‫‪July 2013‬‬
‫‪Chapter 7‬‬
‫‪Response of First‬‬‫‪Order RL and RC‬‬
‫‪Circuits‬‬
‫ِزوشاد ششذ ‪ٚ‬رّبس‪ِ ٓ٠‬سٍ‪ٌٛ‬خ‬
‫اِزسبٔبد عبثمخ ٌٍؼذ‪٠‬ذ ِٓ اٌّ‪ٛ‬اد أدٔبٖ‬
‫ِزبزخ ِدبٔب ًا ػٍ‪ ٝ‬اٌّ‪ٛ‬لؼ‪ ٓ١‬اٌّزو‪ٛ‬س‪ ٓ٠‬أدٔبٖ‬
‫رغزّش أغٍت اٌصذالبد‬
‫طبٌّب وبٔذ اٌؼاللخ ِغٍ‪١‬خ‪.‬‬
‫إٌ‪ٛ‬ربد ِدبٔ‪١‬خ ٌٍٕفغ اٌؼبَ ف‪١‬شخ‪ ٝ‬اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‪ ٞ‬خطأ أ‪ِ ٚ‬الزظبد رشا٘ب ضش‪ٚ‬س‪٠‬خ ثشعبٌخ ٔص‪١‬خ ‪ 9 4444 260‬أ‪ ٚ‬ثبٌجش‪٠‬ذ اإلٌىزش‪ٟٔٚ‬‬
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July 2013
 In this chapter, we will focus on circuits that consist only of sources, resistors, and
either (but not both) inductors or capacitors. For brevity, such configurations are
called RL (resistor-inductor) and RC (resistor-capacitor) circuits.
 Our analysis of RL and RC circuits
will be divided into three phases.
In the first phase, we consider the
currents and voltages that arise when
stored energy in an inductor or
capacitor is suddenly released to a resistive network. This happens when the
inductor or capacitor is abruptly disconnected from its dc source. Thus we can
reduce the circuit to one of the two equivalent forms shown in Fig. 7.1.The
currents and voltages are called natural response to emphasize that the nature of
the circuit itself, not external sources.
 In the second phase we consider the currents and voltages that arise when energy is
being acquired by an inductor or capacitor due to the sudden application of a dc
voltage or current source. This response is referred to as the step response. The
process for finding both the natural and step responses is the same.
 In the third phase of our analysis, we develop a general method that can be used to
find the response of RL and RC circuits to any abrupt change in a dc voltage or
current source.
ٓ‫ٌئ‬ٚ ،‫ػه‬ِٛ‫غزسك د‬٠ ‫ال أزذ‬
.‫ب‬ٙ‫ذػه رزسف‬٠ ٍٓ‫ب أزذ ف‬ٙ‫اعزسم‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Figure 7.2 shows the four possibilities for
the general configuration of RL and RC
circuits. Note that when there are no
independent sources in the circuit, the
Thévenin voltage or Norton current is zero,
and we have a natural-response problem.
 RL and RC circuits are also known as firstorder circuits, because their voltages and
currents
are
described
by
first-order
differential equations. No matter how
complex a circuit may appear, if it can be
reduced to a Thévenin or Norton equivalent
connected to the terminals of an equivalent
inductor or capacitor, it is a first-order
circuit. (If multiple inductors or capacitors
exist in the original circuit, they must be
replaced by a single equivalent element.)
7.1 The Natural Response of an RL Circuit
 We assume that the independent current source generates a constant current of 𝐼𝑠
for a long time. Therefore the inductor appears as a short circuit (𝐿 𝑑𝑖 𝑑𝑡 = 0)
prior to the release of the stored energy.
،‫ك‬٠‫ ِٓ أخً صذ‬ٟ‫ظ ِٓ اٌصؼت أْ رُضَّس‬١ٌ
.‫خ‬١‫غزسك اٌزضس‬٠ ٞ‫ٌىٓ ِٓ اٌصؼت أْ ردذ اٌز‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Because the inductor appears as a short
circuit, the voltage across the inductive
branch is zero, and there can be no
current in either 𝑅𝑜 or 𝑅. Therefore, all
the source current 𝐼𝑠 appears in the
inductive branch. If switch is opened the
circuit shown in Fig. 7.3 reduces to the
one shown in Fig. 7.4.
 Deriving the Expression for the Current:
To find 𝑖 𝑡 , we use Kirchhoff's voltage law
𝐿
𝑑𝑖
+ 𝑅𝑖 = 0
𝑑𝑡
Solving,
Natural response of an RL circuit → 𝑖 𝑡 = 𝐼𝑜 𝑒 − 𝑅 𝐿 𝑡 ,
𝑡≥0
Which shows that the current starts from an initial value 𝐼𝑜 and decreases toward zero
𝑣 0− = 0
𝑣 0+ = 𝐼𝑜 𝑅
𝑝 = 𝑖 2 𝑅 = 𝐼𝑜2 𝑅𝑒 −2
𝑡
𝑤=
𝑝 𝑑𝑥 =
0
𝑅 𝐿 𝑡
𝑡 ≥ 0+
,
1 2
𝐿𝐼0 1 − 𝑒 −2
2
𝑅 𝐿 𝑡
,
𝑡≥0
As 𝑡 becomes infinite, the energy dissipated in the resistor approaches the initial
energy stored in the inductor.
‫لغ‬ٚ ٞ‫مه اٌز‬٠‫لبزخ أْ رغأي صذ‬ٌٛ‫ِٓ ا‬
.‫سزبج ٌّغبػذره‬٠ ْ‫سطخ ئْ وب‬ٚ ٟ‫ف‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 The time constant 𝛕 of the circuit:
𝜏 = time constant =
𝐿
𝑅
Using the time-constant concept
𝑖 𝑡 = 𝐼0 𝑒 −𝑡 𝜏 ,
𝑡≥0
𝑣 𝑡 = 𝐼0 𝑅𝑒 −𝑡 𝜏 ,
𝑡 ≥ 0+
𝑝 = 𝐼02 𝑅𝑒 −2𝑡 𝜏 ,
1
𝑤 = 𝐿𝐼02 1 − 𝑒 −2𝑡
2
𝑡 ≥ 0+
𝜏
,
𝑡≥0
 Table 7.1 gives the value of 𝑒 −𝑡 𝜏 for
integral multiples of 𝜏 from 1 to 10. We
say that five time constants after
switching has occurred, the currents and
voltages have reached their final values.
 Calculating the natural response of an 𝑅𝐿 circuit can be summarized as follows:
1. Find the initial current, 𝐼𝑜 , through the inductor.
2. Find the time constant of the circuit, 𝜏 = 𝐿 𝑅.
3. Use Eq. 7.15, 𝑖 𝑡 = 𝐼𝑜 𝑒 −𝑡 𝜏 .t0 generate 𝑖 𝑡 from 𝐼𝑜 and 𝜏.
4. All other calculations of interest follow from knowing 𝑖 𝑡 .
ٜ‫ اٌّشء أْ رش‬ٍٝ‫ب ػ‬١ٔ‫ِٓ ٔىذ اٌذ‬
.‫ا ٌه ِب ِٓ صذالزٗ ثذ‬ٚ‫ػذ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Example 7.1: Determining the Natural Response of an RL Circuit
The switch in the circuit shown in Fig. 7.7 has been closed for a long time before it is
opened at 𝑡 = 0. Find
a) 𝑖𝐿 𝑡 for 𝑡 ≥ 0.
b) 𝑖0 𝑡 for 𝑡 ≥ 0+ .
c) 𝑣0 𝑡 for 𝑡 ≥ 0+.
d) The percentage of the total energy stored in the 2 H inductor that is dissipated in the
10 Ω resistor.
 Solution:
a) @ 𝑡 = 0−,
𝑣 0 = 0,
𝑖𝐿 0− = 𝑖𝐿 0+ = 20 A
Because an instantaneous change in the current cannot occur in an inductor. We
replace the resistive circuit connected to the terminals of the inductor with a single
resistor
𝑅𝑒𝑞 = 2 + 40 ∥ 10 = 10 Ω
1
2
𝜏=
= = 0.2 s
𝑅𝑒𝑞 𝑇0
𝑖𝐿 𝑡 = 20 𝑒 −5𝑡 A, 𝑡 ≥ 0
b) Looking at bottom side:
10
𝑖0 = − 𝑖𝐿
= − 4 𝑒 −5𝑡 A, 𝑡 ≥ 0+
10 + 40
c) 𝑣0 𝑡 = 40𝑖0 = − 160 𝑒 −5𝑡 V,
d) 𝑝10Ω
𝑡 ≥ 0+
𝑣02
𝑡 =
= 2560 𝑒 −10𝑡 W,
10
𝑡 ≥ 0+
∞
2560 𝑒 −10𝑡 𝑑𝑡 = 256 J
𝑤10Ω 𝑡 =
0
1
1
𝑤 0 = 𝐿 𝑖 2 0 = 2 20
2
2
2
= 400 J
256
100 = 64%
400
‫ّب ًا ال‬١‫ْ االعزؼذاد ػظ‬ٛ‫ى‬٠ ‫ث‬١‫ز‬
.‫ّخ‬١‫ثبد ػظ‬ٛ‫ْ اٌصؼ‬ٛ‫ّىٓ أْ رى‬٠
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Example 7.2: Determining the Natural Response of an RL Circuit with Parallel Inductors
In the circuit shown in Fig. 7.8, the initial currents in inductors 𝐿1 and 𝐿2 have been
established by sources not shown. The switch is opened at 𝑡 = 0.
a) Find 𝑖1 , 𝑖2 , and 𝑖3 for 𝑡 ≥ 0.
b) Calculate the initial energy stored in the parallel inductors.
c) Determine how much energy is stored in the inductors as 𝑡 → ∞.
d) Show that the total energy delivered to the resistive network equals the difference
between the results obtained in (b) and (c).
Fig. 7.8
 Solution:
a) We can easily find 𝑣 𝑡 if we reduce the circuit shown in Fig. 7.8 to the equivalent
form shown in Fig. 7.9. The parallel inductors simplify to an equivalent inductance
of 4 H, carrying an initial current of 12 A. The resistive network reduces to a single
resistance of 8 Ω. So the initial value of 𝑖 𝑡 is 12 A and the time constant is 4 8, or
0.5 s.
𝑖 𝑡 = 12 𝑒 −2𝑡 A,
𝑡≥0
𝑣 𝑡 = 96 𝑒 −2𝑡 V,
𝑡 ≥ 0+
The circuit shows that 𝑣 𝑡 = 0 at 𝑡 = 0−, so the expression for 𝑣 𝑡 is valid for
𝑡 ≥ 0+ .
1
𝑖1 =
5
1
𝑖2 =
20
𝑖3 =
𝑡
96 𝑒 −2𝑥 𝑑𝑥 − 8 = 1.6 − 9.6 𝑒 −2𝑡 A,
𝑡≥0
0
𝑡
96 𝑒 −2𝑥 𝑑𝑥 − 4 = − 1.6 − 2.4 𝑒 −2𝑡 A, 𝑡 ≥ 0
0
𝑣(𝑡) 15
= 5.76 𝑒 −2𝑡 A,
10 25
𝑡 ≥ 0+
ٟ‫أر‬٠ ،ً‫خطظ ٌٍفش‬٠ ‫ال أزذ‬
.‫اٌفشً ػٕذِب ال ٔخطظ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Example 7.2):
The expressions for the inductor currents 𝑖1 and 𝑖2 are valid for 𝑡 ≥ 0, where the
expression for the resistor current 𝑖3 is valid for 𝑡 ≥ 0+.
Fig. 7.9
b) 𝑤 =
1
5 64 +
2
1
2
20 16 = 320 J
c) As 𝑡 → ∞, 𝑖1 → 1.6 A
𝑤=
1
5 1.6
2
∞
d) 𝑤 =
2
+
1
20 − 1.6
2
∞
𝑃𝑑𝑡 =
0
and
1152 𝑒
−4𝑡
0
𝑖2 → − 1.6 A
2
= 32 J
𝑒 −4𝑡 ∞
𝑑𝑡 = 1152
= 288 J
−4 0
This is the difference between the initially stored energy 320 J and the energy in
the parallel inductors 32 J .
