8. basic rl and rc circuits -...

1K. A. Saaifan, Jacobs University, Bremen

8. Basic RL and RC Circuits

This chapter deals with the solutions of the responses of RL and RC circuits

The analysis of RC and RL circuits leads to a linear differential equation

This chapter covers the following topics

The Source-Free RL Circuit

The Source-Free RC Circuit

The Unit-Step Function

Driven RL Circuit

Driven RC Circuit

First Order Linear Differential Equations

A first order homogeneous linear differential equation is one of the form

where "First Order'' indicates that both dy/dt and y occur to the first power and "homogeneous'' refers to the zero on the right hand side of the equation

In circuit analysis, y can either be the voltage v or the current i of the circuit

The solution of a homogeneous linear differential equation is called a complementary function

In circuit analysis, we refer to the solution of the circuit as a natural response or a transient response

dydt

py=0

K. A. Saaifan, Jacobs University, Bremen

A Direct Approach

Since the variables can be separated, the differential equation can be rewritten as

We integrate both sides as

Then, we have

The constant of integration must be selected to satisfy the initial condition y(0)=Y0

Thus, we obtain

dyy

=−pdt

∫ dyy =−p∫d t


ln yt=−ptA

lnY 0=A

ln yt=−ptlnY 0

ln yt−lnY 0=−pt

ln y tY 0

=−pt

y t=Y 0e−pt for t0

A General Solution Approach


We assume a general solution of y(t) in exponential form

where A and s are constant to be determined

Substitute the exponential solution into the differential equation

where (s +p) is the characteristic equation

Determine the value of s

Invoke the initial condition to determine the remaining constant A

The final form is

y t=Aest

sp =0

dydt

py=0 AsestpAes t=0 sp Aes t=0

s=−p

y t=Y 0e−pt for t0

y 0=Y 0=Aes0

A=Y 0

dydt

py=0

sypy=spy=0

The characteristic equation


8.1 The Source-Free RL Circuit

RivL=0

+

-didt

RLi=0

We assume a series RL circuit for which i(t) to be determined

Apply KVL

We will solve the natural response


Using the initial condition i(0)=I0, we have

The natural response is given as

i t=Aest

sRLi=0 s=−

RL

i t=I0e−RLt

for t0

i 0=I0=Aes 0

A=I 0


Example: RL with a switch

We have two different circuits: one with the switch closed and one with the switch open

We are asked to find v(0.2) for the circuit shown in Figure (c)

From Figure (b), we compute the current iL

this current is used as initial conditions when the switch is opened

Write the differential equation of the circuit (c)

The general solution of i(t) in exponential form

Since iL(0)=2.4 A, the solution is

iL=2410

=2.4 A

40 iL10 iL5d iLdt

=0

diLdt

10 iL=0

i t=Aest

i t=2.4e−10t for t0

iL


The Exponential Response

The expression for the current in an RL series circuit describes the natural response of the inductor

The current decreases exponentially with time

The L/R term in the above equation is known commonly as the time constant, τ, of the RL series circuit

i t=I0e−RLt

i t=I0e−t

Power in The RL Series Circuit

The power being dissipated in the resistor is

The total energy is found as

pR=vRi t=Ri2 t=RI 0

2e−2 t

wR=∫0

∞pRdt=RI 0

2∫0

∞e−2 tdt

=12LI 0

2

The time constant is τ=L/R

+

-


dvdt

vRC

=0

8.2 The Source-Free RC Circuit

We assume a series RL circuit for which i(t) to be determined

Apply KCL

We will solve the natural response


Using the initial condition v(0)=V0, we have

The natural response is given as

v t=Aest

s1RC

v=0 s=−1RC

v t=V 0e−1RC

t

for t0

v 0=V 0=Aes 0

A=V 0

Cdvdt

=−vR

iC


v t=V 0e−

1RC

t

v t=V 0e−t

pR=viR t=v2tR

=V 0

2

Re−2 t

RC

wR=∫0

∞pRdt=

V 02

R ∫0

∞e−2 t

RC dt

=12CV 0

2

The time constant is τ=RC

The Exponential Response

The expression for the current in an RC series circuit describes the natural response of the inductor

The voltage decreases exponentially with time

The 1/RC term in the above equation is known commonly as the time constant, τ, of the RC series circuit

Power in The RC Series Circuit

The power being dissipated in the resistor is

The total energy is found as

10

Determine the inductor voltage v in the circuit for t > 0.

