur when
Since t > 0 and ωd is positive only the positive terms on the right side are possible solutions. For this problem it is
given that one such peak occurs at t = 20π/3μs. This requirement leads to the conclusion that ωd is one of the
following values:
(1)
Also note that at the peaks sin(ωd t) = 0 because tan(ωd t) = 0 and cos(ωd t) = ±1. Thus, the peak values of the
capacitor voltage are on the curves:
At tp = 20π/3μs, vC =0.6VS. Thus:
Problem 5.74
For the circuit in Problem 5.70, what are the values of R and L so that the peak at t = 20π/3μs is vC =0.7VS? Assume
C=1.6μF.
Solution:
Known quantities:
The capacitance C and the peak value at t = 20π/3μs.
Find:
The values of R and L.
Assumptions:
For t < 0, the circuit is in DC steady-state. For t > 0, the circuit is underdamped. Variables observe the passive sign
convention.
Analysis:
For t < 0, the capacitor acts as an open-circuit such that the initial conditions on the state variables are
For t > 0, KVL around the RLC series loop yields
or
KCL applied to the upper right node yields
or
This expression can be substituted into the previous expression to yield
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Compare to the generalized second-order ODE to find
The circuit is assumed to be underdamped for t > 0, so
and the solution for every variable in the circuit is a decaying sinusoid. Thus:
where
and
The initial conditions on vC and its first derivative are
and
or
The peaks of the decaying sinusoid occur when:
Collect coefficients of the sinusoidal functions to find:
Or
Plug in for the constants C1 and C2 to express the right side as
The result is that peaks of the decaying sinusoid occur when
Since t > 0 and ωd is positive only the positive terms on the right side are possible solutions. For this problem it is
given that one such peak occurs at t = 20π/3μs. This requirement leads to the conclusion that ωd is one of the
following values:
(1)
Also note that at the peaks sin(ωd t) = 0 because tan(ωd t) = 0 and cos(ωd t) = ±1. Thus, the peak values of the
capacitor voltage are on the curves:
At tp = 20π/3μs, vC =0.7VS. Then:
(2)
Assume that the peak time is the first of the decaying sinusoid. Express (1) in terms of ωn and ζ and square both
sides of (1) and (2) to yield two equations.
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Thus:
Or
and
Finally, use the values to find
Problem 5.75
Determine iL for t > 0 in Figure P5.75, assuming i(0) = 2.5 A and vC (0) = 10 V.
Solution:
Known quantities:
All parameter values.
Find:
iL(t) for t > 0.
Assumptions:
Variables observe the passive sign convention.
Analysis:
For t > 0, KCL at the upper left node yields:
or
(1)
KVL around the right mesh yields:
or
(2)
Plug (2) into (1) to find a 2nd-order ODE in the inductor current.
Plug in for L and C and divide both sides by the coefficient of the inductor current to find:
(3)
Compare to the generalized 2nd-order ODE to find:
The result is
The damped natural frequency is
The transient response is underdamped such that the solution for the inductor current is
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or
Since i(0) = 2.5 A
Also, since vC (0) = 10 V, equation (2) yields:
where
Thus:
Problem 5.76
Find the maximum value of vC for t > 0 in Figure P5.76, assuming DC steady-state at t = 0−.
Solution:
Known quantities:
As described in Figure P5.76.
Find:
The maximum value of V .
Assumptions:
The circuit is in steady state at t = 0− .
Analysis:
V (0− ) = V (0+ ) = 0
Applying KVL:
d 2V
dt
−2t
2
dV
+ 4V = 48
dt
−2t
+4
Solving the differential equation: V = k1e + k 2te
From the initial condition:
V (0) = 0⇒ k1 = −12
k
dV (0)
i L (0) = C
⇒ 6 + 2 = 3⇒ k 2 = −12
4
dt
+ 12
V (t) = −12e−2t − 12te−2t + 12V
for t > 0
The maximum value of V is: Vmax = V (∞) = 12V
Problem 5.77
For t > 0, determine the time t at which i2 = 2.5A
in Figure P5.77, assuming DC steady-state at t = 0−.
Solution:
Known quantities:
As described in Figure P5.77.
Find:
The value of t such that i2 = 2.5A.
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Assumptions:
The circuit is in steady state at t = 0− . Variables observe the passive sign convention.