‫ زممذ‬ٟ‫ازذح ػّاللخ اٌز‬ٚ ‫ح‬ٛ‫ظ ٕ٘بن خط‬١ٌ
.‫شح‬١‫اد صغ‬ٛ‫ػخ خط‬ّٛ‫ ئّٔب ِد‬،‫اإلٔدبص‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Assessment problem 7.1:
The switch in the circuit shown has been closed for a long time and is opened at 𝑡 = 0.
a) Calculate the initial value of 𝑖.
b) Calculate the initial energy stored in the inductor.
c) What is the time constant of the circuit for 𝑡 > 0?
d) What is the numerical expression for 𝑖 𝑡 for 𝑡 ≥ 0?
e) What percentage of the initial energy stored has been dissipated in the 2 Ω resistor
5 ms after the switch has been opened?
 Solution:
a) The circuit for 𝑡 < 0 is shown below. Note that the inductor behaves like a short
circuit, effectively eliminating the 2 Ω resistor from the circuit.
First combine the 30 Ω and 6 Ω resistors in parallel:
30 6 = 5 Ω
Use voltage division to find the voltage drop across the parallel resistors:
5
𝑣=
120 = 75 V
5+3
Now find the current using Ohm's law:
𝑣
75
𝑖 0− = − = −
= − 12.5 A
6
6
1
1
c) 𝑤 0 = 𝐿𝑖 2 0 = 8 × 10−3 12.5
2
2
2
= 625 mJ
،‫شح‬١‫ِٓ اٌّّىٓ اٌفشً ثطشق وث‬
.‫ازذح‬ٚ ‫مخ‬٠‫ٌىٓ إٌدبذ ِّىٓ ثطش‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Example 7.1):
c) To find the time constant, we need to find the equivalent resistance seen by the
inductor for 𝑡 > 0.When the switch opens, only the 2 Ω resistor remains
connected to the inductor. Thus,
𝐿 8 × 10−3
𝜏= =
= 4 ms
𝑅
2
d) 𝑖 𝑡 = 𝑖 0− 𝑒 𝑡
e)
𝜏
= − 12.5 𝑒 −𝑡
𝑖 5ms = − 12.5 𝑒 −250
So 𝑤 5 ms =
0.004
0.005
= − 12.5 𝑒 −250𝑡 A,
𝑡≥0
= − 12.5 𝑒 −1.25 = − 3.58 A
1 2
1
𝐿𝑖 5 ms = 8 × 10−3 3.58
2
2
2
= 51.3 mJ
𝑤 𝑑𝑖𝑠 = 625 − 51.3 = 573.7 mJ
%dissipated =
573.7
100 = 91.8%
625
َ‫غذ األزال‬١ٌ ‫ أزالِه‬ٍٝ‫ؼزّذ ػ‬٠ ‫ٔدبزه‬
.‫مظخ‬١ٌ‫ ا‬ٟ‫ ف‬ٟ‫ئّٔب اٌز‬ٚ ‫ِه‬ٛٔ ٟ‫ رشا٘ب ف‬ٟ‫اٌز‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Assessment problem 7.2:
At 𝑡 = 0, the switch in the circuit shown moves instantaneously from position a to
position b.
a) Calculate 𝑣𝑜 for 𝑡 ≥ 0+.
b) What percentage of the initial energy stored in the inductor is eventually dissipated in
the 4 Ω resistor?
 Solution:
a) First, use the circuit for 𝑡 < 0 to find the initial current in the inductor:
Using current division,
𝑖 0− =
10
6.4 = 4 A
10 + 6
Now use the circuit for 𝑡 > 0 to find the equivalent resistance seen by the inductor,
and use this value to find the time constant:
‫ك اٌضؼفبء‬٠‫س رغذ طش‬ٛ‫ئْ اٌصخ‬
.‫بء‬٠ٛ‫ب األل‬ٙ١ٍ‫شرىض ػ‬٠ ‫ّٕب‬١‫ث‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Assessment problem 7.2):
𝑅𝑒𝑞 = 4 (6 + 10) = 3.2 Ω,
∴𝜏=
𝐿
0.32
=
= 0.1 s
𝑅𝑒𝑞
3.2
Use the initial inductor current and the time constant to find the current in the
inductor:
𝑖 𝑡 = 𝑖 0− 𝑒 −𝑡
𝜏
= 4 𝑒 −𝑡
0.1
= 4 𝑒 −10𝑡 A,
𝑡≥0
Use current division to find the current in the 10 Ω resistor:
𝑖𝑜 𝑡 =
4
4
−𝑖 =
− 4 𝑒 −10𝑡 = − 0.8 𝑒 −10𝑡 A,
4 + 10 + 6
20
t ≥ 0+
Finally, use Ohm's law to find the voltage drop across the 10 Ω resistor:
𝑣𝑜 𝑡 = 10𝑖𝑜 = 10 − 0.8 𝑒 −10𝑡 = − 8 𝑒 −10𝑡 V,
t ≥ 0+
b) The initial energy stored in the inductor is
𝑤 0 =
1 2 −
1
𝐿𝑖 0 = 0.32 4
2
2
2
= 2.56 J
Find the energy dissipated in the 4 Ω resistor by integrating the power over all
time:
𝑣4Ω 𝑡 = 𝐿
𝑝4Ω
𝑑𝑖
= 0.32 − 10 4 𝑒 −10𝑡 = − 12.8 𝑒 −10𝑡 V, t ≥ 0+
𝑑𝑡
𝑣4Ω 2
𝑡 =
= 40.96 𝑒 −20𝑡 W,
4
t ≥ 0+
∞
40.96 𝑒 −20𝑡 𝑑𝑡 = 2.048 J
𝑤4Ω 𝑡 =
0
Find the percentage of the initial energy in the inductor dissipated in the 4 Ω
resistor:
%𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 =
2.048
100 = 80%
2.56
‫ األشخبص‬ٌٝ‫بد رٕدزة ئ‬١ٌٛ‫ فبٌّغئ‬،‫اززس‬
.‫ب‬ٍّٙ‫ٓ ٌزس‬٠‫اٌّغزؼذ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.7:
In the circuit shown in Fig. P7.7, the switch makes contact with position b just before
breaking contact with position a. As already mentioned, this is known as a make-beforebreak switch and is designed so that the switch does not interrupt the current in an
inductive circuit. The interval of time between "making" and "breaking" is assumed to
be negligible. The switch has been in the a position for a long time. At 𝑡 = 0 the switch
is thrown from position a to position b.
a) Determine the initial current in the inductor.
b) Determine the time constant of the circuit for 𝑡 > 0.
c) Find 𝑖,𝑣1 and 𝑣2 for 𝑡 ≥ 0.
d) What percentage of the initial energy stored in the inductor is dissipated in the 20 Ω
resistor 12 ms after the switch is thrown from position a to position b?
 Solution:
24
=2A
12
𝐿 1.6
b) 𝜏 = =
= 20 ms
𝑅 80
a) 𝑖 0 =
c) 𝑖 = 2 𝑒 −50𝑡 A, 𝑡 ≥ 0
𝑑𝑖
𝑣1 = 𝐿 = 1.6 − 100 𝑒 −50𝑡 = − 160 𝑒 −50𝑡 V 𝑡 ≥ 0+
𝑑𝑡
𝑣2 = − 72𝑖 = −144 𝑒 −50𝑡 V
𝑡≥0
d)
𝑃𝑑𝑖𝑠𝑠 = 𝑖 2 72 = 4 𝑒 −100𝑡 72 = 288 𝑒 −100𝑡 W
𝑡
𝑒 −100𝑥 𝑡
−100𝑥
𝑤72 Ω = 288 𝑒
𝑑𝑥 = 288
= 288(1 − 𝑒 −100𝑡 )J
−100
0
0
−1.5
𝑤72 Ω 15 ms = 288 − 288 𝑒
= 2.24 J
1
𝑤 0 = 1.6 4 = 3.2 J
2
2.24
%𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 =
100 = 69.92 %
ٟ‫ٍه ٔفظ اٌطبلخ اٌز‬ٙ‫غز‬٠ ّٟٕ‫اٌز‬
3.2
.‫ظ‬١‫ب اٌزخط‬ٙ‫ٍى‬ٙ‫غز‬٠
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.4:
The switch in the circuit in Fig. P7.4 has been closed for a long time before opening at
𝑡 = 0.
c) Find 𝑖1 0− and 𝑖2 0− .
d) Find 𝑖1 0+ and 𝑖2 0+ .
e) Find 𝑖1 𝑡 for 𝑡 ≥ 0.
f) Find 𝑖2 𝑡 for 𝑡 ≥ 0+.
g) Explain why 𝑖2 0− ≠ 𝑖2 0+ .
 Solution:
a) 𝑡 < 0
2 𝑘Ω ∥ 6 kΩ = 1.5 kΩ
40
𝑖𝑔 0− =
= 20 mA
1500 + 500
2000
𝑖1 0− = (20 × 10−3 )
= 5 mA
8000
6000
𝑖2 0− = 20 × 10−3
= 15 mA
8000
b) 𝑖1 0+ = 𝑖1 0− = 5 mA
𝑖2 0+ = − 𝑖1 0+ = − 5 mA (when switch is open)
𝐿 400 × 10−3
c)
𝜏= =
= 5 × 10−5 ;
𝑅
8 × 103
𝑖1 𝑡 = 𝑖1 0+ 𝑒 −𝑡 𝜏
𝑖1 𝑡 = 5 𝑒 −20,000𝑡 mA,
𝑡≥0
1
= 20,000
𝜏
𝑖2 𝑡 = − 𝑖1 𝑡
when
𝑡 ≥ 0+
∴ 𝑖2 𝑡 = − 5 𝑒 −20,000𝑡 mA,
𝑡 ≥ 0+
e) The current in a resistor can change instantaneously. The switching operation
forces 𝑖2 (0−) to equal 15 mA and 𝑖2 0+ = −5 mA.
d)
.ً‫ فبش‬ٞ‫اعأي أ‬ٚ ،‫خ زظ‬١ٍّ‫إٌدبذ ِدشد ػ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.5:
The switch shown in Fig. P7.5 has been open a long time before closing at 𝑡 = 0.
a) Find 𝑖𝑜 0− .
b) Find 𝑖𝐿 0− .
c) Find 𝑖𝑜 0+ .
d) Find 𝑖𝐿 0+ .
e) Find 𝑖𝑜 ∞ .
f) Find 𝑖𝐿 ∞ .
g) Write the expression for 𝑖𝐿 𝑡 for 𝑡 ≥ 0.
h) Find 𝑣𝐿 0− .
i) Find 𝑣𝐿 0+ .
j) Find 𝑣𝐿 ∞ .
k) Write the expression for 𝑣𝐿 𝑡 for 𝑡 ≥ 0+.
l) Write the expression for 𝑖𝑜 𝑡 for 𝑡 ≥ 0+.
 Solution:
a) 𝑖𝑜 0− = 0
since the switch is open for 𝑡 < 0.
For 𝑡 = 0− the circuit is:
b)
120Ω 60Ω = 40Ω
∴
12
= 240 mA
10 + 40
120
=
𝑖 = 160 mA
180 𝑔
𝑖𝑔 =
𝑖𝐿 0−
For 𝑡 = 0+ the circuit is:
c)
120Ω 40Ω = 30 Ω
∴
12
= 300 mA
10 + 30
120
𝑖𝑎 =
300 = 225 mA
160
𝑖𝑔 =
∴ 𝑖𝑜 0+ = 225 − 160 = 65 mA
‫ش‬١‫لذ غ‬ٌٛ‫ ا‬ٟ‫اٌمشاس إٌّبعت ف‬
.‫ظ ثّٕبعت أثذا‬١ٌ ‫إٌّبعت‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Problem 7.5):
d) 𝑖𝐿 0+ = 𝑖𝐿 0− = 160 mA
e) 𝑖𝑜 ∞ = 𝑖𝑎 = 225 mA
f) 𝑖𝐿 ∞ = 0, since the switch short circuits the branch containing the 20 Ω, resistor
and the 100 mH inductor.
