Response of First- Order RL RC Circuits - eng-hs.net of-First-Order-RL... · Response of...

July 2013

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رغزش أغت اصذالبد

.طبب وبذ اؼاللخ غ١خ

Chapter 7

Response of First-

Order RL and RC

Circuits

زوشاد ششذ ربس٠ سخ

ااد أدب ازسببد عبثمخ ؼذ٠ذ

ا ػ أدبالؼ١ ازوس٠ زبزخ دبب

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July 2013




ال أزذ ٠غزسك دػه، ئ

.اعزسمب أزذ ف ٠ذػه رزسفب

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.

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١ظ اصؼت أ رضس أخ صذ٠ك،

.ى اصؼت أ ردذ از ٠غزسك ازضس١خ

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 first-

order 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.

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البزخ أ رغأي صذ٠مه از لغ

.ف سطخ ئ وب ٠سزبج غبػذره

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

𝐿𝑑𝑖

𝑑𝑡+ 𝑅𝑖 = 0

Solving,

Natural response of an RL circuit → 𝑖 𝑡 = 𝐼𝑜𝑒− 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+ = 𝐼𝑜𝑅

𝑝 = 𝑖2𝑅 = 𝐼𝑜2𝑅𝑒−2 𝑅 𝐿 𝑡 , 𝑡 ≥ 0+

𝑤 = 𝑝 𝑑𝑥𝑡

0

=1

2𝐿𝐼0

2 1 − 𝑒−2 𝑅 𝐿 𝑡 , 𝑡 ≥ 0

As 𝑡 becomes infinite, the energy dissipated in the resistor approaches the initial

energy stored in the inductor.

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ىذ اذ١ب ػ اشء أ رش

.ػذا ه ب صذالز ثذ

The time constant 𝛕 of the circuit:

𝜏 = time constant =𝐿

𝑅

Using the time-constant concept

𝑖 𝑡 = 𝐼0𝑒−𝑡 𝜏 , 𝑡 ≥ 0

𝑣 𝑡 = 𝐼0𝑅𝑒−𝑡 𝜏 , 𝑡 ≥ 0+

𝑝 = 𝐼02𝑅𝑒−2𝑡 𝜏 , 𝑡 ≥ 0+

𝑤 =1

2𝐿𝐼0

2 1 − 𝑒−2𝑡 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 = 𝐼𝑜𝑒−𝑡 𝜏 .t0 generate 𝑖 𝑡 from 𝐼𝑜 and 𝜏.

4. All other calculations of interest follow from knowing 𝑖 𝑡 .

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𝑅𝑒𝑞 = 2 + 40 ∥ 10 = 10 Ω

𝜏 =1

𝑅𝑒𝑞=

2

𝑇0= 0.2 s

𝑖𝐿 𝑡 = 20 𝑒−5𝑡 A, 𝑡 ≥ 0

𝑖0 = − 𝑖𝐿10

10 + 40= − 4 𝑒−5𝑡 A, 𝑡 ≥ 0+

d) 𝑝10Ω 𝑡 =𝑣0

2

10= 2560 𝑒−10𝑡 W, 𝑡 ≥ 0+

𝑤10Ω 𝑡 = 2560 𝑒−10𝑡 𝑑𝑡∞

0

= 256 J

𝑤 0 =1

2𝐿 𝑖2 0 =

1

2 2 20 2 = 400 J

256

400 100 = 64%

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

b) Looking at bottom side:

c) 𝑣0 𝑡 = 40𝑖0 = − 160 𝑒−5𝑡 V, 𝑡 ≥ 0+

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.

ا ال ز١ث ٠ى االعزؼذاد ػظ١ب

.٠ى أ رى اصؼثبد ػظ١خ

Solution:

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𝑖 𝑡 = 12 𝑒−2𝑡 A, 𝑡 ≥ 0

𝑣 𝑡 = 96 𝑒−2𝑡 V, 𝑡 ≥ 0+

𝑖1 =1

5 96 𝑒−2𝑥 𝑑𝑥

𝑡

0

− 8 = 1.6 − 9.6 𝑒−2𝑡 A, 𝑡 ≥ 0

𝑖2 =1

20 96 𝑒−2𝑥 𝑑𝑥

𝑡

0

− 4 = − 1.6 − 2.4 𝑒−2𝑡 A, 𝑡 ≥ 0

𝑖3 =𝑣(𝑡)

10 15

25= 5.76 𝑒−2𝑡 A, 𝑡 ≥ 0+

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.

The circuit shows that 𝑣 𝑡 = 0 at 𝑡 = 0−, so the expression for 𝑣 𝑡 is valid for

𝑡 ≥ 0+.

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).

ال أزذ ٠خطظ فش، ٠أر

.افش ػذب ال خطظ

Fig. 7.8

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July 2013




𝑤 =1

2 5 1.6 2 +

1

2 20 − 1.6 2 = 32 J

d) 𝑤 = 𝑃𝑑𝑡∞

0

= 1152 𝑒−4𝑡𝑑𝑡∞

0

= 1152𝑒−4𝑡

−4 ∞ 0

= 288 J

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+.

b) 𝑤 =1

2 5 64 +

1

2 20 16 = 320 J

c) As 𝑡 → ∞, 𝑖1 → 1.6 A and 𝑖2 → − 1.6 A

This is the difference between the initially stored energy 320 J and the energy in

the parallel inductors 32 J .

١ظ بن خطح ازذح ػاللخ از زممذ

.اإلدبص، ئب دػخ خطاد صغ١شح

Fig. 7.9

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30 6 = 5 Ω

𝑣 =5

5 + 3 120 = 75 V

𝑖 0− = − 𝑣

6= −

75

6= − 12.5 A

c) 𝑤 0 =1

2𝐿𝑖2 0 =

1

2 8 × 10−3 12.5 2 = 625 mJ

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:

Use voltage division to find the voltage drop across the parallel resistors:

Now find the current using Ohm's law:

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?