For t<0, l0 V appears across the 4 Ω resistor, so a dc current of iL= l0/4 = 2.5 A flows through the inductor (which acts as a short circuit)

For t>0, the battery is removed so we write the simple KVL equation:

Thus, we can represent the circuit with the equation

Finally, the voltage is

6 iL4 iL5diLdt

=0diLdt

2 iL=0

i t=iL0e−2 t

=2.5e−2 t A

vt=Ldi td t

=52.5−2e−2t=−25e−2t V


11

Noting carefully how the circuit changes once the switch in the circuit is thrown, determine v(t) at t = 0 and at t = 160 μs

vt=v0e−t

=50e−6250 t V


With no current flow permitted through the capacitor, we know v(0)=50 V since the capacitor voltage cannot change in zero time

After the switch is thrown, the only remaining circuit is a simple source-free RC circuit. With τ = RC =160 μs

Finally, the voltage isv 160s=50e−1=18.39 V

Before the switch is thrown, the 80Ω resistor is connected only by one of its terminals and therefore may be ignored (i=0)


8.3 A More General Perspective

General RL Circuits

The time constant of a single-inductor circuit will be τ=L/Req where Req is the resistance seen by the inductor

Example: Req=R3+R4+R1R2/ (R1+R2)

General RC Circuits

The time constant of a single-capacitor circuit will be τ=ReqC , where Req is the resistance seen by the capacitor

Example: Req=R2+R1R3/ (R1+R3)


At t=0.15 s in the circuit, find the value of (a) iL; (b) i1; (c) i2

For t < 0, (the switch is open)

For t>0, 100% of the 2-A source contributes to i2, The 8-Ω resistor is shorted out so i1=0

Thus

where , , and

Finally, the currents are

i2 0−=0

i10−=2 2

28=0.4 A

i2 t=2−iL t

iLt=iL0e−t =

LReq

=0.42

=0.2 s

iL0.15=1.6e−0.15

0.2 =755.6 mA

i2 0.15=2−iL0.15=1.244 A

iL0−=2 8

28=1.6 A

iL0=iL0−=iL0

=1.6 A


The voltage on a capacitor or the current through an inductor is the same prior to and after a switch at t=0

Resistor current (or voltage) prior to the switch i(0-) can be different from the voltage after the switch i(0+)

All voltages and currents in an RC or RL circuit follow the same natural response e-t/τ

1st Order Response Observations


8.4 The Unit-Step Function

So far, we have been studying the natural response of RL and RC circuits (when no sources or forcing functions were present)

In other words, we have been solving problems in which energy sources are suddenly removed from the circuit

We shall consider that type of response which results when energy sources are suddenly applied to a circuit

The unit-step function u(t) is a convenient notation to represent change:

ut=0 t01 t0

ut−t0=0 tt0

1 tt0


Switches and Steps

In order to obtain an exact equivalent for the voltage-step forcing function, we may provide a single-pole double-throw switch

The exact equivalent for the current-step forcing function, we may replace this circuit by a dc source in series with a switch


Modeling Pulses Using u(t)

By manipulating the unit-step forcing function, we can generate many useful forcing functions

A rectangular voltage pulse

The two unit steps u(t −t0) and −u(t −t1) are needed to obtain the rectangular voltage pulse

A pulsed sinewave

v t= 0V 0

0

tt0t0tt1tt1

v t=V m ut−t0−u t−t1sin 2f t


8.5 Driven RL Circuit

Now, we consider the behavior of a simple RL network to the sudden application of a dc source

The shown circuits represent an RL circuit subjects to a voltage-step forcing function V0u(t)

Applying Kirchhoff’s voltage law

Since i(t)=0 for t<0, we study the solution for t>0

Then, we have

Integrating both sides

iRLdidt

=V 0ut

iRLdidt

=V 0 t0

Ldi

V 0−iR=dt t0

−LR

ln V 0−Ri =tK


−LR

lnV 0=K

−LR ln V 0−Ri−lnV 0=t

ln V 0−Ri−lnV 0=−RLt ln

V 0−RiV 0

=−RLt

i t=V 0

R−V 0

Re−RLt

t0

i t=V 0

R1−e

−RLtu t

invoke the initial condition to find K (i(t)=0 for t<0)

and hence

Rearranging

Which can be rewritten as

or


The expression for the voltage in an RL series circuit describes the energizing characteristics of the inductor