𝐿 100 × 10−3
g) 𝜏 = =
= 5 ms:
𝑅
20
1
= 200
𝜏
∴ 𝑖𝐿 = 0 + 160 − 0 𝑒 −200𝑡 = 160 𝑒 −200𝑡 mA,
h) 𝑣𝐿 0− = 0
𝑡≥0
since for 𝑡 < 0 the current in the inductor is constant
i) Refer to the circuit at 𝑡 = 0+ and note:
20 0.16 + 𝑣𝐿 0+ = 0;
j) 𝑣𝐿 ∞ = 0,
∴ 𝑣𝐿 0+ = − 3.2 V
since the current in the inductor is a constant at 𝑡 = ∞.
k) 𝑣𝐿 𝑡 = 0 + − 3.2 − 0 𝑒 −200𝑡 = − 3.2 𝑒 −200𝑡 V, 𝑡 ≥ 0+
l) 𝑖𝑜 (𝑡) = 𝑖𝑎 − 𝑖𝐿 = 225 − 160 𝑒 −200𝑡 mA,
𝑡 ≥ 0+
ٜ‫ا ِذ‬ٛ‫ذسو‬٠ ٌُ ‫ا‬ٍٛ‫ش ِّٓ فش‬١‫اٌىث‬
.‫ا‬ٍّٛ‫ُ ِٓ إٌدبذ ػٕذِب اعزغ‬ٙ‫لشث‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.1:
In the circuit in Fig. P7.1, the voltage and current expressions are
𝑣 = 160𝑒 −10𝑡 V,
𝑡 ≥ 0+
𝑖 = 8𝑒 −10𝑡 A,
𝑡≥0
Find
a) 𝑅.
b) 𝜏 (In milliseconds).
c) 𝐿.
d) The initial energy stored in the inductor.
e) The time (in milliseconds) it takes to dissipate 60% of the initial stored energy.
 Solution:
𝑣
160 𝑒 −10𝑡
a) = 𝑅 =
= 20 Ω
𝑖
8 𝑒 −10𝑡
b) 𝜏 =
1
= 100 ms
10
c) 𝜏 =
𝐿
= 0.1
𝑅
𝐿 = 100 20 × 10−3 = 2 H
1
d) 𝑤 0 = 𝐿 𝑖(0)
2
2
=
1
2 64 = 64 J
2
𝑡
1280 𝑒 −20𝑥 𝑑𝑥 = 64 − 64 𝑒 −20𝑡
e) 𝑤𝑑𝑖𝑠𝑠 =
0
100 − 100 𝑒 −20𝑡 = 60
solving ,
∴ 𝑒 −20𝑡 = 0.4
𝑡 = 45.81 ms
،‫ب‬ٙ‫م‬١‫ ٌزسم‬ٝ‫غؼ‬٠ ‫ع ٌٗ أ٘ذاف‬ٛٔ :ْ‫ػب‬ٛٔ ‫إٌبط‬
.‫ أرُ االعزؼذاد ٌّغبػذح ِٓ ٌٗ أ٘ذاف‬ٍٝ‫ع ػ‬ٛٔٚ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.6:
The switch in the circuit in Fig. P7.6 has been closed a long time. At 𝑡 = 0 it is opened.
Find 𝑣𝑜 𝑡 for 𝑡 ≥ 0.
 Solution:
For 𝑡 < 0
80
=2A
40
2 50
=
= 1 A = 𝑖𝐿 (0+)
100
𝑖𝑔 =
𝑖𝐿 0−
For 𝑡 > 0
𝑖𝐿 𝑡 = 𝑖𝐿 (0+)𝑒 −𝑡
𝜏=
𝜏
A
𝐿
0.20
1
=
=
= 0.01 s
𝑅 5 + 15 100
𝑖𝐿 0+ = 1 A
𝑖𝐿 𝑡 = 𝑒 −100𝑡 A,
𝑡≥0
𝑣𝑜 𝑡 = − 15𝑖𝐿 (𝑡)
𝑣𝑜 𝑡 = − 15 𝑒 −100𝑡 V,
𝑡 ≥ 0+
.ّٓ‫ذفغ اٌث‬٠ ٌّٓ ‫إٌدبذ ٍِه‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Problem 7.11:
The switch in the circuit in Fig. P7.l2 has been in position 1 for a long time. At 𝑡 = 0, the
switch moves instantaneously to position 2. Find 𝑣𝑜 𝑡 for 𝑡 ≥ 0+.
 Solution:
𝑡<0∶
240
= 10 A
16 + 8
40
= 10
=8A
50
𝑖𝐿 0+ =
𝑖𝐿 0−
𝑡>0∶
10 40
+ 10 = 18 Ω
50
𝐿
72
𝜏=
=
× 10−3 = 4 ms;
𝑅𝑒 18
𝑅𝑒 =
1
= 250
𝜏
∴ 𝑖𝐿 = 8 𝑒 −250𝑡 A
𝑣𝑜 = 8𝑖𝑜 = 64 𝑒 −250𝑡 V,
𝑡 ≥ 0+
ٞ‫ذ اٌز‬١‫ز‬ٌٛ‫ اٌزٔت ا‬ٛ٘ ‫إٌدبذ‬
.‫غفشٖ سفمبؤٔب‬٠ ‫ال‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.14:
The switch in Fig. P7.14 has been closed for a long time before opening at 𝑡 = 0. Find
a) 𝑖𝐿 𝑡 , 𝑡 ≥ 0.
b) 𝑣𝐿 𝑡 , 𝑡 ≥ 0+.
c) 𝑖∆ 𝑡 , 𝑡 ≥ 0+.
 Solution:
a) 𝑡 < 0 ∶
−72
𝑖𝐿 0 =
= −2.4 A
24 + 6
𝑡>0
𝑖 𝑇 100
5
= − 𝑖𝑇
160
8
100 60
𝑣𝑇 = 20𝑖∆ + 𝑖 𝑇
= −12.5𝑖 𝑇 + 37.5𝑖 𝑇
160
𝑣𝑇
= 𝑅𝑇𝑕 = −12.5 + 37.5 = 25Ω
𝑖𝑇
𝑖∆ = −
𝜏=
𝐿 250
=
× 10−3
𝑅
25
𝑖𝐿 = −2.4𝑒 −100𝑡 A,
1
= 100
𝜏
𝑡≥0
b) 𝑣𝐿 = 250 × 10−3 240 𝑒 −100𝑡 = 60 𝑒 −100𝑡 V, 𝑡 ≥ 0+
c) 𝑖∆ =
100𝑖 𝐿
160
= − 1.5 𝑒 −100𝑡 A 𝑡 ≥ 0+
.ً‫ شؼبس وً فبش‬ٛ٘ ‫ ٘زا‬،‫غبد‬ٚ‫أٔزُ أ‬ٚ ٓ٠‫أٔب زض‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.19:
The two switches shown in the circuit in Fig. P7.19 operate simultaneously. Prior to
𝑡 = 0 each switch has been in its indicated position for a long time. At 𝑡 = 0 the two
switches move instantaneously to their new positions. Find
a) 𝑣𝑜 𝑡 , 𝑡 ≥ 0+.
b) 𝑖𝑜 𝑡 , 𝑡 ≥ 0.
 Solution:
a) 𝑡 < 0 ∶
𝑡 = 0+ ∶
‫ه أزذ‬٠‫ أس‬ٟ‫ شخصب ثال أ٘ذاف و‬ٟٔ‫أ َ ِس‬
.ُ٘‫ٓ رزسشن أخغبد‬٠‫ اٌز‬ٟ‫ر‬ٌّٛ‫اع ا‬ٛٔ‫أ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Problem 7.19):
𝑡>0∶
𝑖𝑅 = 2 𝑒 −𝑡
𝜏
mA ;
𝜏=
𝐿
= 0.66 × 10−3
𝑅
𝑖𝑅 = 2 𝑒 −1500𝑡 A
𝑣𝑅 = 7.5 × 103 −2 𝑒 −1500𝑡 = −15,000 𝑒 −1500𝑡 V
𝑣1 = 1.25 −2 −1500 𝑒 −1500𝑡 = 3750 𝑒 −1500𝑡 V
𝑣𝑜 = −𝑣1 − 𝑣𝑅 = − 3750𝑒 −1500𝑡 + 15,000 𝑒 −1500𝑡 = 11,250 𝑒 −1500𝑡 V
1
b) 𝑖𝑜 =
6
𝑡
11,250 𝑒 −1500𝑡 𝑑𝑡 + 0 = −1.25 𝑒 −1500𝑡 + 1.25 mA
0
ٍٝ‫ْ ػ‬ٚ‫ؼثش‬٠ ‫أغٍت إٌبط‬
.ٕٗ‫ِب ُ٘ ثصذد اٌجسث ػ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
7.2
The Natural Response of an RC Circuit:
 Assume that the switch has been in position a for a long time, allowing the loop
made up of the dc voltage source 𝑉𝑔 to reach a steady-state condition.
 Remember that a capacitor behaves
as an open circuit in the presence of a
constant voltage. Thus the voltage
source cannot sustain a current, and
so the source voltage appears across
the capacitor terminals.
 When the switch is moved from
position a to position b (at 𝑡 = 0), the
voltage on the capacitor is 𝑉𝑔 . Because
there can be no instantaneous change
in the voltage at the terminals of a
capacitor, the problem reduces to solving the circuit shown in Fig. 7.11.
 Deriving the Expression for the Voltage:
We can find the voltage 𝑣 𝑡 by thinking in terms of node voltages. Using any lower
point between 𝑅 and 𝐶 as the reference node gives
𝐶
𝑑𝑣 𝑣
+ =0
𝑑𝑡 𝑅
Mathematical techniques can be used to obtain
𝑣 𝑡 = 𝑣 0 𝑒 −𝑡
𝑅𝐶
,
𝑡≥0
𝑣 0− = 𝑣 0 = 𝑣 0+ = 𝑣𝑔 = 𝑣0
𝜏 = 𝑅𝐶
𝑣 𝑡 = 𝑣0 𝑒 −𝑡 𝜏 ,
𝑡≥0
‫ظ واطالق‬١‫ْ رخط‬ٚ‫ػًّ ثذ‬
.‫ت‬٠ٛ‫ْ رص‬ٚ‫اٌشصبص ثذ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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Which indicates that the natural
response of an 𝑅𝐶 circuit is an
exponential decrease of the initial
voltage.
After determining 𝑣 𝑡 , we easily derive the expressions for 𝑖, 𝑝, and 𝑤:
𝑖 𝑡 =
𝑣 𝑡
𝑣0
= 𝑒 −𝑡 𝜏 ,
𝑅
𝑅
𝑡 ≥ 0+
𝑣02 −2𝑡 𝜏
𝑝 = 𝑣𝑖 =
𝑒
,
𝑅
𝑡
𝑤=
𝑡
𝑝 𝑑𝑥 =
0
=
0
𝑡 ≥ 0+
𝑣02 −2𝑥
𝑒
𝑅
1 2
𝐶𝑣0 1 − 𝑒 −2𝑡
2
𝜏
𝜏
𝑑𝑥
,
𝑡≥0
 Calculating the natural response of an 𝑅𝐶 circuit can be summarized as follows:
1. Find the initial voltage, 𝑣0 , across the capacitor.
2. Find the time constant of the circuit, 𝜏 = 𝑅𝐶.
3. Use 𝑣 𝑡 = 𝑣0 𝑒 −𝑡 𝜏 , to generate 𝑣 𝑡 from 𝑣0 and 𝜏.
4. All other calculations of interest follow from knowing 𝑣 𝑡 .
‫الدٔب أوثش‬١ِ ‫ اإلػذاد ٌسفالد‬ٟ‫ٕب ف‬١‫لض‬
.‫برٕب‬١‫ظ ٌس‬١‫ اٌزخط‬ٟ‫ٕبٖ ف‬١‫ِّب لض‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Example 7.3:
The switch in the circuit shown in Fig.7.13 has been in position x for a long time. At 𝑡 =
0, the switch moves instantaneously to position y. Find
a) 𝑣𝐶 (𝑡) for 𝑡 ≥ 0.
b) 𝑣𝑜 (𝑡) for 𝑡 ≥ 0+.
c) 𝑖𝑜 (𝑡) for 𝑡 ≥ 0+.
d) The total energy dissipated in the 60 kΩ resistor.
 Solution:
a) Because the switch has been in position 𝑥 for a long time, the 0.5 𝜇F capacitor will
charge to 100 V and be positive at the upper terminal. We can replace the resistive
network connected to the capacitor at 𝑡 = 0+ with an equivalent resistance of 80 kΩ.
