source-free circuits 2014 ch 10...10.2. source-free rc circuits definition 10.2.1. a source-free rc...

CHAPTER 10

First-Order Circuits

We have been assuming that the circuits under our consideration havereached their steady-states. This assumption has helped us analyze circuitsunder dc conditions in Chapter 6 (see Examples 6.2.11, 6.4.13, and 6.4.14)and perform ac analysis in Chapters 7, 8, and 9.

In this chapter and the next chapter, we relax the steady-state assump-tion. This means we will need to return to solving differential equations.We will start with circuits that contain only one (equivalent) capacitor orinductor. This is enough to see how the voltage or current behaves duringthe charging/discharging of these storage elements.

10.1. Introduction and a Mathematical Fact

10.1.1. In this chapter, we will examine two types of simple circuitswith a storage element:

(a) A circuit with a resistor and one capacitor (called an RC circuit);and

(b) A circuit with a resistor and an inductor (called an RL circuit).

These circuits may look simple but they find applications in electronics,communications, and control systems.

10.1.2. Applying Kirchhoff’s laws to the RC and RL circuits producefirst order differential equations. Hence, the circuits are collectivelyknown as first-order circuits.

10.1.3. There are two ways to excite the circuits.

(a) By initial conditions of the storage elements in the circuit.• Also known as source-free circuits• Assume that energy is initially stored in the capacitive or in-

ductive element.• This is the discharging process.

(b) By using independent sources• This is the charging process• For this chapter, we will consider independent dc sources.

125

Din

Text Box

ECS 203 - Part C - For CPE2 Asst. Prof. Dr.Prapun Suksompong April 22, 2015

126 10. FIRST-ORDER CIRCUITS

Before we start our circuit analysis, it is helpful to consider one math-ematical fact which we will use throughout this chapter:

10.1.4. The solution of the first-order differential equation

d

dtx(t) = ax+ b

is given by

(10.10) x(t) = ea(t−t0)

(x(t0) +

b

a

)− b

a.

This result takes a bit of effort to proof. However, if you have MATLAB, youcan get 10.10 using one line of code:

dsolve(’Dx = a*x+b’,’x(t0) = x0’, ’t’)

In addition, if a < 0, the solution is given by

(10.11) x(t) = ea(t−t0) (x(t0)− x(∞)) + x(∞).

where x(∞) = limt→∞ x(t).When a < 0, from (10.10), x(∞) = −b/a. We can then get (10.11) by

substituting b/a in (10.10) with x(∞).

10.1.5. It turns out that you only have two kinds of plots to worryabout when dealing with (10.11):

10.2. SOURCE-FREE RC CIRCUITS 127

10.2. Source-Free RC Circuits

Definition 10.2.1. A source-free RC circuit occurs when its dc sourceis suddenly disconnected. The energy already stored in the capacitor isreleased to the resistor.

RC

iRiC+

–

v

10.2.2. Consider a series combination of a resistor an initially chargedcapacitor. We assume that at time t = 0, the initial voltage is v(0) = Vo.(Hence, the initial stored energy is w(0) = 1

2CV2o .)

Applying KCL, we get

iC + iR = 0

Cdv

dt+v

R= 0

dv

dt=

(− 1

RC

)v


Hence,

(10.12) v(t) = Voe− tRC = Voe

− tτ ,

where τ = RC.This shows that the voltage response of RC circuit is an exponential

decay of the initial voltage.

• Since the response is due to the initial energy stored and the physicalcharacteristics of the circuit and not due to some external voltageor current source, it is called the natural response of the circuit.

10.2.3. Remarks:

(a) τ = RC is the time constant which is the time required for theresponse to decay to 36.8 percent of its initial value. (e ≈ 2.718 andhence 1/e ≈ 0.368.)

(b) The smaller the time constant, the more rapidly the voltage de-creases, that is, the faster the response.


(c) Energy Consideration:(i) Capacitor:

wC (t) =1

2Cv2 (t) =

1

2C(V0e

−tτ

)2

=1

2CV 2

0 e−2tτ = wC(0)e

−2tτ

(ii) Resistor:

iR(t) =v(t)

R=VoRe

−tτ

pR(t) = v(t)× iR(t) =V 2o

Re

−2tτ

wR(t) =

∫ t

0

p(µ)dµ =1

2CV 2

o (1− e−2tτ ) = wC(0)− wC(t)

• Notice that as t → ∞, wR → 12CV

20 , which is the same as

wC(0), the energy initially stored in the capacitor.• The energy that was initially stored in the capacitor is eventu-

ally dissipated in the resistor.(d) The voltage v(t) is less than 1 percent of V0 after 5τ (five time

constants). Thus, it is customary to assume that the capacitor isfully discharged (or charged) after five time constants. In otherwords, it takes 5τ for the circuit to reach its final state or steadystate when no changes take place with time.