اى افش ثطشق وث١شح،

.ى ادبذ ى ثطش٠مخ ازذح

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𝜏 =𝐿

𝑅=

8 × 10−3

2= 4 ms

d) 𝑖 𝑡 = 𝑖 0− 𝑒𝑡 𝜏 = − 12.5 𝑒−𝑡 0.004 = − 12.5 𝑒−250𝑡 A, 𝑡 ≥ 0

So 𝑤 5 ms =12

𝐿𝑖2 5 ms =

12

8 × 10−3 3.58 2 = 51.3 mJ

𝑤 𝑑𝑖𝑠 = 625 − 51.3 = 573.7 mJ

%dissipated = 573.7

625 100 = 91.8%

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,

e) 𝑖 5ms = − 12.5 𝑒−250 0.005 = − 12.5 𝑒−1.25 = − 3.58 A

دبزه ٠ؼزذ ػ أزاله ١غذ األزال

.از رشاب ف ه ئب از ف ا١مظخ

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𝑖 0− =10

10 + 6 6.4 = 4 A

Solution:

a) First, use the circuit for 𝑡 < 0 to find the initial current in the inductor:

Using current division,

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:


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?

ئ اصخس رغذ طش٠ك اضؼفبء

.ث١ب ٠شرىض ػ١ب األل٠بء

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𝑅𝑒𝑞 = 4 (6 + 10) = 3.2 Ω, ∴ 𝜏 =𝐿

𝑅𝑒𝑞=

0.32

3.2= 0.1 s

𝑖 𝑡 = 𝑖 0− 𝑒−𝑡 𝜏 = 4 𝑒−𝑡 0.1 = 4 𝑒−10𝑡 A, 𝑡 ≥ 0

𝑖𝑜 𝑡 =4

4 + 10 + 6 − 𝑖 =

4

20 − 4 𝑒−10𝑡 = − 0.8 𝑒−10𝑡 A, t ≥ 0+

𝑤 0 =1

2𝐿𝑖2 0− =

1

2 0.32 4 2 = 2.56 J

𝑣4Ω 𝑡 = 𝐿𝑑𝑖

𝑑𝑡= 0.32 − 10 4 𝑒−10𝑡 = − 12.8 𝑒−10𝑡 V, t ≥ 0+

𝑝4Ω 𝑡 =𝑣4Ω

2

4= 40.96 𝑒−20𝑡 W, t ≥ 0+

𝑤4Ω 𝑡 = 40.96 𝑒−20𝑡𝑑𝑡∞

0

= 2.048 J

%𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 = 2.048

2.56 100 = 80%

Continued (Assessment problem 7.2):

Use the initial inductor current and the time constant to find the current in the

inductor:

Use current division to find the current in the 10 Ω resistor:

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

Find the energy dissipated in the 4 Ω resistor by integrating the power over all

time:

Find the percentage of the initial energy in the inductor dissipated in the 4 Ω

resistor:

اززس، فبغئ١بد ردزة ئ األشخبص

.اغزؼذ٠ زسب

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a) 𝑖 0 =24

12= 2 A

b) 𝜏 =𝐿

𝑅=

1.6

80= 20 ms

𝑣1 = 𝐿𝑑𝑖

𝑑𝑡= 1.6 − 100 𝑒−50𝑡 = − 160 𝑒−50𝑡 V 𝑡 ≥ 0+

𝑣2 = − 72𝑖 = −144 𝑒−50𝑡 V 𝑡 ≥ 0

𝑤72 Ω = 288 𝑒−100𝑥 𝑑𝑥𝑡

0

= 288𝑒−100𝑥

−100 𝑡 0

= 288(1 − 𝑒−100𝑡 )J

𝑤 0 =1

2 1.6 4 = 3.2 J

%𝑑𝑖𝑠𝑠𝑖𝑝𝑎𝑡𝑒𝑑 =2.24

3.2 100 = 69.92 %

Solution:

c) 𝑖 = 2 𝑒−50𝑡 A, 𝑡 ≥ 0

d) 𝑃𝑑𝑖𝑠𝑠 = 𝑖2 72 = 4 𝑒−100𝑡 72 = 288 𝑒−100𝑡 W

𝑤72 Ω 15 ms = 288 − 288 𝑒−1.5 = 2.24 J

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-before-

break 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?

از ٠غزه فظ اطبلخ از

.٠غزىب ازخط١ظ

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2 𝑘Ω ∥ 6 kΩ = 1.5 kΩ

𝑖𝑔 0− =40

1500 + 500 = 20 mA

𝑖1 0− = (20 × 10−3) 2000

8000 = 5 mA

𝑖2 0− = 20 × 10−3 6000

8000 = 15 mA

𝑖2 0+ = − 𝑖1 0+ = − 5 mA (when switch is open)

c) 𝜏 =𝐿

𝑅=

400 × 10−3

8 × 103= 5 × 10−5;

1

𝜏= 20,000

𝑖1 𝑡 = 𝑖1 0+ 𝑒−𝑡 𝜏

𝑖1 𝑡 = 5 𝑒−20,000𝑡 mA, 𝑡 ≥ 0

∴ 𝑖2 𝑡 = − 5 𝑒−20,000𝑡 mA, 𝑡 ≥ 0+

Solution:

a) 𝑡 < 0

b) 𝑖1 0+ = 𝑖1 0− = 5 mA

d) 𝑖2 𝑡 = − 𝑖1 𝑡 when 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.

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+ .

.ادبذ دشد ػ١خ زظ، اعأي أ فبش

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120Ω 60Ω = 40Ω

∴ 𝑖𝑔 =12

10 + 40= 240 mA

𝑖𝐿 0− = 120

180 𝑖𝑔 = 160 mA

120Ω 40Ω = 30 Ω

∴ 𝑖𝑔 =12

10 + 30= 300 mA

𝑖𝑎 = 120

160 300 = 225 mA

∴ 𝑖𝑜 0+ = 225 − 160 = 65 mA

Solution:

a) 𝑖𝑜 0− = 0 since the switch is open for 𝑡 < 0.

b) For 𝑡 = 0− the circuit is:

c) For 𝑡 = 0+ the circuit is:

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+.