𝜏 = 𝑅𝐶 = 80 × 103 × 0.5 × 10−6 = 40 ms
𝑣𝑐 𝑡 = 100 𝑒 −25𝑡 V,
t≥0
b) Using voltage dividing:
𝑣𝑜 𝑡 =
48
𝑣𝑐 𝑡 = 60 𝑒 −25𝑡 V,
80
t ≥ 0+
c) Using Ohm’s law:
𝑖𝑜 𝑡 =
𝑣𝑜 (𝑡)
= 𝑒 −25𝑡 mA,
3
60 × 10
d) 𝑝60 𝑘Ω 𝑡 = 𝑖𝑜2 𝑡
t ≥ 0+
60 × 103 = 60 𝑒 −50𝑡 mW, t ≥ 0+
∞
𝑖𝑜2 𝑡
𝑤60𝑘Ω =
60 × 103 𝑑𝑡 = 1.2 mJ
0
،ٟ‫ؼ‬١‫بٖ ٌّدشا٘ب اٌطج‬١ٌّ‫وّب رٕدزة ا‬
.ٌٗ ٓ٠‫ٕدزة إٌدبذ ٌٍّغزؼذ‬٠
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Example 7.4:
The initial voltages on capacitors 𝐶1 and 𝐶2 in the circuit shown in Fig. 7.14 have been
established by sources not shown. The switch is closed at 𝑡 = 0.
a) Find 𝑣1 𝑡 , 𝑣2 (𝑡), and 𝑣(𝑡) for 𝑡 ≥ 0 and 𝑖(𝑡) for ≥ 0+ .
b) Calculate the initial energy stored in the capacitors 𝐶1 and 𝐶2 .
c) Determine how much energy is stored in the capacitor as 𝑡 → ∞.
d) Show that the total energy delivered to the 250 kΩ resistor is the difference between
results obtained in (b) and (c).
 Solution:
a) To find 𝑣 𝑡 , we replace the series-connected capacitors with an equivalent capacitor.
It has a capacitance of 4 𝜇F and is charged to a voltage of 20 V.
𝜏 = 𝑅𝐶 = 250 × 103 × 4 × 10−6 = 1 s
𝑣 𝑡 = 20 𝑒 −𝑡 V,
t ≥ 0.
𝑣(𝑡)
𝑖 𝑡 =
= 80 𝑒 −𝑡 𝜇A,
t ≥ 0+
250,000
Knowing 𝑖 𝑡 , we calculate the expressions for 𝑣1 𝑡 and 𝑣2 𝑡 :
106 𝑡
𝑣1 𝑡 = −
80 × 10−6 𝑒 −𝑡 𝑑𝑡 − 4 = 16 𝑒 −𝑡 − 20 V,
t ≥ 0,
5 0
106 𝑡
𝑣2 𝑡 = −
80 × 10−6 𝑒 −𝑡 𝑑𝑡 + 24 = 4 𝑒 −𝑡 + 20 V,
t ≥ 0.
20 0
1
1
b) 𝑤1 = 5 × 10−6 16 = 40 𝜇J,
𝑤2 = 20 × 10−6 576 = 5760 𝜇J
2
2
The total initial energy stored in the two capacitors is: 𝑤𝑜 = 40 + 5760 = 5800 𝜇J
c) As 𝑡 → ∞,
𝑣1 → − 20 V and
𝑣2 → + 20 V
1
𝑤∞ = 5 + 20 × 10−6 400 = 5000 𝜇J
2
∞
∞
400 𝑒 −2𝑡
d)
𝑤=
𝑝 𝑑𝑡 =
𝑑𝑡 = 800 𝜇J
0
0 250,000
‫ش‬١١‫ رغ‬ٍٝ‫ؼىف أغٍت إٌبط ػ‬٠
800 𝜇J = 5800 − 5000 𝜇J
.‫ش األعجبة‬١١‫ظ رغ‬١ٌٚ ‫إٌزبئح‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Assessment problem 7.3:
The switch in the circuit shown has been closed for a long time and is opened at 𝑡 = 0.
Find
a) The initial value of 𝑣(𝑡).
b) The time constant for 𝑡 > 0.
c) The numerical expression for 𝑣 𝑡 after the switch bas been opened.
d) The initial energy stored in the capacitor.
e) The length of time required to dissipate 75% of the initially stored energy.
 Solution:
a) The circuit for 𝑡 < 0 is shown below. Note that the capacitor behaves like an open
circuit.
Find the voltage drop across the open circuit by finding the voltage drop across the
50 kΩ resistor. First use current division to find the current through the 50 kΩ
resistor:
𝑖50𝑘
80 × 103
=
7.5 × 10−3 = 4 mA
80 × 103 + 20 × 103 + 50 × 103
Use Ohm's law to find the voltage drop:
𝑣 0− = 50 × 103 𝑖50𝑘 = 50 × 103 0.004 = 200 V
‫ث‬١‫ّىٕه أْ رجذأ ِٓ ز‬٠
.ْٚ‫خش‬٢‫ ا‬ٝٙ‫أز‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Assessment problem 7.3):
b) To find the time constant, we need to find the equivalent resistance seen by the
capacitor for 𝑡 > 0. When the switch opens, only the 50 kΩ resistor remains
connected to the capacitor. Thus,
𝜏 = 𝑅𝐶 = 50 × 103 0.4 × 10−6 = 20 ms
c) 𝑣 𝑡 = 𝑣 0− 𝑒 −𝑡
𝜏
= 200 𝑒 −𝑡
0.02
= 200 𝑒 −50𝑡 V,
𝑡≥0
1
1
𝑑) 𝑤 0 = 𝐶𝑣 2 = 0.4 × 10−6 (200)2 = 8 mJ
2
2
1
1
e) 𝑤 𝑡 = 𝐶𝑣 2 𝑡 = 0.4 × 10−6 (200 𝑒 −50𝑡 )2 = 8 𝑒 −100𝑡 mJ
2
2
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:
8 × 10−3 𝑒 −100𝑡 = 2 × 10−3 ,
𝑒 100𝑡 = 4,
𝑡 = (ln 4) 100 = 13.86 ms
‫ال رغٍُّ لٍجه‬ٚ ‫اج‬ِٛ‫عٍُّ لبسثه ٌأل‬
.‫ ِب صاي اٌجسش أوثش إِٔب‬،‫الِشأح‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Assessment problem 7.4:
The switch in the circuit shown has been closed for a long time before being opened at
𝑡 = 0.
a) Find 𝑣𝑜 (𝑡) for 𝑡 ≥ 0.
b) What percentage of the initial energy stored in the circuit has been dissipated after the
switch has been open for 60 ms?
 Solution:
a) This circuit is actually two 𝑅𝐶 circuits in series, and the requested voltage, 𝑣𝑜 is the
sum of the voltage drops for the two 𝑅𝐶 circuits. The circuit for 𝑡 < 0 is shown
below:
Find the current in the loop and use it to find the initial voltage drops across the two
𝑅𝐶 circuits:
15
𝑖=
= 0.2 mA,
𝑣5 0− = 4 V,
𝑣1 0− = 8 V
75,000
There are two time constants in the circuit, one for each 𝑅𝐶 subcircuit. 𝜏5 is the time
constant for the 5 𝜇F − 20 kΩ subcircuit, and 𝜏1 is the time constant for the
1 𝜇F − 40 kΩ subcircuit:
𝜏5 = 20 × 103 5 × 10−6 = 100 ms;
τ1 = 40 × 103 1 × 10−6 = 40 ms
‫ فبٌٕغبء‬،ٖ‫ش‬١‫ ِص‬ٟ‫زسىُ اٌشخً ف‬٠ ‫ال‬
.ٕٗ‫بثخ ػ‬١ٔ ‫فؼٍٓ رٌه‬٠ ٗ‫بر‬١‫ ز‬ٟ‫داد ف‬ٛ‫خ‬ٌّٛ‫ا‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Assessment problem 7.4) :
Therefore,
𝑣5 𝑡 = 𝑣5 0− 𝑒 −𝑡
𝜏5
= 4 𝑒 −𝑡
0.1
= 4 𝑒 −10𝑡 V,
𝑡≥0
𝑣1 𝑡 = 𝑣1 0− 𝑒 −𝑡
𝜏1
= 8 𝑒 −𝑡
0.04
= 8 𝑒 −25𝑡 V,
t≥0
𝑣𝑜 𝑡 = 𝑣1 𝑡 + 𝑣5 𝑡 = 8 𝑒 −25𝑡 + 4 𝑒 −10𝑡 V,
𝑡≥0
Finally,
b) Find the value of the voltage at 60 ms for each subcircuit and use the voltage to
find the energy at 60 ms :
𝑣1 60 ms = 8 e−25(0.06) ≅ 1.79 V,
1
1
2
2
𝑤1 60 ms = 𝐶𝑣12 60 ms =
𝑣5 60 ms = 4 e−10(0.06) ≅ 2.20 V
1 × 10−6 (1.79)2 ≅ 1.59 μJ
1
1
𝑤5 60 ms = 𝐶𝑣52 60 ms = 5 × 10−6 2.20
2
2
2
≅ 12.05 μJ
𝑤 60 ms = 1.59 + 12.05 = 13.64 μJ
Find the initial energy from the initial voltage:
𝑤 0 = 𝑤1 0 + 𝑤5 0 =
1
1
1 × 10−6 (8)2 + 5 × 10−6 (4)2 = 72 μJ
2
2
Now calculate the energy dissipated at 60 ms and compare it to the initial energy:
𝑤𝑑𝑖𝑠𝑠 = 𝑤 0 − 𝑤 60 ms = 72 − 13.64 = 58.36 μJ
58.36 × 10−6
% dissipated =
72 × 10−6
100 = 81.05%
،‫ب‬ٙ‫ي اِشأح أزج‬ٚ‫ أ‬ٝ‫ٕغ‬٠ ‫اٌشخً ال‬
.‫ب‬ٙٔ‫ي سخً خب‬ٚ‫ أ‬ٝ‫اٌّشأح ال رٕغ‬ٚ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.23:
The switch in the circuit in Fig. P7.23 has been in position a for a long time and 𝑣2 = 0 V.
At 𝑡 = 0, the switch is thrown to position b. Calculate:
a) 𝑖, 𝑣1 , and 𝑣2 for 𝑡 ≥ 0+.
b) The energy stored in the capacitor at 𝑡 = 0.
c) The energy trapped in the circuit and the total energy dissipated in the 5 kΩ resistor if
the switch remains in position b indefinitely.

Solution:
a) 𝑣1 0− = 𝑣1 0+ = 75 V
2×8
𝐶𝑒𝑞 =
= 1.6 µF
10
𝑣2 0+ = 0
1
= 125
𝜏
𝜏 = 5 1.6 × 10−3 = 8 ms;
𝑖=
75
× 10−3 𝑒 −125𝑡 mA,
5
𝑡 ≥ 0+
− 106 𝑡
𝑣1 =
15 × 10−3 𝑒 −125𝑥 𝑑𝑥 + 75 = 60 𝑒 −125𝑡 + 15 V, 𝑡 ≥ 0
2
0
6 𝑡
10
𝑣2 =
15 × 10−3 𝑒 −125𝑥 𝑑𝑥 + 0 = − 15 𝑒 −125𝑡 + 15 V,
𝑡≥0
8 0
1
b)
𝑤 0 = 2 × 10−6 5625 = 5625 µJ
2
1
1
c) 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 = 2 × 10−6 225 + 8 × 10−6 225 = 1125 µJ
2
2
1
𝑤𝑑𝑖𝑠𝑠 = 1.6 × 10−6 5625 = 4500 µJ
2
check ∶ 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 + 𝑤𝑑𝑖𝑠𝑠 = 1125 + 4500 = 5625 µJ;
𝑤 0 = 5625µJ
.‫ز اٌّشأح‬١ٍّ‫ر‬ٚ ً‫طبْ أعزبر اٌشخ‬١‫اٌش‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.24:
The switch in the circuit in Fig. P7.24 is closed at 𝑡 = 0 after being open for a long
time.
a) Find 𝑖1 0− and 𝑖2 0− .
b) Find 𝑖1 0+ and 𝑖2 0+ .
c) Explain why 𝑖1 0− = 𝑖1 0+ .
d) Explain why 𝑖2 0− ≠ 𝑖2 0+ .
e) Find 𝑖1 𝑡 for 𝑡 > 0.
f) Find 𝑖2 𝑡 for 𝑡 ≥ 0+.