Analysis:
In steady state, the inductors behave as short circuits. Apply voltage division and Ohm’s law to find the initial
conditions on the inductor currents.
After the switch is closed, the inductors are part of a circuit comprised of two meshes with no independent source.
Apply KCL at the upper left node of that circuit to find:
or
(1)
Apply KVL around the right mesh of that circuit to find:
or
(2)
Combine (1) and (2) to find:
Compare to the generalized 2nd-order ODE to determine:
Thus, the transient solution is overdamped and the solution for i2 is
and so
Apply the two initial conditions on the inductor currents to solve for the unknown constants.
By inspection:
(3)
and
Also, use equations (1) and (2) to find:
Thus:
(4)
Solve (3) and (4) to find:
Finally,
Use this result and numerical analysis to find t0 that satisfies:
The result is t0 = 873 msec.
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Problem 5.78
For t > 0, determine the time t at which i1 = 6A in Figure P5.78, assuming DC steady-state at t = 0−.
Solution:
Known quantities:
As described in Figure P5.78.
Find:
The value of t such that i1 = 6A.
Assumptions:
The circuit is in steady state at t = 0− .
Analysis:
In steady state, the inductors behave as short circuits. Apply voltage division and Ohm’s law to find the initial
conditions on the inductor currents.
After the switch is closed, the inductors are part of a circuit comprised of two meshes with no independent source.
Apply KCL at the upper left node of that circuit to find:
or
(1)
Apply KVL around the right mesh of that circuit to find:
or
(2)
Combine (1) and (2) to find:
Compare to the generalized 2nd-order ODE to determine:
Thus, the transient solution is overdamped and the solution for i1 is
and so
Apply the two initial conditions on the inductor currents to solve for the unknown constants.
By inspection:
(3)
and
Also, use equation (1) to find:
Thus:
(4)
Solve (3) and (4) to find:
Finally,
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Use this result and numerical analysis to find t0 that satisfies:
The result is t0 = 694 msec.
Problem 5.79
For 𝑡𝑡 > 0, determine the time, 𝑡𝑡, at which 𝑣𝑣 = 7.5 V in Figure P5.79, assuming DC steady state at 𝑡𝑡 = 0− .
Known quantities:
Those in Figure P5.79.
Find:
The time, 𝑡𝑡, at which 𝑣𝑣 = 7.5 V.
Analysis:
Follow the steps stated in Section 5.4.
Step 1— Steady-state response for 𝒕𝒕 > 𝟎𝟎:
After the switch has been open for a long time, there is no source connected to the circuit. Therefore, there is no
energy in the circuit:
𝑣𝑣(∞) = 0
Step 2— Initial conditions at 𝒕𝒕 = 𝟎𝟎:
To solve the eventual differential equation, initial conditions must be provided. Because our desired state variable
is 𝑣𝑣, the initial conditions that must be found are 𝑣𝑣(0+ ) and 𝑑𝑑𝑑𝑑(0+ )/𝑑𝑑𝑑𝑑.
(Step 2) Find 𝒗𝒗(𝟎𝟎+ ):
To find the initial voltage, examine the circuit in steady-state. The capacitors are open-circuits, so 𝑣𝑣 will be the
voltage drop across the 2 Ω resistor. Combine the resistors to determine the voltage drop across the 2 Ω
resistor:
𝑉𝑉 = 𝐼𝐼 ∗ (3 + 2) ∥ 3
= 20 ∗
5∗3
5+3
= 37.5 𝑉𝑉
Use voltage division to determine the voltage drop across the 2 Ω resistor:
𝑣𝑣 =
(Step 2) Find 𝒅𝒅𝒅𝒅(𝟎𝟎+ )/𝒅𝒅𝒅𝒅:
2
∗ 37.5
2+3
= 15 𝑉𝑉
To find the change in voltage across the capacitor at 𝑡𝑡 = 0+ , use KCL on the first resistor:
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𝑉𝑉𝑅𝑅1 (0+ )
𝑑𝑑𝑑𝑑(0+ ) 𝑣𝑣(0+ )
= 𝐶𝐶
+
𝑑𝑑𝑑𝑑
𝑅𝑅2
𝑅𝑅1
Substitute known values:
𝑑𝑑𝑑𝑑(0+ ) 1 (37.5 − 15) 15
= �
− �
𝑑𝑑𝑑𝑑
𝐶𝐶
3
2
= 0 𝑉𝑉
Step 3—Differential Equation:
Now, find the differential equation that represents the circuit at 𝑡𝑡 > 0 by using KCL and KVL.