امشاس ابعت ف الذ غ١ش

.ابعت ١ظ ثبعت أثذا

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g) 𝜏 =𝐿

𝑅=

100 × 10−3

20= 5 ms:

1

𝜏= 200

∴ 𝑖𝐿 = 0 + 160 − 0 𝑒−200𝑡 = 160 𝑒−200𝑡 mA, 𝑡 ≥ 0

20 0.16 + 𝑣𝐿 0+ = 0; ∴ 𝑣𝐿 0+ = − 3.2 V

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.

h) 𝑣𝐿 0− = 0 since for 𝑡 < 0 the current in the inductor is constant

i) Refer to the circuit at 𝑡 = 0+ and note:

j) 𝑣𝐿 ∞ = 0, 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+

اىث١ش فشا ٠ذسوا ذ

.لشث ادبذ ػذب اعزغا

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a) 𝑣

𝑖= 𝑅 =

160 𝑒−10𝑡

8 𝑒−10𝑡= 20 Ω

b) 𝜏 =1

10= 100 ms

c) 𝜏 =𝐿

𝑅= 0.1

𝐿 = 100 20 × 10−3 = 2 H

d) 𝑤 0 =1

2𝐿 𝑖(0) 2 =

1

2 2 64 = 64 J

e) 𝑤𝑑𝑖𝑠𝑠 = 1280 𝑒−20𝑥𝑑𝑥𝑡

0

= 64 − 64 𝑒−20𝑡

solving , 𝑡 = 45.81 ms

Solution:

100 − 100 𝑒−20𝑡 = 60 ∴ 𝑒−20𝑡 = 0.4

𝑣 = 160𝑒−10𝑡 V, 𝑡 ≥ 0+

𝑖 = 8𝑒−10𝑡 A, 𝑡 ≥ 0

Problem 7.1:

In the circuit in Fig. P7.1, the voltage and current expressions are

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.

ع أذاف ٠غؼ زسم١مب، : ابط ػب

.ع ػ أر االعزؼذاد غبػذح أذاف

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𝑖𝑔 =80

40= 2 A

𝑖𝐿 0− =2 50

100 = 1 A = 𝑖𝐿(0+)

𝑖𝐿 𝑡 = 𝑖𝐿(0+)𝑒−𝑡 𝜏 A

𝜏 =𝐿

𝑅=

0.20

5 + 15=

1

100= 0.01 s

𝑖𝐿 0+ = 1 A

𝑖𝐿 𝑡 = 𝑒−100𝑡 A, 𝑡 ≥ 0

𝑣𝑜 𝑡 = − 15𝑖𝐿(𝑡)

𝑣𝑜 𝑡 = − 15 𝑒−100𝑡 V, 𝑡 ≥ 0+

Solution:

For 𝑡 < 0

For 𝑡 > 0

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.

.ادبذ ه ٠ذفغ اث

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𝑡 < 0 ∶

𝑖𝐿 0+ =240

16 + 8 = 10 A

𝑖𝐿 0− = 10 40

50= 8 A

𝑡 > 0 ∶

𝑅𝑒 = 10 40

50+ 10 = 18 Ω

𝜏 =𝐿

𝑅𝑒=

72

18× 10−3 = 4 ms;

1

𝜏= 250

∴ 𝑖𝐿 = 8 𝑒−250𝑡 A

𝑣𝑜 = 8𝑖𝑜 = 64 𝑒−250𝑡 V, 𝑡 ≥ 0+

Solution:

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+.

ادبذ ازت از١ذ از

.ال ٠غفش سفمبؤب

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𝑖𝐿 0 =−72

24 + 6= −2.4 A

𝑡 > 0

𝑖∆ = −𝑖𝑇 100

160= −

5

8𝑖𝑇

𝑣𝑇 = 20𝑖∆ + 𝑖𝑇 100 60

160= −12.5𝑖𝑇 + 37.5𝑖𝑇

𝑣𝑇

𝑖𝑇= 𝑅𝑇𝑕 = −12.5 + 37.5 = 25Ω

𝜏 =𝐿

𝑅=

250

25× 10−3

1

𝜏= 100

𝑖𝐿 = −2.4𝑒−100𝑡 A, 𝑡 ≥ 0

a) 𝑡 < 0 ∶

b) 𝑣𝐿 = 250 × 10−3 240 𝑒−100𝑡 = 60 𝑒−100𝑡 V, 𝑡 ≥ 0+

c) 𝑖∆ =100𝑖𝐿

160= − 1.5 𝑒−100𝑡 A 𝑡 ≥ 0+

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+.

.أب زض٠ أز أغبد، زا شؼبس و فبش

Solution:

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𝑡 = 0+ ∶

Solution:

a) 𝑡 < 0 ∶

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.

أس شخصب ثال أذاف و أس٠ه أزذ

.أاع ار از٠ رزسشن أخغبد

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𝑡 > 0 ∶

𝑖𝑅 = 2 𝑒−𝑡 𝜏 mA ; 𝜏 =𝐿

𝑅= 0.66 × 10−3

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−1500𝑡V

b) 𝑖𝑜 =1

6 11,250 𝑒−1500𝑡𝑑𝑡

𝑡

0

+ 0 = −1.25 𝑒−1500𝑡 + 1.25 mA


أغت ابط ٠ؼثش ػ

.ب ثصذد اجسث ػ

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ػ ثذ رخط١ظ واطالق

.اشصبص ثذ رص٠ت

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

𝐶𝑑𝑣

𝑑𝑡+

𝑣

𝑅= 0

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 = 𝑣0𝑒−𝑡 𝜏 , 𝑡 ≥ 0

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لض١ب ف اإلػذاد سفالد ١الدب أوثش

.ب لض١ب ف ازخط١ظ س١برب

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

𝑅=

𝑣0

𝑅𝑒−𝑡 𝜏 , 𝑡 ≥ 0+

𝑝 = 𝑣𝑖 =𝑣0

2

𝑅𝑒−2𝑡 𝜏 , 𝑡 ≥ 0+

𝑤 = 𝑝 𝑑𝑥𝑡

0

= 𝑣0

2

𝑅

𝑡

0

𝑒−2𝑥 𝜏 𝑑𝑥

=1

2𝐶𝑣0

2 1 − 𝑒−2𝑡 𝜏 , 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 = 𝑣0𝑒−𝑡 𝜏 , to generate 𝑣 𝑡 from 𝑣0 and 𝜏.