 Solution:
a) 𝑡 < 0 :
𝑖1 0− = 𝑖2 0− =
3
= 100 mA
30
ًّ‫ رؼ‬ٟ‫ اٌّبدح اٌخبَ اٌز‬ٛ٘ ً‫اٌشخ‬
.‫شح‬١‫ب اٌّشأح اٌٍّغبد األخ‬ٙ١‫ف‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Problem 7.24):
b) 𝑡 > 0 :
0.2
= 100 mA
2
0.2
=−
= − 25 mA
8
𝑖1 0+ =
𝑖2 0+
c) Capacitor voltage cannot change instantaneously, therefore,
𝑖1 0− = 𝑖1 0+ = 100 mA
d) Switching can cause an instantaneous change in the current in a
resistive branch. In this circuit
𝑖2 0− = 100 mA
e) 𝑣𝑐 = 0.2𝑒 −𝑡
𝜏
𝑉,
𝑖2 0+ = 25 mA
and
𝑡≥0
𝜏 = 𝑅𝑒 𝐶 = 1.6 2 × 10−6 = 3.2µ𝑠
𝑣𝑐 = 0.2 𝑒 −312,500𝑡 V, 𝑡 ≥ 0,
𝑣𝑐
𝑖1 = = 0.1 𝑒 −312,500𝑡 mA,
2
f) 𝑖2 =
1
= 312,500
𝜏
𝑡≥0
− 𝑣𝑐
= − 25 𝑒−312,500𝑡 mA, 𝑡 ≥ 0+
8
‫رش أٔٗ ئرا أخطأ‬ٛ١‫اٌىّج‬ٚ ‫ٓ اٌّشأح‬١‫اٌفشق ث‬
.‫ج‬ٚ‫ اٌض‬ٍٝ‫َ ػ‬ٌٍٛ‫ ثب‬ٟ‫ٍم‬٠ ‫ ال‬ٛٙ‫ش ف‬١‫األخ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.24(‫تحذف هذه المساله بحلها‬:
The switch in the circuit in Fig. P7.24 has been in position a for a long time. At 𝑡 = 0, the
switch is thrown to position b.
a) Find 𝑖𝑜 𝑡 for 𝑡 ≥ 0+.
b) What percentage of the initial energy stored in the capacitor is dissipated in the 4 kΩ
resistor 250 𝜇s after the switch has been thrown?
 Solution:
8 27 33
= 118.80 V
60
3 6
𝑅𝑒 =
= 2 KΩ
9
a) 𝑣 0 =
𝜏 = 𝑅𝑒 𝐶 = 2000 0.25 × 10−6 = 500 µs;
1
= 2000
𝜏
𝑣 = 118.80 𝑒 −2000𝑡 V
𝑡≥0
𝑣
𝑖𝑜 =
= 39.6 𝑒 −2000𝑡 mA
3000
b) 𝑤 0 =
𝑖4𝑘
1
0.25 118.80
2
2
= 1764.18 µJ
118.80 𝑒 −2000𝑡
=
= 19.8 𝑒 −2000𝑡 mA
6
𝑝4𝑘 =
19.8 𝑒 −2000𝑡
2
4000 × 10−6 = 1568.16 × 10−3 𝑒 −4000𝑡
−4000𝑥 250 × 10−6
𝑒
𝑤4𝑘 = 1568.16 × 10−3
= 392.04 1 − 𝑒 −1 µJ
−4000
0
392.04
%=
1 − 𝑒 −1 × 100 = 14.05 %
1764.18
.‫ذح أخجبس‬٠‫ب خش‬ٙ‫ب ال ثاسادر‬ٙ‫ؼز‬١‫اٌّشأح ثطج‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.26:
Both switches in the circuit in Fig. P7.26 have been closed for a long time. At 𝑡 = 0, both
switches open simultaneously.
a) Find 𝑖𝑜 𝑡 for 𝑡 ≥ 0+.
b) Find 𝑣𝑜 𝑡 for 𝑡 ≥ 0.
c) Calculate the energy (in microjoules) trapped in the circuit.
 Solution:
a) 𝑡 < 0 :
6000 0.04
6000 + 4000
= 3000 0.024
= 40 − 24
= 6000 0.016
𝑖𝑜 0− =
= 24 mA
𝑣𝑜 0−
𝑖2 0−
𝑣2 0−
= 72 V
= 16 mA
= 96 V
𝑡>0:
𝜏 = 𝑅𝐶 = 200 µs;
1
= 5000
𝜏
.ٍُ‫زى‬٠ ‫ ال‬ٞ‫اٌّشأح رست اٌشخً اٌز‬
.‫ب‬ٙ١ٌ‫غزّغ ئ‬٠ ٗٔ‫ب رظٓ أ‬ٙٔ‫ئ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Continued (Problem 7.26):
𝑖𝑜 𝑡 =
24
𝑒 −𝑡
1 × 103
𝜏
= 24 𝑒 −5000𝑡 mA,
𝑡 ≥ 0+
b)
106
𝑣𝑜 =
0.6
𝑡
24 × 10−3 𝑒 −5000𝑥 𝑑𝑥 + 72
0
𝑒 −5000𝑥 𝑡
= 40,000
+ 72
− 5000
0
= − 8𝑒 −5000𝑡 + 80 V,
c) 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 =
1
2
𝑡≥0
0.3 × 10−6 80
2
+
1
2
0.6 × 10−6 80
2
𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 = 2880 µJ
Check by combining the capacitors into a single equivalent capacitance of 0.2 𝜇F
with a 24 V initial voltage:
1
1
𝑤𝑑𝑖𝑠𝑠 = 𝐶𝑒𝑞 𝑉𝑜 2 = 0.2 × 10−6 24 2 = 57.6 µJ
2
2
1
1
𝑤 0 = 0.3 × 10−6 96 2 + 0.6 × 10−6 72 2 = 2937.6 µJ
2
2
𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 + 𝑤𝑑𝑖𝑠𝑠 = 𝑤(0)
2880 + 57.6 = 2937.6
OK
،‫ٓ عٕخ‬١‫اٌّشأح رىزُ اٌست أسثؼ‬
.‫ازذح‬ٚ ‫ال رىزُ اٌجغض عبػخ‬ٚ
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 Problem 7.25:
In the circuit shown in Fig. P7.25, both switches operate together; that is, they either open
or close at the same time. The switches arc closed a long time before opening at 𝑡 = 0.
a) How many microjoules of energy have been dissipated in the 12 kΩ resistor 12 ms
after the switches open?
b) How long does it take to dissipate 75% of the initially stored energy?
 Solution:
a) 𝑡 < 0 :
𝑅𝑒𝑞 = 12𝐾 68𝐾 = 10.2kΩ
𝑣𝑜 0 =
−120 10,200
= −102 V
(10,200 + 1800)
𝑡>0
10
12,000 × 10−6 = 40 ms;
3
𝑣𝑜 = −102 𝑒 −25𝑡 V,
𝑡≥0
2
𝑣𝑜
𝑝=
= 867 × 10−3 𝑒 −50𝑡 W
12,000
1
= 25
𝜏
𝜏=
12×10 −3
867 × 10−3 𝑒 −50𝑡 𝑑𝑡
𝑤𝑑𝑖𝑠𝑠 =
0
= 17.34 × 10−3 1 − 𝑒 −0.6 = 7824 µJ
1 10
b)
𝑤 0 =
102 2 × 10−6 = 17.34 µJ
2 3
0.75𝑤 0 = 13 mJ
𝑡𝑜
867 × 10−3 𝑒 −50𝑥 𝑑𝑥 = 13 × 10−3
0
∴ 1 − 𝑒 −50𝑡 𝑜 = 0.75;
𝑒 50𝑡 𝑜 = 4;
so
𝑡𝑜 = 27.73 ms
ٓ‫اٌّشأح لذ رصفر ػ‬
.‫ب ال رٕغب٘ب‬ٕٙ‫ ٌى‬،‫بٔخ‬١‫اٌخ‬
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 Problem 7.30:
The switch in the circuit in Fig. P7.27 has been in position 1 for a long time before moving
to position 2 at 𝑡 = 0. Find 𝑖𝑜 𝑡 for 𝑡 ≥ 0+.
 Solution:
𝑡>0:
𝑡>0:
𝑣𝑇 = −5𝑖𝑜 − 15 𝑖𝑜 = −20𝑖𝑜 = 20𝑖 𝑇
∴
𝑣𝑇
= 𝑅𝑇𝑕 = 20 Ω
𝑖𝑇
𝜏 = 𝑅𝐶 = 40 ms;
1
= 25,000
𝜏
𝑣𝑜 = 15 𝑒 −25,000𝑡 V,
𝑡≥0
−𝑣𝑜
𝑖𝑜 =
= −0.75𝑒 −25,000𝑡 A, 𝑡 ≥ 0+
20
‫ي‬ٛ‫ اٌذخ‬ٟ‫طبْ ف‬١‫ئرا أخفك اٌش‬
.‫فذ اِشأح‬ٚ‫ ِىبْ أ‬ٌٝ‫ئ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Problem 7.33:‫)(مش موجود يمسح‬
The switch in the circuit seen in Fig. P7.33 has been closed for a long time. The switch
opens at 𝑡 = 0. Find the numerical expressions for 𝑖𝑜 𝑡 and 𝑣𝑜 𝑡 when 𝑡 ≥ 0+.
 Solution:
After making a Thévenin equivalent we have:
180
= 12 mA
15
0.25
𝜏=
× 10−3 = 0.125 × 10−4 ;
20
𝑉𝑠 180
𝐼𝑓 = =
= 9 mA
𝑅
20
𝐼𝑜 =
1
= 80,000
𝜏
𝑖𝑜 = 9 + 12 − 9 𝑒 −80,000𝑡 = 9 + 3 𝑒 −80,000𝑡 mA
𝑣𝑜 = 180 − 12 20 𝑒 −80,000𝑡 = − 60 𝑒 −80,000𝑡 V
‫ئرا أسدد فضر عشن‬
.‫ اِشأح‬ٌٟ‫عٍّٗ ئ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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7.3
The Step Response of 𝑹𝑳 and RC Circuits:
 The response of a circuit to the sudden application of a constant voltage or current
source is referred to as the step response of the circuit.
 To explain the step response, we show how the circuit responds when energy is
being stored in the inductor or capacitor. We begin with the step response of an 𝑅𝐿
circuit.
 The Step Response of an 𝑹𝑳 Circuit:
 To begin, we can modify the first-order
circuit shown in Fig. 7.2(a) by adding a
switch. We use the resulting circuit,
shown in Fig. 7.16, in developing the
step response of an 𝑅𝐿 circuit.
 Energy stored in the inductor at the time the switch is closed is given in terms of a
nonzero initial current 𝑖 0 . The task is to find the expressions for the current in
the circuit and for the voltage across the inductor after the switch has been closed.
The procedure is the same as that used in Section 7.1; we use circuit analysis to
derive the differential equation that describes the circuit in terms of the variable of
interest, and then we use elementary calculus to solve the equation.
 After the switch in Fig. 7.16 has been closed, Kirchhoff’s voltage law requires that
𝑉𝑠 = 𝑅𝑖 + 𝐿
𝑑𝑖
𝑑𝑡
Which can be solved for the current
𝑑𝑖 − 𝑅𝑖 + 𝑉𝑠 − 𝑅
𝑉𝑠
=
=
𝑖−
𝑑𝑡
𝐿
𝐿
𝑅
،‫ئبًا‬١‫ذ ِٕه ش‬٠‫ رش‬ٟٙ‫ ف‬،‫ب‬ٙ‫ر‬ٛ‫ئرا خفضذ اٌّشأح ص‬
. ‫ء‬ٟ‫ ٌُ رأخز ٘زا اٌش‬ٟٙ‫ب ف‬ٙ‫ر‬ٛ‫ئرا سفؼذ ص‬ٚ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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−𝑅
𝑉𝑠
𝑖−
𝑑𝑡
𝐿
𝑅
𝑑𝑖
−𝑅
=
𝑑𝑡
𝑖 − 𝑉𝑠 𝑅
𝐿
or
𝑑𝑖 =
or
Integrating both sides and separating:
𝑖 𝑡 =
𝑉𝑠
𝑉𝑠 − 𝑅
+ 𝐼0 −
𝑒
𝑅
𝑅
𝐿 𝑡
(7.35)
When the initial energy in the inductor is zero, 𝐼0 is zero. Equating reduces to
𝑉𝑠 𝑉𝑠 − 𝑅 𝐿 𝑡
− 𝑒
7.36
𝑅 𝑅
Equation indicates that after the switch has been closed, the current increases
𝑖 𝑡 =
exponentially from zero to a final value of 𝑉𝑠 𝑅 . The time constant of the circuit,
𝐿 𝑅, determines the rate of increase. One time constant after the switch has been
closed, the current will have reached approximately 63 % of its final value
𝑉𝑠 𝑉𝑠 −1
𝑉𝑠
− 𝑒 ≈ 0.6321
7.37
𝑅 𝑅
𝑅
If the current were to continue to increase at its initial rate, it would reach its final
𝑖 𝜏 =
value at 𝑡 = 𝜏; that is, because
𝑑𝑖 − 𝑉𝑠 − 1 −𝑡
=
𝑒
𝑑𝑡
𝑅
𝜏
𝜏
𝑉𝑠 −𝑡
= 𝑒
𝐿
𝜏
The initial rate at which 𝑖 𝑡 increases is
𝑑𝑖
𝑉𝑠
0 =
𝑑𝑡
𝐿
If the current were to continue to increase at this rate, the expression for 𝑖 would be
𝑖=
at 𝑡 = 𝜏:
𝑉𝑠
𝑡
𝐿
𝑉𝑠 𝐿 𝑉𝑠
𝑖=
=
𝐿𝑅 𝑅
7.40
(7.41)
.‫أػظ اٌّشأح اٌخشعبء عشا رٕطك‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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Equations 7.36 and 7.40 are plotted in
Fig. 7.17. The values given by Eqs. 7.37
and 7.41 are also shown in this figure.