KCL yields:
𝑖𝑖𝐶𝐶1 = 𝑖𝑖𝐶𝐶2 + 𝑖𝑖𝑅𝑅2
where 𝐶𝐶1 and 𝐶𝐶2 are the left and right capacitors, respectively, in Figure P5.79 and 𝑅𝑅1 and 𝑅𝑅2 are the first and
second resistors, respectively.
Simplify the derived equation:
𝑑𝑑𝑉𝑉𝐶𝐶1
𝑑𝑑𝑑𝑑 𝑣𝑣
𝐶𝐶1
= 𝐶𝐶2
+
𝑑𝑑𝑑𝑑 𝑅𝑅2
𝑑𝑑𝑑𝑑
To eliminate the state variable 𝑉𝑉𝐶𝐶1 , use KVL:
𝑉𝑉𝐶𝐶1 = 𝑖𝑖1 𝑅𝑅1 + 𝑣𝑣
and take the derivative:
𝑑𝑑𝑉𝑉𝐶𝐶1
𝑑𝑑 2 𝑣𝑣 1 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
= 𝑅𝑅1 �𝐶𝐶2 2 +
�+
𝑑𝑑𝑡𝑡
𝑑𝑑𝑑𝑑
𝑅𝑅2 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Substitute this result into the equation obtained using KCL:
𝑑𝑑 2 𝑣𝑣 1 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1
𝑑𝑑𝑑𝑑
𝐶𝐶1 �𝑅𝑅1 �𝐶𝐶2 2 +
� + � = 𝐶𝐶2
+
𝑣𝑣
𝑑𝑑𝑡𝑡
𝑑𝑑𝑑𝑑 𝑅𝑅2
𝑅𝑅2 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
algebraically simplify the equation:
𝑑𝑑 2 𝑣𝑣 𝐶𝐶1 𝑅𝑅1 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 1
0 = 𝐶𝐶1 𝑅𝑅1 𝐶𝐶2 2 +
+ 𝐶𝐶1
− 𝐶𝐶2
−
𝑣𝑣
𝑑𝑑𝑡𝑡
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 𝑅𝑅2
𝑅𝑅2 𝑑𝑑𝑑𝑑
= 𝐶𝐶1 𝑅𝑅1 𝐶𝐶2
Step 4—Solve for 𝝎𝝎𝒏𝒏 and 𝜻𝜻:
=
𝑑𝑑 2 𝑣𝑣
𝐶𝐶1 𝑅𝑅1
𝑑𝑑𝑑𝑑 1
+�
+ 𝐶𝐶1 − 𝐶𝐶2 �
− 𝑣𝑣
𝑑𝑑𝑡𝑡 2
𝑅𝑅2
𝑑𝑑𝑑𝑑 𝑅𝑅2
1 𝑑𝑑 2 𝑣𝑣 1 𝑑𝑑𝑑𝑑 1
+
− 𝑣𝑣
12 𝑑𝑑𝑡𝑡 2 4 𝑑𝑑𝑑𝑑 2
Compare the second-order differential equation obtained in Step 3 to the standard form of equation 5.25:
1 𝑑𝑑 2 𝑥𝑥 2𝜁𝜁 𝑑𝑑𝑑𝑑
+
+ 𝑥𝑥 = 𝐾𝐾𝑆𝑆 𝑓𝑓(𝑡𝑡)
𝜔𝜔𝑛𝑛2 𝑑𝑑𝑡𝑡 2 𝜔𝜔𝑛𝑛 𝑑𝑑𝑑𝑑
Using the obtained equation in Step 3, match coefficients to equation 5.25:
𝜔𝜔𝑛𝑛 = �
1
12
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
= 0.29 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
Use this result to solve for 𝜁𝜁:
1
∗ 𝜔𝜔𝑛𝑛
𝜁𝜁 = 4
2
1
∗ 0.29
=4
2
= 0.036
Step 5—determine response type:
Because 𝜁𝜁 < 1, this is an underdamped response. Therefore, the solution is equivalent to equation 5.39:
where 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �1 − 𝜁𝜁 2 .