10.2.4. In summary: The key in working with a source-free RC circuitis to determine the voltage across the capacitor which can be found from :

1. the initial voltage v(0) = V0 .2. the time constant τ (= RC)3. Equation (10.12)


10.2.5. There are basically three extensions of what we found:

(a) The initial condition v(t0) can be given at non-zero t0. This hasalready been taken care of by (10.11).

(b) There can be more than one resistors. In which case, find the equiv-alent resistance at the capacitor terminals. See Example 10.2.6.

(c) The initial value of the voltage v(t0) might not be given but con-dition(s) of the circuit before time t0 is given instead. See Exam-ple 10.2.8. In which case, we may have to consider the long-term(steady-state) behavior of the circuit before t0. In such situation,recall two properties of capacitors studied in 6.2.7:

(i) For DC, capacitor becomes open circuit.(ii) Voltage across the capacitor cannot change instantaneously.

Example 10.2.6. Let vC(0) = 15 V. Determine vC(t), vx(t), and ix(t)for t ≥ 0.

0.1 F

ix

+

– vC

+

– vx5 Ω

8 Ω

12 Ω


Example 10.2.7. Circuit with switch(es):

Example 10.2.8. The switch in the circuit below has been closed for along time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate theinitial energy stored in the capacitor.

20 mF

t = 0

+

– v

3 Ω 1 Ω

9 Ω 20 V


10.3. Source-Free RL Circuits

10.3.1. Consider the series connection of a resistor and an inductor.

RL

i

+

– vL

+

– vR

Assume that the inductor has an initial current Io or i(0) = Io. (Hence,the energy stored in the inductor is w(0) = 1

2LI2o .)

Applying KVL, we get

vL + vR = 0

Ldi

dt+Ri = 0

di

dt+R

Li = 0

From (10.10),

(10.13) i(t) = Ioe−RtL = Ioe

−tτ .

Note that:

(a) τ = LR is the time constant which is the time required for the

response to decay to 36.8 percent of its initial value.(b) The energy dissipated at the resistor can be found by

vR(t) = i(t)R = IoRe−tτ

pR(t) = vR(t)× i(t) = I2oRe

−2tτ

wR(t) =

∫ t

0

p(µ)dµ =1

2LI2

o (1− e−2tτ ) = wL(0)− wL(t).

10.3. SOURCE-FREE RL CIRCUITS 133

10.3.2. In summary: The key in working with a source free RL circuitis to determine the current through the inductor which can be found from:

(a) the initial current i(0) = Io.(b) the time constant τ (= L

R).(c) Equation (10.13)

10.3.3. Similar extensions of the results to the three cases discussed in(10.2.5) are also possible.

Example 10.3.4. The switch in the circuit below has been closed for along time. At t = 0, the switch is opened. Calculate i(t) for t > 0.

2 H

t = 0

i(t)

2 Ω 4 Ω

12 Ω 40 V 16 Ω

Example 10.3.5. Determine i, io and vo for all t in the circuit. Assumethat the switch was closed for a long time.

18 A 4 Ω

3 Ω

2 Ω

i 1 H

+

– v0

t = 0

i0


10.4. Unit step function

10.4.1. Mathematically, we use the step function(s) to model abruptchange(s) in voltage or current.

Definition 10.4.2. The unit step function u(t) is 0 for negative valuesof t and 1 for positive values of t. In mathematical terms,

u(t) =

0, t < 01, t > 0

In [Alexander and Sadiku], the unit step function is undefined at t = 0.

0 t

u(t)

1

Example 10.4.3. A voltage source of V0u(t)

(b)(a)

a

b

V0u(t) V0

a

b

t = 0

10.4. UNIT STEP FUNCTION 135

Example 10.4.4. A current source of I0u(t)

(b)(a)

a

b

I0u(t) I0

a

b

t = 0 i

10.4.5. We can also use the step functions to represent the gate function(pulse) which may be regarded as a step function that switches on at onevalue of t and switches off at another value of t.