The voltage across an inductor is 𝐿 𝑑𝑖 𝑑𝑡, so from Eq. 7.35, for 𝑡 ≥ 0+,
𝑣=𝐿
−𝑅
𝐿
𝐼0 −
𝑉𝑠 − 𝑅
𝑒
𝑅
𝐿 𝑡
= 𝑉𝑠 − 𝐼0 𝑅 𝑒 − 𝑅
𝐿 𝑡
7.42
The voltage across the inductor is zero before the switch is closed. Equation 7.42
indicates that the inductor voltage jumps to 𝑉𝑠 − 𝐼0 𝑅 at the instant the switch is closed
and then decays exponentially to zero.
When the initial inductor current is zero, Eq. 7.42 simplifies to
𝑣 = 𝑉𝑠 𝑒 − 𝑅
𝐿 𝑡
If the initial current is zero, the voltage
across the inductor jumps to 𝑉𝑠 . We also
expect the inductor voltage to approach
zero as 𝑡 increases, because the current in
the circuit is approaching the constant
value of 𝑉𝑠 𝑅 . Figure 7.18 shows the plot
of Eq. 7.43 and the relationship between the time constant and the initial rate at which
the inductor voltage is decreasing.
If there is an initial current in the inductor, Eq. 7.35 gives the solution for it. The
algebraic sign of 𝐼0 is positive if the initial current is in the same direction as 𝑖;
otherwise, 𝐼0 carries a negative sign.
‫ رجذأ‬ٟٙ‫ٓ ف‬١‫ٓ اِشأر‬١‫خذِد صذالخ ث‬ُٚ ‫ئرا‬
. ‫ ارسبد ضذ اِشأح ثبٌثخ‬ٌٝ‫ي ئ‬ٚ‫ رإ‬ٚ‫أ‬
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 Example 7.5:
The switch in the circuit shown in Fig.7.19 has been in position a for a long time. At
𝑡 = 0, the switch moves from position a to position b. The switch is a make-beforebreak type; that is, the connection at position b is established before the connection at
position a is broken, so there is no interruption of current through the inductor.
a) Find the expression for 𝑖(𝑡) for 𝑡 ≥ 0.
b) What is the initial voltage across the inductor just after the switch has been moved to
position b?
c) How many milliseconds after the switch has been moved does the inductor voltage
equal 24V?
d) Does this initial voltage make sense in terms of circuit behavior?
e) Plot both 𝑖(𝑡) and 𝑣(𝑡) versus 𝑡.
 Solution:
a) The switch has been in position a for a long time, so the 200 mH inductor is a short
circuit across the 8 A current source. Therefore, the inductor carries an initial current
of 8 A. This current is oriented opposite to the reference direction for 𝑖; thus 𝐼0 is
− 8 A. When the switch is in position b, the final value of 𝑖 will be 24 2, or 12 A.
The time constant of the circuit is 200 2 = 100 ms.
𝑖 = 12 + −8 − 12 𝑒 −𝑡
b)
𝑣=𝐿
0.1
= 12 − 20 𝑒 −10𝑡 A,
𝑑𝑖
= 0.2 200𝑒 −10𝑡 = 40 𝑒 −10𝑡 V,
𝑑𝑡
𝑡≥0
t ≥ 0+
𝑣 0+ = 40 V
‫ب اٌشدبػخ‬ٙ‫ئْ اٌّشأح لذ رٕمص‬
‫ب ال رضاي رٍر‬ٕٙ‫خ ٌالٔزسبس ٌى‬١‫اٌىبف‬
.‫ رفؼً أٔذ رٌه‬ٟ‫مه و‬٠‫رضب‬ٚ ‫ه‬١ٍ‫ػ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Continued (Example 7.5) :
c) 24 = 40 𝑒 −10𝑡
For 𝑡:
𝑡=
1 40
ln
= 51.08 × 10−3 = 51.08 ms
10 24
d) Yes; in the instant after the switch has been moved to b, the inductor sustains a
current of 8 A counterclockwise around the closed path. This current causes a 16 V
drop across the 2 Ω resistor. This voltage drop adds to the drop across the source,
producing a 40 V drop across the inductor.
e)
‫ رشرؼذ‬ٟ‫ اٌّشأح اٌز‬ٟ٘ ‫أخًّ اِشأح‬
.ْ‫ وبْ صِب‬،‫ب‬ٙ١‫ شفز‬ٍٝ‫وٍّبد اٌست ػ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Assessment problem 7.5:
Assume that the switch in the circuit shown in Fig.7.19 has been in position b for a long
time and at 𝑡 = 0 it moves to position a. Find
a) 𝑖 0+ .
b) 𝑣(0+).
c) 𝜏,
𝑡 > 0.
d) 𝑖 𝑡 ,
𝑡 ≥ 0.
e) 𝑣 𝑡 , 𝑡 ≥ 0+.
 Solution:
a) Use the circuit at t < 0, shown below, to calculate the initial current in the inductor:
𝑖 0− =
24
= 12 A = 𝑖(0+)
2
Note that 𝑖 0− = 𝑖 0+ because the current in an inductor is continuous.
b) Use the circuit at 𝑡 = 0+, shown below, to calculate the voltage drop across the
inductor at 0+. Note that this is the same as the voltage drop across the 10Ω
resistor, which has current from two sources — 8 A from the current source and 12
A from the initial current through the inductor.
ٝ‫ٓ زز‬٠‫ عٓ اٌؼشش‬ٟ‫رظً اٌّشأح ف‬
.‫ب‬ٙ‫بر‬١‫آخش ٌسظخ ِٓ ز‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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
Contiuned (Assessment problem 7.5) :
𝑣 0+ = − 10 8 + 12 = − 200 V
c) To calculate the time constant we need the equivalent resistance seen by the
inductor for 𝑡 > 0. Only the 10Ω resistor is connected to the inductor for 𝑡 > 0.
Thus,
𝐿
200 × 10−3
𝜏= =
= 20 ms
𝑅
10
d) To find 𝑖(𝑡), we need to find the final value of the current in the inductor. When
the switch has been in position a for a long time, the circuit reduces to the one
below:
Note that the inductor behaves as a short circuit and all of the current from the 8 A
source flows through the short circuit. Thus,
𝑖𝑓 = − 8 A
Now,
𝑖 𝑡 = 𝑖𝑓 + 𝑖 0+ − 𝑖𝑓 𝑒 −𝑡/𝜏 = − 8 + [12 − − 8 ]𝑒 −𝑡/0.02
= −8 + 20 𝑒 −50𝑡 A, t ≥ 0
e) To find 𝑣(𝑡), use the relationship between voltage and current for an inductor:
𝑣 𝑡 =𝐿
𝑑𝑖(𝑡)
= 200 × 10−3 − 50 20 𝑒 −50𝑡 = − 200 𝑒 −50𝑡 V,
𝑑𝑡
t ≥ 0+
‫ب‬ٙ١‫د اٌشخً ف‬ّٛ٠ ْ‫رفضً اٌّشأح أ‬
.‫ب ثؼذ رٌه‬ِٕٙ ‫د‬ّٛ٠ ْ‫أ‬ٚ ،‫ال‬ٚ‫أ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 We can also describe the voltage 𝑣 𝑡
across the inductor in Fig. 7.16 in terms of
the circuit current.
𝑉𝑠 𝑣 𝑡
−
𝑅
𝑅
𝑑𝑖
1 𝑑𝑣
=−
𝑑𝑡
𝑅 𝑑𝑡
𝑖 𝑡 =
Multiply each side by die inductance 𝐿, we get an expression for the voltage
across the inductor on the left-hand side.
𝐿 𝑑𝑣
𝑅 𝑑𝑡
𝑑𝑣 𝑅
+ 𝑣=0
𝑑𝑡 𝐿
𝑣=−
7.47
The solution to Eq. 7.47 is identical to that given in Eq. 7.42.
 General observation about the step response of an 𝑅𝐿 circuit is to be considered.
When we derived the differential equation for the inductor current, we obtained
𝑑𝑖 𝑅
𝑉𝑠
+ 𝑖=
𝑑𝑡 𝐿
𝐿
7.48
Observe that Eqs. 7.47 and 7.48 have the same form. Each equates the sum of the
first derivative of the variable and a constant times the variable to a constant
value. In Eq. 7.47, the constant on the right-hand side is zero; hence this equation
takes on the same form as the natural response equations in Section 7.1. In both
Eq. 7.47 and Eq. 7.48, the constant multiplying the dependent variable is the
reciprocal of the time constant, that is, 𝑅 𝐿 = 1 𝜏. We will see a similar situation
in the derivations for the step response of an 𝑅𝐶 circuit. In Section 7.4, we will
use these observations to develop a general approach to finding the natural and
step responses of 𝑅𝐿 and 𝑅𝐶 circuits.
ْ‫ْ أ‬ٛ‫ؼ‬١‫غزط‬٠ ‫اع فمظ ِٓ اٌشخبي ال‬ٛٔ‫ثالثخ أ‬
.‫ي‬ٛٙ‫اٌى‬ٚ ‫ش‬ٛ١‫اٌش‬ٚ ‫ اٌشجبة‬:‫ا اٌّشأح‬ّٛٙ‫ف‬٠
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 The Step Response of an RC Circuit
Summing the currents away from the top node
𝐶
𝑑𝑣𝐶 𝑣𝐶
+
= 𝐼𝑠
𝑑𝑡
𝑅
𝑑𝑣𝐶 𝑣𝐶
𝐼𝑠
+
=
𝑑𝑡
𝑅𝐶 𝐶
7.49
7.50
Comparing Eq. 7.50 with Eq. 7.48 reveals that the form of the solution for 𝑣𝐶 is the
same as that for the current in the inductive circuit
𝑣𝐶 = 𝐼𝑠 𝑅 + 𝑉0 − 𝐼𝑠 𝑅 𝑒 −𝑡
𝑅𝐶
,
𝑡≥0
A similar derivation for the current in the capacitor yields the differential equation
𝑑𝑖
1
+
𝑖=0
𝑑𝑡 𝑅𝐶
Equation 7.52 has the same form as Eq. 7.47
𝑖 = 𝐼𝑠 −
𝑉0 −𝑡
𝑒
𝑅
𝑅𝐶
,
𝑡 ≥ 0+
Where 𝑉0 is the initial value of 𝑣𝐶 , the voltage across the capacitor. From Eq.
7.51, note that the initial voltage across the capacitor is 𝑉0 , the final voltage across the
capacitor is 𝐼𝑠 𝑅, and the time constant of the circuit is 𝑅𝐶. Also note that the solution
for 𝑣𝐶 is valid for 𝑡 ≥ 0. These observations are consistent with the behavior of a
capacitor in parallel with a resistor when driven by a constant current source.
Equation 7.53 predicts that the current in the capacitor at 𝑡 = 0+ is 𝐼𝑠 − 𝑉0 𝑅.
This prediction makes sense because the capacitor voltage cannot change
instantaneously, and therefore the initial current in the resistor is 𝑉0 𝑅. The capacitor
branch current changes instantaneously from zero at 𝑡 = 0− to 𝐼𝑠 − 𝑉0 𝑅 at 𝑡 = 0+.
The capacitor current is zero at 𝑡 = ∞. Also note that the final value of 𝑣 = 𝐼𝑠 𝑅.
ٟ‫ ٌىٓ رخزف‬،‫اح‬ٚ‫دائّب ًا رطبٌت اٌّشأح ثبٌّغب‬
ٚ‫الة أ‬ٚ‫ُ أسفف اٌذ‬١‫٘زٖ اٌفىشح ػٕذ رمغ‬
.ً‫ ِس‬ٞ‫ أ‬ٟ‫لذ دفغ اٌسغبة ف‬ٚ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Example 7.6: Determining the step Response of an RC Circuit
The switch in the circuit shown in Fig.7.22 has been in position 1 for a long time. At
𝑡 = 0, the switch moves to position 2. Find:
a) 𝑣𝑜 (𝑡) for 𝑡 ≥ 0.
b) 𝑖𝑜 (𝑡) for 𝑡 ≥ 0+.