4. All other calculations of interest follow from knowing 𝑣 𝑡 .

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𝜏 = 𝑅𝐶 = 80 × 103 × 0.5 × 10−6 = 40 ms

𝑣𝑐 𝑡 = 100 𝑒−25𝑡 V, t ≥ 0

𝑣𝑜 𝑡 =48

80𝑣𝑐 𝑡 = 60 𝑒−25𝑡 V, t ≥ 0+

𝑖𝑜 𝑡 =𝑣𝑜(𝑡)

60 × 103= 𝑒−25𝑡 mA, t ≥ 0+

𝑤60𝑘Ω = 𝑖𝑜2 𝑡 60 × 103 𝑑𝑡

∞

0

= 1.2 mJ

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.

b) Using voltage dividing:

c) Using Ohm’s law:

d) 𝑝60 𝑘Ω 𝑡 = 𝑖𝑜2 𝑡 60 × 103 = 60 𝑒−50𝑡 mW, t ≥ 0+

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.

وب ردزة ا١ب دشاب اطج١ؼ،

.٠دزة ادبذ غزؼذ٠

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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).

𝜏 = 𝑅𝐶 = 250 × 103 × 4 × 10−6 = 1 s

𝑣 𝑡 = 20 𝑒−𝑡 V, t ≥ 0.

𝑖 𝑡 =𝑣(𝑡)

250,000= 80 𝑒−𝑡 𝜇A, t ≥ 0+

𝑣1 𝑡 = − 106

5 80 × 10−6𝑒−𝑡 𝑑𝑡

𝑡

0

− 4 = 16 𝑒−𝑡 − 20 V, t ≥ 0,

𝑣2 𝑡 = − 106

20 80 × 10−6𝑒−𝑡 𝑑𝑡

𝑡

0

+ 24 = 4 𝑒−𝑡 + 20 V, t ≥ 0.

b) 𝑤1 =1

2 5 × 10−6 16 = 40 𝜇J, 𝑤2 =

1

2 20 × 10−6 576 = 5760 𝜇J

𝑤∞ =1

2 5 + 20 × 10−6 400 = 5000 𝜇J

d) 𝑤 = 𝑝 𝑑𝑡∞

0

= 400 𝑒−2𝑡

250,000 𝑑𝑡

∞

0

= 800 𝜇J

800 𝜇J = 5800 − 5000 𝜇J

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.

Knowing 𝑖 𝑡 , we calculate the expressions for 𝑣1 𝑡 and 𝑣2 𝑡 :

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

٠ؼىف أغت ابط ػ رغ١١ش

.ازبئح ١ظ رغ١١ش األعجبة

Solution:

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𝑖50𝑘 =80 × 103

80 × 103 + 20 × 103 + 50 × 103 7.5 × 10−3 = 4 mA

𝑣 0− = 50 × 103 𝑖50𝑘 = 50 × 103 0.004 = 200 V

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:

Use Ohm's law to find the voltage drop:


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.

٠ىه أ رجذأ ز١ث

.از ا٢خش

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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

𝑑) 𝑤 0 =1

2𝐶𝑣2 =

1

2 0.4 × 10−6 (200)2 = 8 mJ

e) 𝑤 𝑡 =1

2𝐶𝑣2 𝑡 =

1

2 0.4 × 10−6 (200 𝑒−50𝑡)2 = 8 𝑒−100𝑡 mJ

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,

The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:

8 × 10−3𝑒−100𝑡 = 2 × 10−3, 𝑒100𝑡 = 4, 𝑡 = (ln 4) 100 = 13.86 ms

ع لبسثه ألاج ال رغ لجه

.الشأح، ب صاي اجسش أوثش أب

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𝑖 =15

75,000= 0.2 mA, 𝑣5 0− = 4 V, 𝑣1 0− = 8 V

𝜏5 = 20 × 103 5 × 10−6 = 100 ms; τ1 = 40 × 103 1 × 10−6 = 40 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:

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:


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?

ال ٠زسى اشخ ف ص١ش، فبغبء

.اخداد ف ز١بر ٠فؼ ره ١بثخ ػ

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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

𝑤5 60 ms =1

2𝐶𝑣5

2 60 ms =1

2 5 × 10−6 2.20 2 ≅ 12.05 μJ

𝑤 0 = 𝑤1 0 + 𝑤5 0 =1

2 1 × 10−6 (8)2 +

1

2 5 × 10−6 (4)2 = 72 μJ

% dissipated = 58.36 × 10−6

72 × 10−6 100 = 81.05%

Continued (Assessment problem 7.4) :

Therefore,

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, 𝑣5 60 ms = 4 e−10(0.06) ≅ 2.20 V

𝑤1 60 ms =1

2𝐶𝑣1

2 60 ms =1

2 1 × 10−6 (1.79)2 ≅ 1.59 μJ

𝑤 60 ms = 1.59 + 12.05 = 13.64 μJ

Find the initial energy from the initial voltage:

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

اشخ ال ٠غ أي اشأح أزجب،

.اشأح ال رغ أي سخ خبب

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𝐶𝑒𝑞 =2 × 8

10= 1.6 µF

𝜏 = 5 1.6 × 10−3 = 8 ms; 1

𝜏= 125

𝑖 =75

5× 10−3𝑒−125𝑡 mA, 𝑡 ≥ 0+

𝑣1 =− 106

2 15 × 10−3𝑒−125𝑥 𝑑𝑥

𝑡

0

+ 75 = 60 𝑒−125𝑡 + 15 V, 𝑡 ≥ 0

𝑣2 =106

8 15 × 10−3𝑒−125𝑥 𝑑𝑥

𝑡

0

+ 0 = − 15 𝑒−125𝑡 + 15 V, 𝑡 ≥ 0

b) 𝑤 0 =1

2 2 × 10−6 5625 = 5625 µJ

c) 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 =1

2 2 × 10−6 225 +

1

2 8 × 10−6 225 = 1125 µJ

𝑤𝑑𝑖𝑠𝑠 =1

2 1.6 × 10−6 5625 = 4500 µJ

check ∶ 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 + 𝑤𝑑𝑖𝑠𝑠 = 1125 + 4500 = 5625 µJ; 𝑤 0 = 5625µJ

Solution:

a) 𝑣1 0− = 𝑣1 0+ = 75 V 𝑣2 0+ = 0

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.