Example 10.4.6. The gate function below switches on at t = 2 s andswitches off at t = 5 s.

0 t

v(t)

10

1 2 3 4 5

It consists of the sum of two unit step functions:

v(t) = 10u(t− 2)− 10u(t− 5).

0 t

10u(t – 2)

10

1 20

–10u(t – 5)

10

1 2 3 4 5 t

–10

Definition 10.4.7. When the dc source of an RC circuit is suddenlyapplied (i.e., this happens when the capacitor is being charged), the voltageor current source can be modeled as a step function, and the response isknown as a step response.


10.5. Step Response of an RC Circuit

In the previous sections, we have been considering RC and RL circuitswith no source. Here, we start to look at the cases with dc sources.

10.5.1. Consider an RC circuit with voltage step input below:

t = 0R

VS

+

– vC

The voltage across the capacitor is v. We assume an initial voltage1

v(0) = V0 on the capacitor.For t ≥ 0, applying KCL, we have

Cdv

dt+v − VsR

= 0

dv

dt+

v

RC=

VsRC

dv

dt=

(− 1

RC

)v +

VsRC

From (10.10), the solution is

v(t) = Vs + (Vo − Vs)e−tτ , t ≥ 0.

1Since the voltage of a capacitor cannot change instantaneously v(0−) = v(0+) = V0, where v(0−) isthe voltage across the capacitor just before switching and v(0+) is its voltage immediately after switching.

10.5. STEP RESPONSE OF AN RC CIRCUIT 137

0 t

v(t)

VS

V0

10.5.2. In general, the step response of an RC circuit whose switchchanges position at time t = 0 can be written as

(10.14) v(t) = v(∞) + (v(0)− v(∞))e−tτ , t > 0.

where v(0) is the initial voltage and v(∞) is the final or steady-state value.We obtain

(a) v(0) from the given circuit for t < 0 and(b) v(∞) and τ from the circuit for t > 0.

10.5.3. If the switch changes position at time t = to instead of at t = 0,the response becomes

(10.15) v(t) = v(∞) + (v(to)− v(∞))e−(t−to)

τ , t > t0

where v(to) is the initial value at t = to.

10.5.4. The key in finding the step response of an RC circuit is todetermine the voltage across the capacitor which can be found from:

1. the initial capacitor voltage v(0) or v(t0),2. the final capacitor voltage v(∞),3. the time constant τ ,4. Equations (10.14) or (10.15).

10.5.5. Extension:


Example 10.5.6. The switch in the circuit below has been in positionA for a long time. At t = 0, the switch moves to B. Determine v(t) fort > 0 and calculate its value at t = 1 s and 4 s.

B

24 V

A

30 V

t = 0

3 kΩ 4 kΩ

5 kΩ 0.5 mF+

– v

Example 10.5.7. Find v(t) for t > 0 in the circuit below. Assume theswitch has been open for a long time and is closed at t = 0. Numericallyevaluate v(t) at t = 0.5.

10 V 5 V

2 Ω 6 Ω

- F+

– v

t = 0

1

3

3. [Alexander and Sadiku, 2009, Q7.7] Assuming that the switch in Figure 3 has been in

position A for a long time and is moved to position B at t = 0, find vo(t) for t 0.

Figure 3

4. [F2010] Consider the circuit in Figure 4 below. Assume the switch has been at position 1 for a long time and moves to position 2 at t = 0 sec.

Figure 4

Let

Vs1 = 5 V, Vs2 = 0 V, R1 = 6 , R2 = 3 , and C = 10 F.

(a) (3 pt) Find v(0-). Do not forget to justify your answer. (b) (1 pt) Find v(0). Do not forget to justify your answer. (c) (4 pt) Find v(t) for t > 0.

Vs1 Vs2

R1 Ω R2 Ω

t = 0

C

1

2

+

v(t)

-

5. [F2010] Consider the circuit in Figure 5 below. Assume the switch has been at position 1 for a long time and moves to position 2 at t = 5 sec.

Figure 5

Let

Vs1 = 16 V, Vs2 = 8 V, R1 = 3 , R2 = 5 , and C = 8 F.

(a) (3 pt) Find v(0). (b) (2 pt) Find v(5). (c) (4 pt) Find v(t). (d) (1 pt) Evaluate v(t) at t = 7.