 Solution:
a) We begin by computing the open-circuit voltage, which is given by the − 75 V
source divided across the 40 kΩ and 160 kΩ resistors:
160 × 103
𝑉𝑜𝑐 =
− 75 = − 60 V
40 + 160 × 103
Next, we calculate the Thévenin resistance, as seen to the right of the capacitor, by
shorting the − 75 V source and making series and parallel combinations of the
resistors:
𝑅𝑇𝑕 = 8000 + 40,000 160,000 =
40 kΩ
The value of the Norton current source is the ratio of the open-circuit voltage to the
Thévenin resistance, or − 60 40 × 103 = − 1.5 mA. The resulting Norton
equivalent circuit is shown in Fig. 7.23. From Fig. 7.23, 𝐼𝑠 𝑅 = − 60 V and
𝑅𝐶 = 10 ms. We note that 𝑣0 0 = 30 V, so the solution for 𝑣0 is:
𝑣𝑜 = − 60 + 30 − − 60 𝑒 −100𝑡 = − 60 + 90 𝑒 −100𝑡 V,
𝑡≥0
.‫ب‬ٙ‫ه ئال ئػدبثه ث‬١‫ء ف‬ٟ‫لذ رشفض اٌّشأح وً ش‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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
Contiuned (Example 7.6) :
b) We write the solution for 𝑖𝑜 directly from Eq. 7.53 by noting that 𝐼𝑠 = − 1.5 mA
and 𝑉0 𝑅 = 30 40 × 10−3 = 0.75 mA:
𝑖𝑜 = − 2.25 𝑒 −100𝑡 mA,
𝑡 ≥ 0+
We can check the consistency of the solutions for 𝑣𝑜 and 𝑖𝑜 by noting that
𝑖𝑜 = 𝐶
𝑑𝑣𝑜
= 0.25 × 10−6 − 9000 𝑒 −100𝑡 = − 2.25 𝑒 −100𝑡 mA
𝑑𝑡
Because 𝑑𝑣𝑜 0− 𝑑𝑡 = 0, the expression for 𝑖𝑜 clearly is valid only for 𝑡 ≥ 0+.
،‫ّب ٌٍشخبي‬١‫ْ صػ‬ٛ‫ى‬٠ ْ‫ّىٕٗ أ‬٠ ً‫ٓ أٌف سخ‬١‫ازذ ِٓ ث‬ٚ ً‫سخ‬
.‫ي‬ٚ‫وزٌه اٌشخً األ‬ٚ ،‫ْ إٌغبء‬ٛ‫زجؼ‬١‫خ ف‬١‫ اٌجبل‬999 ‫اي‬
‫أِب ـ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Assessment 7.6:
a) Find the expression for the voltage across the 160 kΩ resistor in the circuit shown in
Fig.7.22. Let this voltage be denoted 𝑣𝐴 , and assume that the reference polarity for the
voltage is positive at the upper terminal of the 160 kΩ resistor.
b) Specify the interval of time for which the expression obtained in (a) is valid.
 Solution:
a)
From Example 7.6,
𝑣𝑜 𝑡 = − 60 + 90 𝑒 −100𝑡 V
Write a KCL equation at the top node and use it to find the relationship between
𝑣𝑜 and 𝑣𝐴 :
𝑣𝐴 − 𝑣𝑜
𝑣𝐴
𝑣𝐴 + 75
+
+
=0
8000
160,000 40,000
20𝑣𝐴 − 20𝑣𝑜 + 𝑣𝐴 + 4𝑣𝐴 + 300 = 0
25𝑣𝐴 = 20𝑣𝑜 − 300 ⟹
𝑣𝐴 = 0.8𝑣𝑜 − 12
Use the above equation for 𝑣𝐴 in terms of 𝑣𝑜 to find the expression for 𝑣𝐴 :
𝑣𝐴 𝑡 = 0.8 − 60 + 90 𝑒 −100𝑡 − 12 = − 60 + 72 𝑒 −100𝑡 V,
t ≥ 0+
b) t ≥ 0+, since there is no requirement that the voltage be continuous in a resistor.
.‫ئرا طجخذ لزٍذ‬ٚ ،‫ئرا اثزغّذ عسشد‬ٚ ،‫ٕذ فزٕذ‬٠‫َ ئرا رض‬ٛ١ٌ‫فزبح ا‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.35:
The switch in the circuit shown in Fig. P7.35 has been closed for a long time before
opening at 𝑡 = 0.
a) Find the numerical expressions for 𝑖𝐿 𝑡 and 𝑣𝑜 𝑡 for 𝑡 ≥ 0.
b) Find the numerical values of 𝑣𝐿 0+ and 𝑣𝑜 0+ .
 Solution
a) 𝑡 < 0 :
:
𝑖𝐿 0− = 6 A
𝑡>0:
32 + 48
𝑖𝐿 ∞ =
=4A
20
𝐿 5 × 10−3
1
𝜏= =
= 250 µs;
= 4000
𝑅
20
𝜏
𝑖𝐿 = 4 + 6 − 4 𝑒 −4000𝑡 = 4 + 2 𝑒 −4000𝑡 A, 𝑡 ≥ 0
𝑣𝑜 = − 8𝑖𝐿 + 48 = − 8 4 + 2 𝑒 −4000𝑡 + 48 = 16 − 16 𝑒 −4000𝑡 V,
b) 𝑣𝐿 = 5 × 10−3
𝑑𝑖𝐿
= 5 × 10−3 − 8000 𝑒 −4000𝑡 = − 40 𝑒 −4000𝑡 V,
𝑑𝑡
𝑡 ≥ 0+
𝑡 ≥ 0+
𝑣𝐿 0+ = − 40 V
𝑣𝑜 0+ = 0 V
check ∶ at 𝑡 = 0+ the circuit is:
𝑣𝐿 0+ = 32 − 72 + 0 = − 40 V,
𝑣𝑜 0+ = 48 − 48 = 0 V
‫ اٌّشأح‬ٍٝ‫ِب أشك ػ‬
.‫أْ رىزُ عشا‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.37:
The switch in the circuit shown in Fig. P7.37 has been in position a for a long time. At
𝑡 = 0. The switch moves instantaneously to position b.
a) Find the numerical expression for 𝑖𝑜 𝑡 when 𝑡 ≥ 0.
b) Find the numerical expression for 𝑣𝑜 𝑡 for 𝑡 ≥ 0+.
 Solution:
a) 𝑡 < 0 :
KCL equation at the top node:
𝑣𝑜 (0−) 𝑣𝑜 (0−) 𝑣𝑜 (0−)
50 =
+
+
8
40
10
Multiply by 40 and solve:
2000 = 5 + 1 + 4 𝑣𝑜 ;
𝑣𝑜 = 200 V
𝑣𝑜
200
∴ 𝑖𝑜 0− =
=
= 20 A
10
10
𝑡>0:
ٓ١‫شزى‬٠ ٓ١‫ عٓ األسثؼ‬ٟ‫ش ِٓ إٌغبء ف‬١‫وث‬
!ٓ١‫ اٌغز‬ٟ‫ٓ ف‬ٙٔ‫اٌغجت أ‬ٚ ‫ش‬ٙ‫ِٓ آالَ اٌظ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Continued (Problem 7.37):
Use voltage division to find the Thévenin voltage:
𝑉𝑇𝑕
40
= 𝑣𝑜 =
800 = 200 V
40 + 120
Remove the voltage source and make series and parallel combinations of
resistors to find the equivalent resistance:
𝑅𝑇𝑕 = 10 + 120 40 = 10 + 30 = 40 Ω
The simplified circuit is:
𝐿 40 × 10−3
𝜏= =
= 1 ms;
𝑅
40
200
𝑖𝑜 ∞ =
=5A
40
∴ 𝑖𝑜 = 𝑖𝑜 ∞ + 𝑖𝑜 0+ − 𝑖𝑜 (∞) 𝑒 −𝑡
1
= 1000
𝜏
𝜏
= 5 + 20 − 5 𝑒 −1000 𝑡 = 5 + 15 𝑒 −1000 𝑡 A,
b) 𝑣𝑜 = 10𝑖𝑜 + (0.04)
𝑡≥0
𝑑𝑖𝑜
𝑑𝑡
= 50 + 150 𝑒 −1000 𝑡 + 0.04 15 −1000𝑒 −1000 𝑡
= 50 + 150 𝑒 −1000 𝑡 − 600 𝑒 −1000 𝑡
𝑣𝑜 = 50 − 450 𝑒 −1000 𝑡 V,
𝑡 ≥ 0+
ّٓ‫ثث‬ٚ ‫ب‬ٕٙ‫زؼٍك ثغ‬٠ ‫ ِب‬ٟ‫ال رىزة اٌّشأح ػبدح ئال ف‬
.‫شح‬١‫ وث‬ٜ‫بء أخش‬١‫أش‬ٚ ،‫ب‬ٙ‫خ‬ٚ‫ثذخً ص‬ٚ ‫ب‬ٙ‫اث‬ٛ‫أث‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Problem 7.36:
After the switch in the circuit of Fig. P7.36 has been open for a long time, it is closed at
𝑡 = 0. Calculate
a) The initial value of 𝑖.
b) The final value of 𝑖.
c) The time constant for 𝑡 ≥ 0.
d) The numerical expression for 𝑖 𝑡 when 𝑡 ≥ 0.
 Solution:
a) For 𝑡 < 0, calculate the Thévenin equivalent for the circuit to the left and right of
the 75 mH inductor. We get
5 − 120
= − 5 mA
15 k + 8 k
𝑖 0− = 𝑖 0+
= − 5 mA
b) For 𝑡 > 0 the circuit reduces to
𝑖 0− =
Therefore 𝑖 ∞ =
5
= 0.333 mA
15,000
𝐿 75 × 10−3
c) 𝜏 = =
= 5 µ𝑠
𝑅
15,000
d) 𝑖 𝑡 = 𝑖 ∞ + 𝑖 0+ − 𝑖(∞) 𝑒 −𝑡 𝜏 = 0.333 + − 5 − 0.333 𝑒 −200,000𝑡
= 0.333 − 5.333 𝑒 −200,000𝑡 mA,
𝑡≥0
.‫ػذ أْ ال رؼذ‬ٌٛ‫ٍخ ٌٍجش ثب‬١‫ع‬ٚ ً‫أفض‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Problem 7.56:
The switch in the circuit of Fig. P7.56 has been in position a for a long time. At 𝑡 = 0 the
switch is moved to position b. Calculate:
a) The initial voltage on the capacitor.
b) The final voltage on the capacitor.
c) The time constant (in microseconds) for 𝑡 > 0.
d) The length of time (in microseconds) required for the capacitor voltage to reach zero
after the switch is moved to position b.
 Solution:
a) Use voltage division to find the initial value of the voltage:
𝑣𝑐 0+ = 𝑣9𝑘 =
9K
120 = 90 V
9K+3K
b) Use Ohm's law to find the final value of voltage:
𝑣𝑐 ∞ = 𝑣40𝑘 = −1.5 × 10−3 40,000 = −60 V
c) Find the Thévenin equivalent with respect to the terminals of the capacitor:
𝑉𝑇𝑕 = −60 V,
𝑅𝑇𝑕 = 40 K + 10 K = 50 KΩ
𝜏 = 𝑅𝑇𝑕 𝐶 = 1000 µs
d) 𝑣𝑐 = 𝑣𝑐 ∞ + 𝑣𝑐 0+ − 𝑣𝑐 (∞) 𝑒 −𝑡
𝜏
= −60 + 90 + 60 𝑒 −1000𝑡 = −60 + 150 𝑒 −100𝑡 V,
𝑡≥0
We want 𝑣𝑐 = −60 + 150 𝑒 −100𝑡 = 0:
Therefore 𝑡 =
ln
150
60 = 916.3 µs
100
،‫ال رٕظ االػززاس ٌٍشخً ئرا وٕذ ِخطئب‬
.‫ ٌُ رىٓ ِخطئب‬ٌٛ ٝ‫ٌٍّشأح زز‬ٚ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Problem 7.54:
The switch in the circuit seen in Fig. P7.54 has been in position a for a long time. At 𝑡 =
0, the switch moves instantaneously to position b. Find 𝑣𝑜 𝑡 and 𝑖𝑜 𝑡 for 𝑡 ≥ 0+.