.اش١طب أعزبر اشخ ر١ز اشأح

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𝑖1 0− = 𝑖2 0− = 3

30 = 100 mA

Solution:

a) 𝑡 < 0 :

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+.

اشخ ابدح اخب از رؼ

.ف١ب اشأح اغبد األخ١شح

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𝑖1 0+ =0.2

2 = 100 mA

𝑖2 0+ = −0.2

8 = − 25 mA

𝑖1 0− = 𝑖1 0+ = 100 mA

𝑖2 0− = 100 mA and 𝑖2 0+ = 25 mA

𝜏 = 𝑅𝑒𝐶 = 1.6 2 × 10−6 = 3.2µ𝑠 1

𝜏= 312,500

𝑣𝑐 = 0.2 𝑒−312,500𝑡 V, 𝑡 ≥ 0,

𝑖1 =𝑣𝑐

2= 0.1 𝑒−312,500𝑡 mA, 𝑡 ≥ 0

f) 𝑖2 =− 𝑣𝑐

8= − 25 𝑒−312,500𝑡 mA, 𝑡 ≥ 0+


b) 𝑡 > 0 :

c) Capacitor voltage cannot change instantaneously, therefore,

d) Switching can cause an instantaneous change in the current in a

resistive branch. In this circuit

e) 𝑣𝑐 = 0.2𝑒−𝑡 𝜏 𝑉, 𝑡 ≥ 0

افشق ث١ اشأح اىج١رش أ ئرا أخطأ

.األخ١ش ف ال ٠م ثب ػ اضج

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a) 𝑣 0 = 8 27 33

60= 118.80 V

𝑅𝑒 = 3 6

9= 2 KΩ

𝜏 = 𝑅𝑒𝐶 = 2000 0.25 × 10−6 = 500 µs; 1

𝜏= 2000

𝑣 = 118.80 𝑒−2000𝑡 V 𝑡 ≥ 0

𝑖𝑜 =𝑣

3000= 39.6 𝑒−2000𝑡 mA

b) 𝑤 0 =1

2 0.25 118.80 2 = 1764.18 µJ

𝑖4𝑘 =118.80 𝑒−2000𝑡

6= 19.8 𝑒−2000𝑡 mA

𝑝4𝑘 = 19.8 𝑒−2000𝑡 2 4000 × 10−6 = 1568.16 × 10−3𝑒−4000𝑡

𝑤4𝑘 = 1568.16 × 10−3𝑒−4000𝑥

−4000

250 × 10−6

0

= 392.04 1 − 𝑒−1 µJ

% =392.04

1764.18 1 − 𝑒−1 × 100 = 14.05 %

Solution:

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?

.اشأح ثطج١ؼزب ال ثاسادرب خش٠ذح أخجبس

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𝑖𝑜 0− = 6000 0.04

6000 + 4000 = 24 mA

𝑣𝑜 0− = 3000 0.024 = 72 V

𝑖2 0− = 40 − 24 = 16 mA

𝑣2 0− = 6000 0.016 = 96 V

𝜏 = 𝑅𝐶 = 200 µs; 1

𝜏= 5000

Solution:

a) 𝑡 < 0 :

𝑡 > 0 :

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.

. اشأح رست اشخ از ال ٠زى

.ئب رظ أ ٠غزغ ئ١ب

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𝑖𝑜 𝑡 =24

1 × 103𝑒−𝑡 𝜏 = 24 𝑒−5000𝑡 mA, 𝑡 ≥ 0+

𝑣𝑜 =106

0.6 24 × 10−3𝑒−5000𝑥 𝑑𝑥

𝑡

0

+ 72

= 40,000𝑒−5000𝑥

− 5000

𝑡 0

+ 72

= − 8𝑒−5000𝑡 + 80 V, 𝑡 ≥ 0

c) 𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 = 1

2 0.3 × 10−6 80 2 +

1

2 0.6 × 10−6 80 2

𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 = 2880 µJ

𝑤𝑑𝑖𝑠𝑠 =1

2𝐶𝑒𝑞 𝑉𝑜 2 =

1

2 0.2 × 10−6 24 2 = 57.6 µJ

𝑤 0 =1

2 0.3 × 10−6 96 2 +

1

2 0.6 × 10−6 72 2 = 2937.6 µJ

𝑤𝑡𝑟𝑎𝑝𝑝𝑒𝑑 + 𝑤𝑑𝑖𝑠𝑠 = 𝑤(0)

2880 + 57.6 = 2937.6 OK

b)

Check by combining the capacitors into a single equivalent capacitance of 0.2 𝜇F

with a 24 V initial voltage:

اشأح رىز است أسثؼ١ عخ،

.ال رىز اجغض عبػخ ازذح


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𝑅𝑒𝑞 = 12𝐾 68𝐾 = 10.2kΩ

𝑣𝑜 0 = −120 10,200

(10,200 + 1800)= −102 V

𝑡 > 0

𝜏 =10

3 12,000 × 10−6 = 40 ms;

1

𝜏= 25

𝑣𝑜 = −102 𝑒−25𝑡 V, 𝑡 ≥ 0

𝑝 =𝑣𝑜

2

12,000= 867 × 10−3𝑒−50𝑡 W

𝑤𝑑𝑖𝑠𝑠 = 867 × 10−3𝑒−50𝑡 𝑑𝑡12×10−3

0

= 17.34 × 10−3 1 − 𝑒−0.6 = 7824 µJ

b) 𝑤 0 = 1

2

10

3 102 2 × 10−6 = 17.34 µJ

0.75𝑤 0 = 13 mJ

867 × 10−3𝑒−50𝑥 𝑑𝑥𝑡𝑜

0

= 13 × 10−3

∴ 1 − 𝑒−50𝑡𝑜 = 0.75; 𝑒50𝑡𝑜 = 4; so 𝑡𝑜 = 27.73 ms

a) 𝑡 < 0 :

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?

