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CHAPTER 7

First-Order Circuits

In Chapter 6, we have studied the relationships between current andvoltage of capacitors/inductors. These elements can be used to store energyand release energy when needed. In this chapter, we will see how thevoltage or current behaves during the charging/discharging of these storageelements.

First, we will examine two types of simple circuits with a storage ele-ment:

(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 continual applications inelectronics, communications, and control systems.

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

7.0.2. 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.

87

Din

Text Box

ECS 203 (ME2) - Part 4A Dr.Prapun Suksompong

88 7. FIRST-ORDER CIRCUITS

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

7.0.3. The solution of the first-order differential equation

d

dtx(t) = ax+ b

is given by

(7.3) x(t) = ea(t−t0)(x(t0) +

b

a

)− b

a.

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

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

where x(∞) = limt→∞ x(t).The results above take a bit of effort to proof. However, if you have

MATLAB, you can get 7.3 using one line of code:

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

When a < 0, from (7.3), x(∞) = −b/a. We can then get (7.4) by substi-tuting b/a in (7.3) with x(∞).

7.0.4. It turns out that you only have two kinds of plots to worry aboutwhen dealing with (7.4):

7.1. SOURCE-FREE RC CIRCUITS 89

7.1. Source-Free RC Circuits

A source-free RC circuit occurs when its dc source is suddenly dis-connected. The energy already stored in the capacitor is released to theresistor.

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

2CV2o .)

Applying KCL, we get

iC + iR = 0

Cdv

dt+v

R= 0

dv

dt+

v

CR= 0

dv

dt=

(− 1

CR

)v

Hence,

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.


7.1.1. 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.

7.1. SOURCE-FREE RC CIRCUITS 91

(c) The voltage v(t) is less than 1 percent of V0 after 5τ (five timeconstants). 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.

(d) iR(t) = v(t)R = Vo

R e−tτ .

(e) pR(t) = viR = V 2o

R e−2tτ

(f) wR(t) =∫ t

0 p(µ)dµ = 12CV

2o (1− e−2t

τ ) = 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.

7.1.2. In summary: The key in working with a source-free RC circuitis to determine:

1. The initial voltage v(0) = V0 across the capacitor.2. The time constant τ (= RC)


7.1.3. There are basically three ways that we can extend what we havefound.

(a) The initial condition v(t0) can be given at non-zero t0. This hasalready been taken care of by (7.4).

(b) There can be more than one resistors. In which case, you first sim-plify the circuit by finding the equivalent resistance at the capacitorterminals. See Example 7.1.4.

(c) The initial value of the voltage v(t0) might not be given but condi-tion(s) of the circuit before time t0 is given instead. In which case,you may find v(t−0 ) by considering the long-term behavior of the cir-cuit before t0. In such situation, you should review two propertiesof capacitors:

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

See Example 7.1.5.

Example 7.1.4. Let vC(0) = 15 V. Determine vC(t), vx(t), and ix(t) fort > 0.

Remark: The answer is the same if we replace t > 0 in the abovequestion by t ≥ 0.

7.2. SOURCE-FREE RL CIRCUITS 93

In Example 7.1.4, you may ask how can we have the initial voltage of15 V across the capacitor in the first place. The next example suggest away to get the initial voltage.

Example 7.1.5. 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.


7.2. Source-Free RL Circuits

Consider the series connection of a resistor and an inductor. Assumethat the inductor has an initial current Io or i(0) = Io.(hence, the energystored in the inductor is w(0) = 1

2LI2o .)

Applying KVL, we get

vL + vR = 0

Ldi

dt+Ri = 0

di

dt+R

Li = 0

Hence,

i(t) = Ioe−RtL = Ioe

−tτ .

Note that:

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

sponse to decay to 36.8 percent of its initial value.(b) vR(t) = iR = IoRe

−tτ .

(c) p(t) = vRi = I2oRe

−2tτ .

(d) wR(t) =∫ t

0 p(µ)dµ = 12LI

2o (1− e−2t

τ ) = wL(0)− wL(t) .

7.2. SOURCE-FREE RL CIRCUITS 95

7.2.1. In summary: The key in working with a source free RL circuitis to determine:

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

R).• In general, R is the Thevenin equivalent resistance at the ter-

minal of the inductor.(We take out the inductor L and findR = RTH at its terminals.)

Example 7.2.2. 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.


Example 7.2.3. Determine i, io and vo for all t in the circuit. Assumethat the switch was closed for a long time.

7.3. UNIT STEP FUNCTION 97

7.3. Unit step function

We use the step function to represent an abrupt change in voltage orcurrent, like the changes that occur in the circuits of control systems anddigital computers.

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

u(t) =

{0, t < 01, t > 0

For example, here is a voltage source of V0u(t)

Similarly, here is a current source of I0u(t)


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 one value oft and switches off at another value of t. The gate function below switcheson at t = 2 s and switches off at t = 5 s.

It consists of the sum of two unit step functions:

v(t) = 10u(t− 2)− 10u(t− 5).

When the dc source of an RC circuit is suddenly applied (i.e., thishappens when the capacitor is being charged), the voltage or current sourcecan be modeled as a step function, and the response is known as a stepresponse.

7.4. STEP RESPONSE OF AN RC CIRCUIT 99

7.4. Step Response of an RC Circuit

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

The voltage across the capacitor is v. We assume an initial voltageV0 on the capacitor. Since the voltage of a capacitor cannot change in-stantaneously v(0−) = v(0+) = V0, where v(0−) is the voltage across thecapacitor just before switching and v(0+) is its voltage immediately afterswitching.

For t > 0, applying KCL, we have

Cdv

dt+v − VsR

= 0

dv

dt+

v

RC=

VsRC

dv

dt=

(− 1

RC

)v +

VsRC

The solution isv(t) = Vs + (Vo − Vs)e

−tτ , t ≥ 0.


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

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.

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

v(t) = v(∞) + (v(to)− v(∞))e−(t−to)

τ , t > t0

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

7.4.4. The key in finding the step response of an RC circuit is to de-termine three values:

1. The initial capacitor voltage v(0) or V (t0).2. The final capacitor voltage v(∞).3. The time constant τ .


Example 7.4.5. The switch in the circuit below has been in position Afor a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0and calculate its value at t = 1 s and 4 s.

Example 7.4.6. 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.


Example 7.4.7. 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.

• 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.


Example 7.4.8. 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.

• 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.


7.5. Step Response of an RL Circuit

Here, our goal is to find the inductor current i as the circuit response.

7.5.1. 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, theresponse 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 7.5.2. Find i(t) in the circuit below for t > 0. Assume thatthe switch has been closed for a long time.

7.5. STEP RESPONSE OF AN RL CIRCUIT 105

Example 7.5.3. The switch in the circuit below has been closed for along time. It opens at t = 0. Find i(t) for t > 0.

Example 7.5.4. At t = 0, switch 1 in the circuit below is closed, andswitch 2 is closed 4 s later. Find i(t) for t > 0.

first-order circuits rc circuit rl circuit - handout 4a.pdf · first-order circuits ... these...

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