module 2 - rlc circuits

8/10/2019 Module 2 - RLC Circuits

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Resistive, Capacitive, and Inductive

AC Circuits

EE 102 Circuits 2

Module 2



The Purely-Resistive AC Circuit

i

Re

= E m sin ωt

v R

where: e = E m sin ωt is the time-dependent AC generator

voltage, volts.

E m = amplitude or maximum value of the AC

generator voltage, volts

ω = angular frequency of the AC generator, rad/s

R = resistance of the load in series with e , ohms.

i = resulting current through the circuit, amps.

v R = voltage drop across the resistive load, volts.

Applying Kirchhoff’s voltage rule over the loop,

0 Rve

ev R

t E v m R sin

+ +

- - -




i

Re

= E m sin ωt

v R

iRv R By Ohm’s law,

t R

E

R

vi m R sin

In a purely-resistive AC

circuit, the current throughthe resistance R is in-phase

with the voltage vR across

it..

t I i m sin R

E I where mm

t

ei

I m

E m




i

Re

= E m sin ωt

v R

The effective or RMS value of the

AC generator voltage and the

resulting currents are;

2

m RMS

E E

R

E I I mm RMS

22

Effective value of e and i :




i

Re

= E m sin ( ωt – φ )

v R

)sin(

t E e m

)sin( t I i m

t

ei

I m

E m

φ

AC generator voltage e with phase angle φ:



Checkpoint 1

If we increase the frequency of the AC generator in a circuit with a

purely resistive load, do (a) amplitude v R (voltage drop across theresistor) and (b) amplitude i R (current thru the resistor), decrease, or

remain the same?



Illustrative Problem 1

A purely resistance AC circuit has resistance R = 200 Ω and the

sinusoidal alternating emf (voltage) device operates at amplitudeE m = 36.0 V and frequency f = 60.0 Hz .

a) What is sinusoidal driving voltage e and its effective value E ?

b) What is the sinusoidal current i and its effective value I ?

c) Power dissipated by the resistive load.



The Purely-Capacitive AC Circuit

i

Ce

= E m sin ωt v C

where: e = E m sin ωt is the time-dependent AC generator

voltage, volts.




C = capacitance of the capacitor, farad


v C = voltage drop across the capacitive load, volts.


0 C ve

evC

t E v mC sin



C C Cvq

In a capacitor, the amount of

charge on a capacitor is

proportional to voltage across its

terminals. That is,


i

Ce

= E m sin ωt v C

Likewise,

t CE i

t E dt

d C

dt

dvC

dt

dqi

m

mC C

cos

]sin[

where: I m = ωCE m .

q = capacitor charge in coulombs

In a purely-capacitive AC circuit, the current through a capacitor is

leading the voltage vC across it by 90o.

t

ei

I m

E m

90 o )90sin(cos o

mm t I t I i




i

Ce

= E m sin ωt v C

t CE i m cos

where: is called the capacitive

reactance of the capacitor, ohms.

t

ei

I m

E m

90 o

Rearranging,

t

C

E

i

m

cos1

)90sin(cos o

C

m

C

m t X

E t

X

E i

C X C

1




i

Ce

= E m sin ωt v C

Power : The instantaneous power dissipated

by the capacitor is p C = v C i = ei .

Plotting p C vs. time t yields,

t

e

i

I m

E m

90 o




i

Ce

= E m sin ωt v C

Power :

t

e

i

I m

E m

90 o

The instantaneous power dissipated






i

Ce

= E m sin ωt v C




The energy dissipated in one cycle is the area of the curve in one cycle




i

Ce

= E m sin ωt v C




The energy dissipated in one cycle is zero, therefore the power

dissipated by the capacitor is zero.




A purely capacitive AC circuit has a capacitance of C = 15 μF and

the sinusoidal alternating emf (voltage) device operates at amplitudeE m = 36.0 V and frequency f = 60.0 Hz .

a) What is the voltage v C of the capacitor and its effective value?

b) The capacitive reactance of the capacitor.

c) What is the current i and effective value of the current throughthe capacitor?



The Purely-Inductive AC Circuit

where: e = E m sin ωt is the time-dependent AC generatorvoltage, volts.




L = inductance of the inductor, henry


v L = voltage drop across the inductive load, volts.


0 Lve

ev L

t E v m L sin

i

L e

= E m sin ωt v L



dt

di Lv L

In an inductor, the voltage drop

across the inductor is proportional

to the rate of change of the current

i through it.

. Thus,

where: I m = E m /ωL

t

e

i

I m

E m

90 o


i

e

= E m sin ωt v L

t E dt

di L m sin

)90sin( om t I i

)90sin(cos omm t L

E t

L

E i

In a purely-inductive AC circuit, the current through the inductor is

lagging the voltage vL across it by 90o.

L



t

e

i

I m

E m

90 o


i

L e

= E m sin ωt v L

where: XL = wL is called the inductive

reactance of the inductor, ohms.

t X

E i

L

m cos

)90sin( om

t L

E

i




i

L e

= E m sin ωt v L

Power : As in the case of the purely-

capacitive circuit, the graph of the

power p = v L i will result in an area

for one cycle of zero. Therefore,

the power dissipated by theinductor will be zero.



Checkpoint

If we increase the driving frequency of the AC generator in a circuit

with a purely capacitive load, do (a) amplitude of v C (voltage dropacross the capacitor) and (b) amplitude i C (current thru the capacitor),

increase, decrease, or remain the same? If instead, the circuit has a

purely inductive load, do (c) amplitude v L and (d) amplitude i L

increase, decrease, or remain the same?