اشأح لذ رصفر ػ

.اخ١بخ، ىب ال رغبب

Solution:

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𝑣𝑇 = −5𝑖𝑜 − 15 𝑖𝑜 = −20𝑖𝑜 = 20𝑖𝑇

∴𝑣𝑇

𝑖𝑇= 𝑅𝑇𝑕 = 20 Ω

𝜏 = 𝑅𝐶 = 40 ms; 1

𝜏= 25,000

𝑣𝑜 = 15 𝑒−25,000𝑡 V, 𝑡 ≥ 0

𝑖𝑜 =−𝑣𝑜

20= −0.75𝑒−25,000𝑡A, 𝑡 ≥ 0+

𝑡 > 0 :

𝑡 > 0 :

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+.

ئرا أخفك اش١طب ف اذخي

.ئ ىب أفذ اشأح

Solution:

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𝐼𝑜 =180

15= 12 mA

𝜏 = 0.25

20 × 10−3 = 0.125 × 10−4;

1

𝜏= 80,000

𝐼𝑓 =𝑉𝑠

𝑅=

180

20= 9 mA

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

Solution:

After making a Thévenin equivalent we have:

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+.

ئرا أسدد فضر عشن

.ع ئ اشأح

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ا، ئرا خفضذ اشأح صرب، ف رش٠ذ ه ش١ئب

.ئرا سفؼذ صرب ف رأخز زا اشء

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 + 𝐿𝑑𝑖

𝑑𝑡

Which can be solved for the current

𝑑𝑖

𝑑𝑡=

− 𝑅𝑖 + 𝑉𝑠

𝐿=

− 𝑅

𝐿 𝑖 −

𝑉𝑠

𝑅

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.أػظ اشأح اخشعبء عشا رطك

or 𝑑𝑖 =− 𝑅

𝐿 𝑖 −

𝑉𝑠

𝑅 𝑑𝑡

or 𝑑𝑖

𝑖 − 𝑉𝑠 𝑅 =

− 𝑅

𝐿𝑑𝑡

Integrating both sides and separating:

𝑖 𝑡 =𝑉𝑠

𝑅+ 𝐼0 −

𝑉𝑠

𝑅 𝑒− 𝑅 𝐿 𝑡 (7.35)

When the initial energy in the inductor is zero, 𝐼0 is zero. Equating reduces to

𝑖 𝑡 =𝑉𝑠

𝑅−

𝑉𝑠

𝑅𝑒− 𝑅 𝐿 𝑡 7.36

Equation indicates that after the switch has been closed, the current increases

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

𝑖 𝜏 =𝑉𝑠

𝑅−

𝑉𝑠

𝑅𝑒−1 ≈ 0.6321

𝑉𝑠

𝑅 7.37

If the current were to continue to increase at its initial rate, it would reach its final

value at 𝑡 = 𝜏; that is, because

𝑑𝑖

𝑑𝑡=

− 𝑉𝑠

𝑅 − 1

𝜏 𝑒−𝑡 𝜏 =

𝑉𝑠

𝐿𝑒−𝑡 𝜏

The initial rate at which 𝑖 𝑡 increases is

𝑑𝑖

𝑑𝑡 0 =

𝑉𝑠

𝐿

If the current were to continue to increase at this rate, the expression for 𝑖 would be

𝑖 =𝑉𝑠

𝐿𝑡 7.40

at 𝑡 = 𝜏: 𝑖 =𝑉𝑠

𝐿

𝐿

𝑅=

𝑉𝑠

𝑅 (7.41)

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خذد صذالخ ث١ اشأر١ ف رجذأ ئرا

.أ رإي ئ ارسبد ضذ اشأح ثبثخ

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 − 𝐼0𝑅 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 − 𝐼0𝑅 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 = 𝑉𝑠𝑒− 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.

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𝑖 = 12 + −8 − 12 𝑒−𝑡 0.1 = 12 − 20 𝑒−10𝑡 A, 𝑡 ≥ 0

b) 𝑣 = 𝐿𝑑𝑖

𝑑𝑡= 0.2 200𝑒−10𝑡 = 40 𝑒−10𝑡 V, t ≥ 0+

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.

𝑣 0+ = 40 V

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-before-

break 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.

ئ اشأح لذ رمصب اشدبػخ

اىبف١خ الزسبس ىب ال رضاي رر

.ػ١ه رضب٠مه و رفؼ أذ ره

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𝑡 =1

10ln

40

24= 51.08 × 10−3 = 51.08 ms

Continued (Example 7.5) :

c) 24 = 40 𝑒−10𝑡

For 𝑡:

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)

أخ اشأح اشأح از رشرؼذ

.وبد است ػ شفز١ب، وب صب

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𝑖 0− =24

2= 12 A = 𝑖(0+)

Solution:

a) Use the circuit at t < 0, shown below, to calculate the initial current in the inductor:

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.


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+.

رظ اشأح ف ع اؼشش٠ زز

.آخش سظخ ز١برب

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𝑣 0+ = − 10 8 + 12 = − 200 V

𝜏 =𝐿

𝑅=

200 × 10−3

10 = 20 ms

𝑖𝑓 = − 8 A

= −8 + 20 𝑒−50𝑡 A, t ≥ 0

𝑣 𝑡 = 𝐿𝑑𝑖(𝑡)

𝑑𝑡= 200 × 10−3 − 50 20 𝑒−50𝑡 = − 200 𝑒−50𝑡 V, t ≥ 0+

Contiuned (Assessment problem 7.5) :

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,

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,

Now,

𝑖 𝑡 = 𝑖𝑓 + 𝑖 0+ − 𝑖𝑓 𝑒−𝑡/𝜏 = − 8 + [12 − − 8 ]𝑒−𝑡/0.02

e) To find 𝑣(𝑡), use the relationship between voltage and current for an inductor:

رفض اشأح أ ٠د اشخ ف١ب

.أال، أ ٠د ب ثؼذ ره

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ثالثخ أاع فمظ اشخبي ال ٠غزط١ؼ أ

.اشجبة اش١ش اىي: ٠فا اشأح

We can also describe the voltage 𝑣 𝑡

across the inductor in Fig. 7.16 in terms of

the circuit current.

𝑖 𝑡 =𝑉𝑠

𝑅−

𝑣 𝑡

𝑅

𝑑𝑖

𝑑𝑡= −

1

𝑅

𝑑𝑣

𝑑𝑡

Multiply each side by die inductance 𝐿, we get an expression for the voltage

across the inductor on the left-hand side.