𝑣𝑣(𝑡𝑡) = 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 sin(𝜔𝜔𝑑𝑑 𝑡𝑡) + 𝛼𝛼2 cos(𝜔𝜔𝑑𝑑 𝑡𝑡)]
Use the initial conditions obtained in Step 2 to solve for 𝛼𝛼1 and 𝛼𝛼2 :
Step 6—Solve for constants 𝜶𝜶𝟏𝟏 and 𝜶𝜶𝟐𝟐 :
(Step 6) Find 𝜶𝜶𝟏𝟏 :
(Step 6) Find𝜶𝜶𝟐𝟐 :
𝑣𝑣(0) = 15 = 𝛼𝛼2
Now, take the derivative of 𝑣𝑣(𝑡𝑡):
𝑑𝑑𝑑𝑑(𝑡𝑡)
= −𝜁𝜁𝜔𝜔𝑛𝑛 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 sin(𝜔𝜔𝑑𝑑 𝑡𝑡) + 𝛼𝛼2 cos(𝜔𝜔𝑑𝑑 𝑡𝑡)] + 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 𝜔𝜔𝑑𝑑 cos(𝜔𝜔𝑑𝑑 𝑡𝑡) − 𝛼𝛼2 𝜔𝜔𝑑𝑑 sin(𝜔𝜔𝑑𝑑 𝑡𝑡)]
𝑑𝑑𝑑𝑑
Evaluate the derivative at 𝑡𝑡 = 0 and apply the initial conditions:
Trivially calculate 𝜔𝜔𝑑𝑑 :
0 = −𝜁𝜁𝜔𝜔𝑛𝑛 𝛼𝛼2 + 𝛼𝛼1 𝜔𝜔𝑑𝑑
𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �1 − 𝜁𝜁 2
= 0.29 ∗ �1 − 0.0362
Plug 𝜔𝜔𝑑𝑑 into the equation just derived:
= 0.29 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
0 = −𝜁𝜁𝜔𝜔𝑛𝑛 𝛼𝛼2 + 𝛼𝛼1 𝜔𝜔𝑑𝑑
𝛼𝛼1 =
𝜁𝜁𝜔𝜔𝑛𝑛 𝛼𝛼2
𝜔𝜔𝑑𝑑
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
=
0.036 ∗ 0.29 ∗ 15
0.29
= 0.54
The resulting equation for 𝑣𝑣(𝑡𝑡) is:
𝑣𝑣(𝑡𝑡) = 𝑒𝑒 −0.01𝑡𝑡 [0.54 sin(0.29𝑡𝑡) + 15 cos(0.29𝑡𝑡)]
Final step:
The equation obtained in the previous steps is plotted below. The time at which the voltage first equals 7.5V can be
solved graphically or numerically. The result is t = 3.68 seconds. Of course, the same voltage will recur at later
times as the solution oscillates until the exponentially decaying envelope falls below 7.5V. One can calculate the
latest time at which the voltage could be as large as 7.5V.
Answer: 𝒗𝒗 = 𝟕𝟕. 𝟓𝟓V at 3.68, 18.72, 24.91, 41.28, 45.88, 64.79, and 65.89 seconds.
Plot of Differential Equation
20
15
Amplitude of v (V)
10
5
0
-5
-10
-15
0
10
20
30
40
50
Time (s)
60
70
80
90
100
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.80
Assume the circuit in Figure P5.80 is in DC steady-state at t = 0− and L = 3H. Find the maximum value of v for t > 0.
Solution:
Known quantities:
As described in Figure P5.80.
Find:
The maximum value of V and the maximum voltage between the contacts of
the switches.
Assumptions:
The circuit is in steady state at t = 0− . L = 3 H.
Analysis:
At t = 0− :
i(0− ) = i (0+ ) =
10
=2A
5
V (0− ) = V (0+ ) = 0 V
After the switch is closed:
Applying KVL:
1 t
1 dV V
+ =0
∫ Vdt +
L −∞
12 dt 3
d 2V
dV 12
+ V =0
dt
L
dt
The particular response is zero for
⇒
2
+4
t > 0 because the circuit is source-free.