Vs1 Vs2

R1 Ω

t = 5

1

2

+

v(t)

-R2 Ω

C

10.5. STEP RESPONSE OF AN RC CIRCUIT 139

Example 10.5.8. An electronic flash unit provides a common exampleof an RC circuit. This application exploits the ability of the capacitorto oppose any abrupt change in voltage. A simplified circuit is shownbelow. It consists essentially of a high-voltage dc supply, a current-limitinglarge resistor R1, and a capacitor C in parallel with the flashlamp of lowresistance R2.

1R1

vS

+

–

vC

2

R2

High

voltage

dc supply

i

• When the switch is in position 1, the capacitor charges slowly dueto the large time constant (τ = R1C). The capacitor voltage risesgradually from zero to Vs , while its current decreases gradually fromI1 = Vs/R1 to zero.• With the switch in position 2, the capacitor voltage is discharged.

The low resistance R2 of the photolamp permits a high dischargecurrent with peak I2 = Vs/R2 in a short duration.

This simple RC circuit provides a short-duration, high current pulse.Such a circuit also finds applications in electric spot welding and the radartransmitter tube.

9. Consider the circuit in Figure 9 below. Let

Vs = 10 V, R1 = 30 k, R2 = 10 k, and C = 4 F.

Figure 9

Assume that the switch has been in position 1 during time t < 0. Then, during time t 0 the

switch changes its position five times: at t1 = 0 ms, t2 = 25 ms, t3 = 50 ms, t4 = 75 ms, t5 = 100

ms.

(At time t1, the switch changes to position 2. At time t2, the switch changes back to position 1.

At time t3, the switch changes again to position 2….)

Plot the voltage v(t) for time t > 0.

Hint: You should have v(t5) 4.59 V.


Example 10.5.9. An RC circuit can be used to provide various timedelays. Consider the circuit below. It basically consists of an RC circuitwith the capacitor connected in parallel with a neon lamp. The voltagesource can provide enough voltage to fire the lamp.

R2

C110 V

70 V

Neon

lamp

+

–

S

0.1 µF

R1

• When the switch is closed, the capacitor voltage increases graduallytoward 110 V at a rate determined by the circuit’s time constant,(R1 +R2)C. The lamp will act as an open circuit and not emit lightuntil the voltage across it exceeds a particular level, say 70 V.• When the voltage level is reached, the lamp fires (goes on), and

the capacitor discharges through it. Due to the low resistance ofthe lamp when on, the capacitor voltage drops fast and the lampturns off. The lamp acts again as an open circuit and the capacitorrecharges.• By adjusting R2, we can introduce either short or long time delays

into the circuit and make the lamp fire, recharge, and fire repeatedlyevery time constant τ = (R1 +R2)C, because it takes a time periodτ to get the capacitor voltage high enough to fire or low enough toturn off.• The warning blinkers commonly found on road construction sites

are one example of the usefulness of such an RC delay circuit.

10.6. STEP RESPONSE OF AN RL CIRCUIT 141

10.6. Step Response of an RL Circuit

Finding step response of an RL circuit follows almost the same proce-dure as finding step response of an RC circuit. However, here, we use theinductor current i as the circuit response.

10.6.1. In general, the step response of an RL circuit whose switchchanges position at time t = 0 can be written as

i(t) = i(∞) + (i(0)− i(∞))e−tτ , t > 0.

where i(0) is the initial current and i(∞) is the final or steady-state value.If the switch changes position at time t = to instead of at t = 0, the

response becomes

i(t) = i(∞) + (i(to)− i(∞))e−(t−to)

τ , t > t0

where v(to) is the initial inductor current value at time t = to.

Example 10.6.2. Find i(t) in the circuit below for t > 0. Assume thatthe switch has been closed for a long time.

- H

t = 0

i

2 Ω 3 Ω

10 V1

3


Example 10.6.3. The switch in the circuit below has been closed for along time. It opens at t = 0. Find i(t) for t > 0.

10 Ω 9 A5 Ω

1.5 Hi

t = 0

Example 10.6.4. At t = 0, switch 1 in the circuit below is closed, andswitch 2 is closed 4 s later. Find i(t) for t > 0.

5 H

t = 0

i

4 Ω 6 Ω

40 V

t = 4

2 Ω

10 V

S1

S2

P

source-free circuits 2014 ch 10...10.2. source-free rc circuits definition 10.2.1. a source-free rc...

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