𝑣 = − 𝐿

𝑅

𝑑𝑣

𝑑𝑡

𝑑𝑣

𝑑𝑡+

𝑅

𝐿𝑣 = 0 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

𝑑𝑡+

𝑅

𝐿𝑖 =

𝑉𝑠

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.

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ا رطبت اشأح ثبغباح، ى رخزف دائب

ز افىشح ػذ رمغ١ أسفف اذالة أ

.لذ دفغ اسغبة ف أ س

The Step Response of an RC Circuit

Summing the currents away from the top node

𝐶𝑑𝑣𝐶

𝑑𝑡+

𝑣𝐶

𝑅= 𝐼𝑠 7.49

𝑑𝑣𝐶

𝑑𝑡+

𝑣𝐶

𝑅𝐶=

𝐼𝑠𝐶

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 = 𝐼𝑠𝑅 + 𝑉0 − 𝐼𝑠𝑅 𝑒−𝑡 𝑅𝐶 , 𝑡 ≥ 0

A similar derivation for the current in the capacitor yields the differential equation

𝑑𝑖

𝑑𝑡+

1

𝑅𝐶𝑖 = 0

Equation 7.52 has the same form as Eq. 7.47

𝑖 = 𝐼𝑠 −𝑉0

𝑅 𝑒−𝑡 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 𝐼𝑠𝑅, 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 = 𝐼𝑠𝑅.

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July 2013




𝑉𝑜𝑐 =160 × 103

40 + 160 × 103 − 75 = − 60 V

𝑅𝑇𝑕 = 8000 + 40,000 160,000 = 40 kΩ

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:

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:

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, 𝐼𝑠𝑅 = − 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

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+.

.لذ رشفض اشأح و شء ف١ه ئال ئػدبثه ثب

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July 2013




𝑖𝑜 = 𝐶𝑑𝑣𝑜

𝑑𝑡= 0.25 × 10−6 − 9000 𝑒−100𝑡 = − 2.25 𝑒−100𝑡 mA

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

Because 𝑑𝑣𝑜 0− 𝑑𝑡 = 0, the expression for 𝑖𝑜 clearly is valid only for 𝑡 ≥ 0+.

سخ ازذ ث١ أف سخ ٠ى أ ٠ى صػ١ب شخبي،

. اجبل١خ ف١زجؼ اغبء، وزه اشخ األي999ـ أب اي

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July 2013




𝑣𝐴 − 𝑣𝑜

8000+

𝑣𝐴

160,000+

𝑣𝐴 + 75

40,000= 0

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:

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.

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.

.فزبح ا١ ئرا رض٠ذ فزذ، ئرا اثزغذ عسشد، ئرا طجخذ لزذ

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July 2013




𝑖𝐿 0− = 6 A

𝑖𝐿 ∞ =32 + 48

20= 4 A

𝜏 =𝐿

𝑅=

5 × 10−3

20= 250 µs;

1

𝜏= 4000

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, 𝑡 ≥ 0+

b) 𝑣𝐿 = 5 × 10−3𝑑𝑖𝐿𝑑𝑡

= 5 × 10−3 − 8000 𝑒−4000𝑡 = − 40 𝑒−4000𝑡 V, 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

a) 𝑡 < 0 :

𝑡 > 0 :

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+ .

ب أشك ػ اشأح

.أ رىز عشا

Solution

:

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July 2013




50 =𝑣𝑜(0−)

8+

𝑣𝑜(0−)

40+

𝑣𝑜(0−)

10

2000 = 5 + 1 + 4 𝑣𝑜 ; 𝑣𝑜 = 200 V

∴ 𝑖𝑜 0− =𝑣𝑜

10=

200

10= 20 A

Solution:

a) 𝑡 < 0 :

KCL equation at the top node:

Multiply by 40 and solve:

𝑡 > 0 :

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+.

وث١ش اغبء ف ع األسثؼ١ ٠شزى١

! آال اظش اغجت أ ف اغز١

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July 2013




𝑉𝑇𝑕 = 𝑣𝑜 =40

40 + 120 800 = 200 V

𝑅𝑇𝑕 = 10 + 120 40 = 10 + 30 = 40 Ω

𝜏 =𝐿

𝑅=

40 × 10−3

40= 1 ms;

1

𝜏= 1000

𝑖𝑜 ∞ =200

40= 5 A

∴ 𝑖𝑜 = 𝑖𝑜 ∞ + 𝑖𝑜 0+ − 𝑖𝑜(∞) 𝑒−𝑡 𝜏

= 5 + 20 − 5 𝑒−1000𝑡 = 5 + 15 𝑒−1000𝑡 A, 𝑡 ≥ 0

b) 𝑣𝑜 = 10𝑖𝑜 + (0.04)𝑑𝑖𝑜𝑑𝑡

𝑣𝑜 = 50 − 450 𝑒−1000𝑡 V, 𝑡 ≥ 0+


Use voltage division to find the Thévenin voltage:

Remove the voltage source and make series and parallel combinations of

resistors to find the equivalent resistance:

The simplified circuit is:

= 50 + 150 𝑒−1000𝑡 + 0.04 15 −1000𝑒−1000𝑡

= 50 + 150 𝑒−1000𝑡 − 600 𝑒−1000𝑡

ال رىزة اشأح ػبدح ئال ف ب ٠زؼك ثغب ثث

.أثاثب ثذخ صخب، أش١بء أخش وث١شح

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July 2013




𝑖 0− =5 − 120

15 k + 8 k= − 5 mA

𝑖 0− = 𝑖 0+ = − 5 mA

Therefore 𝑖 ∞ =5

15,000= 0.333 mA

c) 𝜏 =𝐿

𝑅=

75 × 10−3

15,000= 5 µ𝑠

= 0.333 − 5.333 𝑒−200,000𝑡 mA, 𝑡 ≥ 0

a) For 𝑡 < 0, calculate the Thévenin equivalent for the circuit to the left and right of

the 75 mH inductor. We get

b) For 𝑡 > 0 the circuit reduces to

d) 𝑖 𝑡 = 𝑖 ∞ + 𝑖 0+ − 𝑖(∞) 𝑒−𝑡 𝜏 = 0.333 + − 5 − 0.333 𝑒−200,000𝑡

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.