L = 3 H ⇒ s + 4s + 4 = 0
s1 = s2 = −2
2
V (t) = e−2t (A + Bt)
From the initial condition:
V (0+ ) = 0 = e 0 (A + B(0)) ⇒ A = 0
for t > 0
Substitute the solution into the original KCL equation and evaluate at t = 0+ :
1
1 d −2t
(0) +
[e (A + Bt)] |t=0 +2 = 0
3
12 dt
1
0 + [Be−2t − 2Bte−2t )] |t=0 +2 = 0⇒ B = −24
12
V (t) = −24te−2t V
for t > 0
The maximum absolute value of V is:
12
V ≅ 4.414V
Vmax = V (t = 0.5s) = −
e
The maximum voltage between the contacts of the switches is:
MAX
Vswitch
= VS = 10V
since the voltage between the contacts of the switches is a constant.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.81
Assume the circuit in Figure P5.80 is in DC steady-state at 𝑡𝑡 = 0− . Find the
value of the inductance, 𝐿𝐿, that makes the circuit critically damped for 𝑡𝑡 > 0.
Find the maximum value of 𝑣𝑣 for 𝑡𝑡 > 0.
Known quantities:
Those in Figure P5.80.
Find:
The value of 𝐿𝐿 that makes the circuit critically damped.
The maximum value of 𝑣𝑣 for 𝑡𝑡 > 0.
Analysis:
To determine the value of 𝐿𝐿, determine the differential equation for the circuit at 𝑡𝑡 > 0. Then, use equation 5.25
to solve for the appropriate constants. Determine the differential equation.
Step 1—DC steady-state response:
Before the switch is thrown, the inductor will be in steady-state (i.e., a short-circuit). Therefore, there is no voltage
across the capacitor:
𝑣𝑣𝐶𝐶 (0− ) = 0 𝑉𝑉
After the switch is thrown and a long time passes, there is no source attached to the circuit. Therefore, the voltage
across the capacitor is, again:
𝑣𝑣𝐶𝐶 (0− ) = 0 𝑉𝑉
Step 2—Initial conditions:
Find the initial conditions for the voltage across the capacitor: 𝑣𝑣𝐶𝐶 (0+ ) and 𝑑𝑑𝑣𝑣𝐶𝐶 (0+ )/𝑑𝑑𝑑𝑑.
The voltage across a capacitor cannot change instantaneously. Therefore:
𝑣𝑣𝐶𝐶 (0+ ) = 𝑣𝑣𝐶𝐶 (0− ) = 0 𝑉𝑉
+
To find 𝑑𝑑𝑣𝑣𝐶𝐶 (0 )/𝑑𝑑𝑑𝑑, use KCL:
𝑣𝑣𝐶𝐶
𝑑𝑑𝑣𝑣𝐶𝐶
− 𝐶𝐶
− 𝑖𝑖𝐿𝐿 = 0 𝐴𝐴
3 Ω
𝑑𝑑𝑑𝑑
+
We know that 𝑣𝑣𝐶𝐶 (0 ) = 0 V. Additionally, the inductor current may be trivially solved using Ohm’s law:
10
𝑖𝑖𝐿𝐿 =
2+3
= 2 𝐴𝐴
So the change in voltage across the capacitor may be solved as:
𝑑𝑑𝑣𝑣𝐶𝐶
2
=−
𝑑𝑑𝑑𝑑
𝐶𝐶
= −24 𝑉𝑉
Step 3—Find a second-order differential equation in one state variable:
Use KVL and KCL to find equations. From KCL, we obtained:
𝑣𝑣𝐶𝐶
𝑑𝑑𝑣𝑣𝐶𝐶
− 𝐶𝐶
− 𝑖𝑖𝐿𝐿 = 0
3
𝑑𝑑𝑑𝑑
Use KVL between the capacitor and inductor to obtain:
0 = 𝑣𝑣𝐶𝐶 − 𝑣𝑣𝐿𝐿
= 𝑣𝑣𝐶𝐶 − 𝐿𝐿
5.112
𝑑𝑑𝑖𝑖𝐿𝐿
𝑑𝑑𝑑𝑑
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Now, solve for 𝑖𝑖𝐿𝐿 in the equation obtained from KCL and take the derivative:
𝑑𝑑𝑖𝑖𝐿𝐿 1 𝑑𝑑𝑣𝑣𝐶𝐶
𝑑𝑑 2 𝑣𝑣𝐶𝐶
=
− 𝐶𝐶
𝑑𝑑𝑑𝑑
𝑑𝑑𝑡𝑡 2
3 𝑑𝑑𝑑𝑑
Plug this result into the KVL equation:
1 𝑑𝑑𝑣𝑣𝐶𝐶
𝑑𝑑 2 𝑣𝑣𝐶𝐶
− 𝐶𝐶
�=0
𝑣𝑣𝐶𝐶 − 𝐿𝐿 �
𝑑𝑑𝑡𝑡 2
3 𝑑𝑑𝑑𝑑
Simplify this equation into the standard from in equation 5.40:
𝑣𝑣𝐶𝐶 −
Plug in known values:
𝐿𝐿 𝑑𝑑𝑣𝑣𝐶𝐶
𝑑𝑑 2 𝑣𝑣𝐶𝐶
− 𝐿𝐿𝐿𝐿
=0
𝑑𝑑𝑡𝑡 2
3 𝑑𝑑𝑑𝑑
𝐿𝐿 𝑑𝑑 2 𝑣𝑣𝐶𝐶 𝐿𝐿 𝑑𝑑𝑣𝑣𝐶𝐶
+
− 𝑣𝑣𝐶𝐶 = 0
12 𝑑𝑑𝑡𝑡 2
3 𝑑𝑑𝑑𝑑
Step 4—Determine 𝜻𝜻 and 𝝎𝝎𝒏𝒏 :
To make this system critically damped, 𝜁𝜁 = 1. Use this condition when solving for 𝜁𝜁 and 𝜔𝜔𝑛𝑛 .
The standard
from in equation 5.25 is:
1 𝑑𝑑 2 𝑥𝑥 2𝜁𝜁 𝑑𝑑𝑑𝑑
+
+ 𝑥𝑥 = 𝑘𝑘𝑆𝑆 𝑓𝑓(𝑡𝑡)
𝜔𝜔𝑛𝑛2 𝑑𝑑𝑡𝑡 2 𝜔𝜔𝑛𝑛 𝑑𝑑𝑑𝑑
Match coefficients to solve for 𝜁𝜁 and 𝜔𝜔𝑛𝑛 .
Using the second equation,
If 𝜁𝜁 = 1, then two equations are obtained:
𝐿𝐿
1
=
12 𝜔𝜔𝑛𝑛2
Plugging this result into the first equation yields:
Simplify:
2
𝐿𝐿
=
𝜔𝜔𝑛𝑛 3
𝜔𝜔𝑛𝑛 =
𝐿𝐿
1
=
12
6 2
� �
𝐿𝐿
𝐿𝐿 =
which means 𝜔𝜔𝑛𝑛 = 2 rad/s.
6
𝐿𝐿
36
12
= 3 𝐻𝐻
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Step 5—Determine the type of the transient response:
The response is critically damped, because the equation was solved for 𝜁𝜁 = 1.
Step 6—Solve for the constants in the differential equation solution:
Because this circuit is critically damped, the equation can be written as:
where 𝜏𝜏 =
1
𝜔𝜔𝑛𝑛
𝑡𝑡
𝑣𝑣𝐶𝐶 (𝑡𝑡) = 𝑒𝑒 −𝜏𝜏 (𝛼𝛼1 + 𝛼𝛼2 𝑡𝑡)
.
Use the initial condition on 𝑣𝑣𝐶𝐶 (0+ ):
𝛼𝛼1 = 0
Now, take the derivative:
𝑡𝑡
𝑡𝑡
𝑑𝑑𝑣𝑣𝐶𝐶
1
= 𝑒𝑒 −𝜏𝜏 𝛼𝛼2 − 𝑒𝑒 −𝜏𝜏 𝛼𝛼2 𝑡𝑡
𝑑𝑑𝑑𝑑
𝜏𝜏
Evaluate at 𝑡𝑡 = 0:
𝛼𝛼2 = −24
The complete solution is:
Find the maximum value of 𝒗𝒗𝑪𝑪 for 𝒕𝒕 > 𝟎𝟎:
𝑣𝑣𝐶𝐶 (𝑡𝑡) = −24𝑒𝑒 −2𝑡𝑡 𝑡𝑡
To accomplish this result, plot the function from the previous portion. Figure P5.81.1 displays the result.