The Natural response: The exponential term has the functional form of the natural response of the RL circuit; it is a negative exponential, it approaches zero as time increases, and it is characterized by the time constant L/R

The Forced response: It is the response that is present a long time after the switch is closed

i t=V 0

R

Forced response

−V 0

Re−RLt

Natural response


A General Solution Approach

The solution of any linear differential equation may be expressed as the sum of two parts: the complementary solution (natural response) and the particular solution (forced response)

didt

RLi=V 0

Lut

i t=V 0

R

Forced response

−V 0

Re−RLt

Natural response

ynt=Ae−pt

dydt

py=f t

dypydt=f tdt

y=e−pt∫f teptdtparticular solution

Ae−ptcomplementary solution

yf t=e−pt∫F eptdt =

Fp

The inhomogeneous linear differential equation has the form of

or

The solution is given as

We note that, when f(t) is zero (a source-free circuit), the solution is the natural response

Since f(t)=F, the particular solution leads to the following forced response


Applying KVL yields

First, we evaluate the natural response by solving

We assume , where A and s1 are constant to be determined


Then, we have

Next, we determine the forced response

Finally, the complete solution is i(t)=in(t)+if(t)

whereA=−

V 0

R

Example: RL with a forcing function

didt

RLi=V 0

Lut

didt

RLi=0

int=Aes1t

int=Ae−RL t

if t=V 0

R

i t=V 0

RAe

−RL t

i 0=0=V 0

RA

sRL

i=0 s1=−RL


Determine i(t) for all values of time in the circuit

idc t=502=25 A

We note that the circuit contains a dc voltage source as well as a step-voltage source

Using superposition, we solve the circuit for each source alone

We compute the current due to only a dc voltage source


We compute the current due to only a a step-voltage source

The complete response

istep voltaget=if tint

=25−25e−2t

i t=idc tistep voltaget=50−25e−2 t


Find the current response in a simple series RL circuit when the forcing function is a rectangular voltage pulse of amplitude V0 and duration t0

We represent the forcing function as the sum of two step-voltage sources V0u(t) and -V0u(t-t0)Using superposition, assume i1(t) designate the part of i(t) due to V0u(t) acting alone, and i2(t) represents that part due to -V0u(t-t0) acting alone

We solve the response i1(t) as

We solve the response i2(t) as

We now add the two solutions, but do so carefully, since each is valid over a different interval of time

i t=i1ti2t

i1t=V 0

R1−e

−RL t

t0

i2t=−V 0

R1−e

−RL t−t0

t00


i t= i1ti1ti2t

0tt0tt0


8.6 Driven RC Circuit

Similar to RL circuits, the complete response of an RC circuit consists of the natural and the forced response

Since the procedure is virtually identical to what we have already discussed in detail for RL circuits, we consider a relevant example for a driven RC circuit

Find the capacitor voltage vC(t) and the current i(t) in the 200 Ω resistor for all time


vC 0−=120 50

1050=100 V

For t<0, the circuit has two separate loops

The capacitor acts as open circuits

The current in the 200 Ω resistor

For t>0

The complete response of the capacitor is

The natural response vCn(t)

where

The forced response is

i 0−=50 160200

=192 mA

vC t=vCf tvCnt

vCnt=Ae−tReqC

Req=50∣∣200∣∣60=24

vCf t=50 50∣∣2006050∣∣200

=20 V


The complete response is

We use , then

Thus, we have

and

The current in the 200 Ω resistor

vC t=20Ae−t1.2

vC 0−=vC 0=vC 0

=100 V

100=20A A=80

vC t=2080e−t1.2 t0

vC t=100 t0

i t=vC t200

=100400e−t1.2 mA t0

i t=192 mA t0


Homework Assignment 7P8.7, P8.10, P8.20, P8.22, P8.24, P8.27, P8.29, P8.35, P8.38, P8.40, P8.44, P8.46, P8.48, P8.52, P8.54, P8.57, P8.61 and 8.65

8. basic rl and rc circuits -...

Documents