.أفض ع١خ جش ثبػذ أ ال رؼذ

Solution:

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July 2013




𝑣𝑐 0+ = 𝑣9𝑘 =9 K

9 K + 3 K 120 = 90 V

𝑣𝑐 ∞ = 𝑣40𝑘 = −1.5 × 10−3 40,000 = −60 V

𝑉𝑇𝑕 = −60 V, 𝑅𝑇𝑕 = 40 K + 10 K = 50 KΩ

𝜏 = 𝑅𝑇𝑕𝐶 = 1000 µs

= −60 + 90 + 60 𝑒−1000𝑡 = −60 + 150 𝑒−100𝑡 V, 𝑡 ≥ 0

We want 𝑣𝑐 = −60 + 150 𝑒−100𝑡 = 0:

Therefore 𝑡 =ln

15060

100= 916.3 µs

Solution:

a) Use voltage division to find the initial value of the voltage:

b) Use Ohm's law to find the final value of voltage:

c) Find the Thévenin equivalent with respect to the terminals of the capacitor:

d) 𝑣𝑐 = 𝑣𝑐 ∞ + 𝑣𝑐 0+ − 𝑣𝑐(∞) 𝑒−𝑡 𝜏

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.

ال رظ االػززاس شخ ئرا وذ خطئب،

.شأح زز رى خطئب

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July 2013




𝑖𝑜 0− = 10 20

100= 2 mA; 𝑣𝑜 0− = 2 50 = 100 V

𝑖𝑜 ∞ = − 5 20

100 = − 1 mA; 𝑣𝑜 ∞ = 𝑖𝑜 ∞ 50 = − 50 V

𝑅𝑇𝑕 = 50 KΩ 50 KΩ = 25 KΩ; 𝐶 = 16 nF

𝜏 = 25 0.016 = 0.4 ms; 1

𝜏= 2500

∴ 𝑣𝑜 𝑡 = − 50 + 150 𝑒−2500𝑡 V, 𝑡 ≥ 0+

𝑖𝑐 = 𝐶𝑑𝑣𝑜

𝑑𝑡= − 6 𝑒−2500𝑡 mA, 𝑡 ≥ 0+

𝑖50 =𝑣𝑜

50 = − 1 + 3 𝑒−2500𝑡 mA, 𝑡 ≥ 0+

𝑖𝑜 = 𝑖𝑐 + 𝑖50 = − 1 + 3 𝑒−2500𝑡 mA, 𝑡 ≥ 0+

𝑡 < 0;

𝑡 = ∞:

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+.

ال ٠ضب٠ك اشأح أ رزه ام١، ى

.أ رش اشأح أخش رزه أوثش ب

Solution:

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July 2013




.ال ٠ى رس ثشبػخ اضخ، زه غ١شب و عزخ أشش

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

𝑑𝑡+

𝑥

𝜏= 𝐾 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

𝑑𝑥

𝑑𝑡=

− 𝑥

𝜏+ 𝐾 =

− 𝑥 − 𝐾𝜏

𝜏=

− 𝑥 − 𝑥𝑓

𝜏

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July 2013




ذ٠ب أد٠خ دؼ اغبء رزى، ى

.١ظ ذ٠ب شء دؼ ٠صز

𝑑𝑥

𝑥 − 𝑥𝑓=

− 1

𝜏𝑑𝑡

𝑑𝑥

𝑥 − 𝑥𝑓

𝑥 𝑡

𝑥 𝑡0

=− 1

𝜏 𝑑𝑡

𝑡

𝑡0

𝑥 𝑡 = 𝑥𝑓 + 𝑥 𝑡0 − 𝑥𝑓 𝑒−(𝑡−𝑡0) 𝜏

the unknownvariable as a

function of time=

the finalvalue of the

variable+

the initialvalue of the

variable−

the finalvalue of the

variable × 𝑒

− 𝑡− time of switching time constant

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.

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July 2013




40 × 60 60 + 20 , or 30 V

So, 𝑣𝐶 0 = − 30 V

𝜏 = 𝑅𝐶 = 400 × 103 0.5 × 10−6 = 0.2 s

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:

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:

d) 𝑣𝐶 𝑡 = 90 + − 30 − 90 𝑒−5𝑡 = 90 − 120 𝑒−5𝑡 V, 𝑡 ≥ 0

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.

ألي أ اغبء ١ظ ذ٠ شخص١خ،

.ث ذ٠ و ٠ شخص١خ خذ٠ذح

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July 2013




𝑖(𝑡) = 0 + 300 − 0 𝑒−5𝑡 = 300 𝑒−5𝑡 𝜇𝐴, 𝑡 ≥ 0+

120 𝑒−5𝑡 = 90 or 𝑒5𝑡 =120

90

so 𝑡 =1

5𝑙𝑛

4

3 = 57.54 ms

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

f) Solve the equation derived in (d) for the time when 𝑣𝐶 𝑡 = 0:

Note that when 𝑣𝐶 = 0, 𝑖 = 225 𝜇A and the voltage drop across the 400 kΩ

resistor is 90 V.

g)

ب اخزؼذ اشأرب رزسذثب ئال

.صذ اش١طب ٠غزف١ذ خجشرب

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July 2013




𝑖 0+ = 7.5 20

50= 3 mA

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+

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+

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:

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

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.

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+.

.اـشف١ـك لجـ اطـش٠ـك

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𝑣 𝑡 = 0 + 15 − 0 𝑒−𝑡 80×10−3 = 15 𝑒−12.5𝑡 V, 𝑡 ≥ 0+

𝑖 𝑡 = 20 + 5 − 20 𝑒−12.5𝑡 = 20 − 15 𝑒−12.5𝑡 A, 𝑡 ≥ 0

𝑣 𝑡 = 𝐿𝑑𝑖

𝑑𝑡= 80 × 10−3 15 × 12.5 × 𝑒−12.5𝑡 = 15 𝑒−12.5𝑡 V, 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.

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,

We determine that the solutions for 𝑣 𝑡 and 𝑖 𝑡 agree by noting that

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.

اسز أ ساء سخ در اشأح، ى

.اإوذ أ ساء و سخ در ث فش اشأح

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