Note
that the voltage across the capacitor is flipped due to the notation in Figure P5.81.
Final answer: 𝑳𝑳 = 𝟑𝟑 H, 𝒗𝒗𝑪𝑪𝑪𝑪𝑪𝑪𝑪𝑪 = 𝟎𝟎 V
Plot of Differential Equation
0
-0.5
-1.5
C
Amplitude of v (V)
-1
-2
-2.5
-3
-3.5
-4
-4.5
0
0.5
1
1.5
2
2.5
Time (s)
3
3.5
4
4.5
5
Evaluating the graph gives 𝑣𝑣𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = 0 V.
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G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Problem 5.82
For 𝑡𝑡 > 0, determine 𝑣𝑣 in Figure P5.82, assuming DC steady-state at 𝑡𝑡 = 0− .
Known quantities:
Those in Figure P5.82.
Find:
𝑣𝑣 for 𝑡𝑡 > 0.
Analysis:
Determine and solve a differential equation in terms of 𝑣𝑣 (i.e., 𝑣𝑣 is the state variable).
Step 1—DC steady-state response:
Determine the steady-state value of 𝑣𝑣. The circuit is in steady-state before the switch is thrown. The inductor
acts like a short-circuit. Therefore, the 2 Ω resistor has the full 12 V dropped across it:
𝑣𝑣(0− ) = 12 𝑉𝑉
Step 2—Initial conditions:
(Step 2) Find 𝒗𝒗(𝟎𝟎+ ):
Find the initial conditions for the voltage across the resistor: 𝑣𝑣(0+ ) and 𝑑𝑑𝑑𝑑(0+ )/𝑑𝑑𝑑𝑑. Use KVL and note that the
voltage across a capacitor cannot change instantaneously:
𝑣𝑣(0+ ) = 12 − 𝑣𝑣𝐶𝐶 (0+ )
= 12 − 4
= 8 𝑉𝑉
(Step 2) Find 𝒅𝒅𝒅𝒅(𝟎𝟎+ )/𝒅𝒅𝒅𝒅:
Take the derivative of the equation obtained using KVL:
𝑑𝑑𝑑𝑑(0+ )
𝑑𝑑𝑣𝑣𝐶𝐶 (0+ )
=−
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
Use KCL to derive an equation for 𝑑𝑑𝑣𝑣𝐶𝐶 (0+ )/𝑑𝑑𝑑𝑑.
𝑣𝑣(0+ )
𝑑𝑑𝑣𝑣𝐶𝐶 (0+ )
= 𝑖𝑖𝐿𝐿 + 𝐶𝐶
𝑅𝑅
𝑑𝑑𝑑𝑑
Plug in known values and note that the current through an inductor cannot change simultaneously:
𝑑𝑑𝑣𝑣𝐶𝐶 (0+ )
8
= 4 ∗ � − 6�
𝑑𝑑𝑑𝑑
2
Therefore:
= −8 𝑉𝑉
𝑑𝑑𝑑𝑑(0+ )
= 8 𝑉𝑉
𝑑𝑑𝑑𝑑
Step 3—Find a second-order differential equation in one state variable:
Use KVL on the resistor and inductor:
12 − 𝑣𝑣 − 𝐿𝐿
5.115
𝑑𝑑𝑖𝑖𝐿𝐿
=0
𝑑𝑑𝑑𝑑
PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
Take the derivative of the KCL equation plug in the result from KVL on the resistor and capacitor:
𝑑𝑑𝑖𝑖𝐿𝐿 1 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝑣𝑣
=
+ 𝐶𝐶 2
𝑑𝑑𝑡𝑡
𝑑𝑑𝑑𝑑
𝑅𝑅 𝑑𝑑𝑑𝑑
Plug this result into the equation from KVL with the resistor and inductor:
Simplify:
𝑑𝑑 2 𝑣𝑣
1 𝑑𝑑𝑑𝑑
+ 𝐶𝐶
�=0
12 − 𝑣𝑣 − 𝐿𝐿 �
𝑑𝑑𝑑𝑑
𝑅𝑅 𝑑𝑑𝑑𝑑
𝐿𝐿𝐿𝐿
Step 4—Determine 𝜻𝜻 and 𝝎𝝎𝒏𝒏 :
⇒
𝑑𝑑 2 𝑣𝑣 𝐿𝐿 𝑑𝑑𝑑𝑑
+
+ 𝑣𝑣 = 12
𝑑𝑑𝑑𝑑
𝑅𝑅 𝑑𝑑𝑑𝑑
1 𝑑𝑑 2 𝑣𝑣 2 𝑑𝑑𝑑𝑑
+
+ 𝑣𝑣 = 12
5 𝑑𝑑𝑑𝑑
5 𝑑𝑑𝑑𝑑
Compare the equation derived in Step 3 to the standard from of equation 5.25:
and match coefficients.