 Solution:
𝑡 < 0;
𝑖𝑜 0− = 10
20
= 2 mA;
100
𝑣𝑜 0− = 2 50 = 100 V
𝑡 = ∞:
20
= − 1 mA;
100
= 50 KΩ 50 KΩ = 25 KΩ;
𝑖𝑜 ∞ = − 5
𝑅𝑇𝑕
𝑣𝑜 ∞ = 𝑖𝑜 ∞ 50 = − 50 V
𝐶 = 16 nF
1
= 2500
𝜏
𝜏 = 25 0.016 = 0.4 ms;
∴ 𝑣𝑜 𝑡 = − 50 + 150 𝑒 −2500𝑡 V,
𝑡 ≥ 0+
𝑑𝑣𝑜
𝑖𝑐 = 𝐶
= − 6 𝑒 −2500𝑡 mA,
𝑡 ≥ 0+
𝑑𝑡
𝑣𝑜
𝑖50 =
= − 1 + 3 𝑒 −2500𝑡 mA,
𝑡 ≥ 0+
50
𝑖𝑜 = 𝑖𝑐 + 𝑖50 = − 1 + 3 𝑒 −2500𝑡 mA, 𝑡 ≥ 0+
ٓ‫ٌى‬ٚ ،ً١ٍ‫ك اٌّشأح أْ رّزٍه اٌم‬٠‫ضب‬٠ ‫ال‬
.‫ب‬ِٕٙ ‫ رّزٍه أوثش‬ٜ‫ اِشأح أخش‬ٜ‫أْ رش‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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7.4
A General Solution for Step and Natural Responses:
 The general approach to finding either the
natural response or the step response of
the first-order 𝑅𝐿 and 𝑅𝐶 circuits shown
in Fig. 7.24 is based on their differential
equations having the same form.
 To generalize the solution of these four
possible circuits, we let 𝑥 𝑡 represent the
unknown quantity, giving 𝑥 𝑡
four
possible values. It can represent the
current or voltage at the terminals of an
inductor or the current or voltage at the
terminals of a capacitor.
 We know that the differential equation
describing any one of the four circuits in
Fig. 7.24 takes the form
𝑑𝑥 𝑥
+ =𝐾
𝑑𝑡 𝜏
7.54
Where the value of the constant 𝐾 can be zero. Because the sources in the circuit
are constant voltages and/or currents, the final value of 𝑥 will be constant; that is,
the final value must satisfy Eq. 7.54, and, when 𝑥 reaches its final value, the
derivative 𝑑𝑥 𝑑𝑡 must be zero. Hence
𝑥𝑓 = 𝐾𝜏
− 𝑥 − 𝑥𝑓
𝑑𝑥 − 𝑥
− 𝑥 − 𝐾𝜏
=
+𝐾 =
=
𝑑𝑡
𝜏
𝜏
𝜏
.‫ش‬ٙ‫ش٘ب وً عزخ أش‬١‫ٌزٌه ٔغ‬ٚ ،‫ضخ‬ٌّٛ‫ّىٓ رسًّ ثشبػخ ا‬٠ ‫ال‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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𝑑𝑥
−1
=
𝑑𝑡
𝑥 − 𝑥𝑓
𝜏
𝑥 𝑡
𝑥 𝑡0
𝑑𝑥
−1
=
𝑥 − 𝑥𝑓
𝜏
𝑡
𝑑𝑡
𝑡0
𝑥 𝑡 = 𝑥𝑓 + 𝑥 𝑡0 − 𝑥𝑓 𝑒 −(𝑡−𝑡 0 )
𝜏
the unknown
the final
the initial
the final
− 𝑡− time of switching
time constant
variable as a = value of the + value of the − value of the × 𝑒
function of time
variable
variable
variable
In many cases, the time of switching—that is, 𝑡0 —is zero.
 When computing the step and natural responses of circuits, it may help to
follow these steps:
1. Identify the variable of interest for the circuit. For 𝑅𝐶 circuits, it is most
convenient to choose the capacitive voltage; for 𝑅𝐿 circuits, it is best to choose
the inductive current.
2. Determine the initial value of the variable, which is its value at 𝑡0 . Note that if
you choose capacitive voltage or inductive current as your variable of interest, it
is not necessary to distinguish between 𝑡 = 𝑡0− and 𝑡 = 𝑡0+ This is because they
both are continuous variables. If you choose another variable, you need to
remember that its initial value is defined at 𝑡 = 𝑡0+ .
3. Calculate the final value of the variable, which is its value as 𝑡 → ∞.
4. Calculate the time constant for the circuit.
ٓ‫ ٌى‬،ٍُ‫خ ٌدؼً إٌغبء رزى‬٠ٚ‫ٕب أد‬٠‫ٌذ‬
.ٓ‫صّز‬٠ ٍٓٙ‫ء ٌدؼ‬ٟ‫ٕب ش‬٠‫ظ ٌذ‬١ٌ
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Example 7.7: Using the General Solution Method to Find an 𝑹𝑪 Circuit's Step Response
The switch in the circuit shown in Fig. 7.25 has been in position a for a long time. At
𝑡 = 0 the switch is moved to position b.
a) What is the initial value of 𝑣𝐶 ?
b) What is the final value of 𝑣𝐶 ?
c) What is the time constant of the circuit when the switch is in position b?
d) What is the expression for 𝑣𝐶 𝑡 when 𝑡 ≥ 0?
e) What is the expression for 𝑖 𝑡 when 𝑡 ≥ 0+?
f) How long after the switch is in position b does the capacitor voltage equal zero?
g) Plot 𝑣𝐶 𝑡 and 𝑖 𝑡 versus 𝑡.
 Solution:
a) The switch has been in position a for a long time, so the capacitor looks like an
open circuit. From the voltage divider rule, the voltage across the 60 Ω resistor is:
40 × 60 60 + 20 ,
So,
or 30 V
𝑣𝐶 0 = − 30 V
b) After the switch has been in position b for a Long time, the capacitor will look like
an open circuit. Thus the final value of the capacitor voltage is: + 90 V.
c) The time constant is:
𝜏 = 𝑅𝐶 = 400 × 103 0.5 × 10−6 = 0.2 s
d) 𝑣𝐶 𝑡 = 90 + − 30 − 90 𝑒 −5𝑡 = 90 − 120 𝑒 −5𝑡 V,
𝑡≥0
،‫خ‬١‫ٓ شخص‬ٙ٠‫ظ ٌذ‬١ٌ ‫ي أْ إٌغبء‬ٛ‫ٌٓ أل‬
.‫ذح‬٠‫خ خذ‬١‫َ شخص‬ٛ٠ ً‫ٓ و‬ٙ٠‫ثً ٌذ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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
Contiuned (Example 7.7) :
e) Here the value for 𝜏 doesn't change. Thus we need to find only the initial and final
values for the current in the capacitor. When obtaining the initial value, we must
get the value of 𝑖 0+ , because the current in the capacitor can change
instantaneously. This current is equal to the current in the resistor, which from
Ohm's law is 90 − − 30
400 × 103 = 300 𝜇A. Note that when applying
Ohm's law we recognized that the capacitor voltage cannot change instantaneously.
The final value of 𝑖 𝑡 = 0, so
𝑖(𝑡) = 0 + 300 − 0 𝑒 −5𝑡 = 300 𝑒 −5𝑡 𝜇𝐴,
𝑡 ≥ 0+
f) Solve the equation derived in (d) for the time when 𝑣𝐶 𝑡 = 0:
120 𝑒 −5𝑡 = 90
so
or
𝑒 5𝑡 =
120
90
1
4
𝑡 = 𝑙𝑛
= 57.54 ms
5
3
Note that when 𝑣𝐶 = 0, 𝑖 = 225 𝜇A and the voltage drop across the 400 kΩ
resistor is 90 V.
g)
‫ِب اخزّؼذ اِشأربْ رزسذثبْ ئال‬
.‫ّب‬ٙ‫ذ ِٓ خجشر‬١‫غزف‬٠ ْ‫طب‬١‫صّذ اٌش‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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 Example 7.8: Using the General Solution Method with Zero Initial Conditions
The switch in the circuit shown in Fig. 7.27 has been open for a long time. The initial
charge on the capacitor is zero. At 𝑡 = 0, the switch is closed. Find the expression for
a) 𝑖 𝑡 for 𝑡 ≥ 0+.
b) 𝑣 𝑡 when 𝑡 ≥ 0+.
 Solution:
a) Because the initial voltage on the capacitor is zero, at the instant when the switch
is closed the current in the 30 kΩ branch will be:
𝑖 0+ =
7.5 20
= 3 mA
50
The final value of the capacitor current will be zero because the capacitor
eventually will appear as an open circuit in terms of dc current. Thus 𝑖𝑓 = 0. The
time constant of the circuit will equal the product of the Thévenin resistance (as
seen from the capacitor) and the capacitance. Therefore
𝜏 = 20 + 30 × 103 × 0.1 × 10−6 = 5 ms
𝑖 𝑡 = 0 + 3 − 0 𝑒 −𝑡
5×10 −3
= 3 𝑒 −200𝑡 ,
𝑡 ≥ 0+
b) To find 𝑣 𝑡 , we note from the circuit that it equals the sum of the voltage across
the capacitor and the voltage across the 30 kΩ resistor. To find the capacitor
voltage (which is a drop in the direction of the current), we note that its initial
value is zero and its final value is 7.5 20 , or 150 V. The time constant is the
same as before, or 5 ms.
𝑣𝐶 𝑡 = 150 + 0 − 150 𝑒 −200𝑡 = 150 − 150 𝑒 −200𝑡 V,
𝑡≥0
𝑣 𝑡 = 150 − 150 𝑒 −200𝑡 + 30 3 𝑒 −200𝑡 = 150 − 60 𝑒 −200𝑡 V,
𝑡 ≥ 0+
.‫ـك‬٠‫ـك لجـً اٌطـش‬١‫اٌـشف‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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July 2013
 Example 7.9: Using the General Solution Method to Find an 𝑹𝑳 Circuit's Step Response
The switch in the circuit shown in Fig. 7.28 has been open for a long time. At 𝑡 = 0 the
switch is closed. Find the expression for
a) 𝑣 𝑡 when 𝑡 ≥ 0+.
b) 𝑖 𝑡 when 𝑡 ≥ 0.
 Solution:
a) The switch has been open for a long time, so the initial current in the inductor is
5 A, oriented from top to bottom. Immediately after the switch closes, the current
still is 5 A, and therefore the initial voltage across the inductor becomes 20 − 5 1 ,
or 15 V. The final value of the inductor voltage is 0 V. With the switch closed, the
time constant is 80 1, or 80 ms.
𝑣 𝑡 = 0 + 15 − 0 𝑒 −𝑡
80×10 −3
= 15 𝑒 −12.5𝑡 V,
𝑡 ≥ 0+
b) We have already noted that the initial value of the inductor current is 5 A. After the
switch has been closed for a long time, the inductor current reaches 20 1, or 20 A.
The circuit time constant is 80 ms,
𝑖 𝑡 = 20 + 5 − 20 𝑒 −12.5𝑡 = 20 − 15 𝑒 −12.5𝑡 A,
𝑡≥0
We determine that the solutions for 𝑣 𝑡 and 𝑖 𝑡 agree by noting that
𝑣 𝑡 =𝐿
𝑑𝑖
= 80 × 10−3 15 × 12.5 × 𝑒 −12.5𝑡 = 15 𝑒 −12.5𝑡 V,
𝑑𝑡
𝑡 ≥ 0+
ِٓ ٓ‫ ٌى‬،‫ساء سخً ٔدر اِشأح‬ٚ ْ‫ِٓ اٌّسزًّ أ‬
.‫ساء وً سخً ٔدر ثُ فشً اِشأح‬ٚ ْ‫اٌّإوذ أ‬
ٟٔٚ‫ذ اإلٌىزش‬٠‫ ثبٌجش‬ٚ‫ أ‬9 4444 260 ‫خ‬١‫خ ثشعبٌخ ٔص‬٠‫س‬ٚ‫ ِالزظبد رشا٘ب ضش‬ٚ‫ خطأ أ‬ٞ‫ اٌّغبّ٘خ ثبإلثالؽ ػٓ أ‬ٝ‫شخ‬١‫خ ٌٍٕفغ اٌؼبَ ف‬١ٔ‫ربد ِدب‬ٌٕٛ‫ا‬
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