1 𝑑𝑑 2 𝑥𝑥 2𝜁𝜁 𝑑𝑑𝑑𝑑
+
+ 𝑥𝑥 = 𝑘𝑘𝑆𝑆 𝑓𝑓(𝑡𝑡)
𝜔𝜔𝑛𝑛2 𝑑𝑑𝑡𝑡 2 𝜔𝜔𝑛𝑛 𝑑𝑑𝑑𝑑
(Step 4) Find 𝝎𝝎𝒏𝒏 :
𝜔𝜔𝑛𝑛 = √5
= 2.24 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠
(Step 4) Find 𝜻𝜻:
2
∗ 𝜔𝜔𝑛𝑛
𝜁𝜁 = 5
2
2
∗ 2.24
=5
2
Step 5—Determine the transient response:
= 0.45
The response is underdamped, because 𝜁𝜁 < 1.
Step 6—Solve for the constants in the differential equation:
Because the response is underdamped, the solution is of the form:
𝑣𝑣(𝑡𝑡) = 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 sin(𝜔𝜔𝑑𝑑 𝑡𝑡) + 𝛼𝛼2 cos(𝜔𝜔𝑑𝑑 𝑡𝑡)]
5.116
PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.
G. Rizzoni, Principles and Applications of Electrical Engineering, 6th Edition Problem solutions, Chapter 5
where 𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �1 − 𝜁𝜁 2 = 1.66 rad/s, according to equation 5.39.
Evaluate the solution to the differential equation using the initial condition for 𝑣𝑣(0+ ):
𝛼𝛼2 = 8
Take the derivative and evaluate the equation using the initial condition for 𝑑𝑑𝑑𝑑(0+ )/𝑑𝑑𝑑𝑑.
8 = −𝜁𝜁𝜔𝜔𝑛𝑛 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 sin(𝜔𝜔𝑑𝑑 𝑡𝑡) + 𝛼𝛼2 cos(𝜔𝜔𝑑𝑑 𝑡𝑡)] + 𝑒𝑒 −𝜁𝜁𝜔𝜔𝑛𝑛 𝑡𝑡 [𝛼𝛼1 𝜔𝜔𝑑𝑑 cos(𝜔𝜔𝑑𝑑 𝑡𝑡) − 𝛼𝛼2 𝜔𝜔𝑑𝑑 sin(𝜔𝜔𝑑𝑑 𝑡𝑡)]
Evaluate at 𝑡𝑡 = 0:
Solve for 𝛼𝛼1 :
8 = −𝜁𝜁𝜔𝜔𝑛𝑛 𝛼𝛼2 + 𝛼𝛼1 𝜔𝜔𝑑𝑑
𝛼𝛼1 =
=
Therefore, the equation for 𝑣𝑣 is:
8 + 𝜁𝜁𝜔𝜔𝑛𝑛 𝛼𝛼2
𝜔𝜔𝑑𝑑
8 + 0.45 ∗ 2.24 ∗ 8
1.66
= 9.68
𝑣𝑣(𝑡𝑡) = 𝑒𝑒 −1.01𝑡𝑡 [9.68 sin(1.66𝑡𝑡) + 8 cos(1.66𝑡𝑡)]
Final answer: 𝒗𝒗(𝒕𝒕) = 𝒆𝒆−𝟏𝟏.𝟎𝟎𝟎𝟎𝟎𝟎 [𝟗𝟗. 𝟔𝟔𝟔𝟔 𝐬𝐬𝐬𝐬𝐬𝐬(𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔) + 𝟖𝟖 𝐜𝐜𝐜𝐜𝐜𝐜(𝟏𝟏. 𝟔𝟔𝟔𝟔𝟔𝟔)]
5.117
PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and
educators for course preparation. If you are a student using this Manual, you are using it without permission.