A purely inductive AC circuit has an inductance of L = 230 henrys

and the sinusoidal alternating emf (voltage) device operates atamplitude E m = 36.0 V and frequency f = 60.0 Hz .

a) What is the inductive reactance of the inductor.

b) What is the effective value of the current through the inductor?



Summary of R, L, and C AC Circuits

i R

Re = E m sin ωt v R

R

vi R R

i R is in-phase with the voltage v R .

i C

C e = E m sin ωt v C C

C C

X

vi

i C is leading the voltage v C by 90o.

fC C X C

2

11

i L

Le = E m sin ωt v L

L

L L

X

vi

i L is lagging the voltage v C by 90o.

fL L X L 2

Note in this circuit, v R= E m sin ωt

Note in this circuit, vC = E m sin ωt

Note in this circuit, v L= E m sin ωt



Quiz 1

1. Given the following sinusoidal signals:

e1 = -5 sin(10t + 35o )

e2 = 10 cos(10t – 155o )

find the following; (10 pts)

a) The minimum phase difference between the sinusoids

b) The period of the signals.c) The effective values of e1 and e2.

d) The plot of e1 x e2.

2. In a purely inductive circuit, the AC sinusoidal voltage is

e1 = -100 sin(10t + 35o

), resulting in an effective current of 2.0A. Find; (10 pts)

a) The inductive reactance of the inductor.

b) The inductance of the inductor.

c) The time domain expression of i .



The RC Circuit

0

0

0

C C

C

C R

vdt

dv RC e

viRe

vve

Rearranging and substituting the sinusoid for e ;

t RC

E v

RC dt

dv mC

C sin1

Which is a first-order differential equation with the dependent

variable v C .

i

e = E m sin ωt v C C

v R sw.



The RC Circuit

The solution to this differential equation is composed of two

solutions: the homogenous solution (transient response) and the particular solution (steady-state response). The transient response is

also called the natural response and the steady-state response the

forced response.

RC t

C

mC C et

X R E X v

sin)sin(22

where φ = tan-1(X C /R). Since i = C dv C /dt,

RC t

C

mC

e RC t X R

E CX

i sin

1

)cos(22

RC t C

C

m e R

X t

X R

E i sin)cos(

22



The RC Circuit

RC t C

C

m e R

X t X R

E i sin)cos(22

RC t

C

mC

C

m e X R R

E X t X R

E i

sin)cos(

2222

Steady state Transient

Note: This solution assumes the capacitor is initially not charged.



The RC Circuit

t

e

i

Graph of AC source e (blue) and the response i (red)



The RC Circuit

RC t

C

mC

C

m

e X R R

E X

t X R

E

i

sin)cos( 2222

Steady state Transient

The steady-state component of the response is what is left after a

long time, as the transient part slowly disappears. The radical is called the impedance of the circuit and is

denoted by the letter Z. Thus,

22

C X R

22

C X R Z

In AC circuit analysis, we are interested on the steady state

response of the network.

The product RC is called the time-constant of the circuit usually

designated by τ = RC. The time constant affect how fast or slow

will the transient component stay.



Checkpoint

If we increase the driving

frequency ω of the AC generatorin an RC circuit, do (a) amplitude

of the steady-state current i

(current thru the RC circuit),

increase, decrease, or remain thesame? (b) do the phase φ of

steady-state current i increase,

decrease, or remain the same?

i


v R



Checkpoint

In an RC circuit, the transient

component of the current i disappearsfaster if

a) The driving frequency ω is

increased.

b) The time constant RC is increased.c) The time constant RC is decreased.

i


v R



Checkpoint

In an RC circuit, the steady-state

component of the current is,a) Lagging the source voltage.

b) Leading the source voltage.

c) Remain the same.

i


v R



The RL Circuit

The steady-state response for an RL circuit is;

)sin(22

t X R

E i

L

m

where φ = tan-1(X L/R).

i

e = E m sin ωt v L

v R

L



Checkpoint

In an RL circuit, the steady-state

component of the current is,a) Lagging the source voltage.

b) Leading the source voltage.

c) Remain the same.

i

e = E m sin ωt v L

v R

L




In an RL circuit R = 2000Ω , L = 3.1 henrys and the sinusoidal

alternating emf (voltage) device operates at amplitude E m = 100.0V and frequency f = 60.0 Hz .

a) The inductive reactance X L of the inductor.

b) Phase difference of the circuit current and the emf.

c) The effective value of the current.d) Plot of the current.



The RLC Circuit

The steady-state response for an RLC circuit is;

)sin()(

22

t

X X R

E i

C L

m

where φ = tan-1[(X L-X C )/R] if X L > X C or

φ = tan-1[(X C -X L )/R] if X C > X L.

i

e = E m sin ωtv L

v R

L

C v C

R




An RLC circuit has R = 2000 Ω , L = 3.1 henrys, C = 1.77 μF and the

sinusoidal alternating emf (voltage) device operates at amplitudeE m = 150.0 V and frequency f = 60.0 Hz . Find

a) The impedance Z of the circuit.

b) The phase angle of the circuit current. Is the current leading or

lagging the emf?c) The effective value of the current.

d) The effective value of the voltage across the inductor.

e) At what frequency will X L = X C ?



Euler’s Identity

According to Euler;

t jet jt sincos

t je B At jBt A 22)sincos

Similarly;

where is called an imaginary number.1 j

This means that e jωt has a real part (cos ωt) and an imaginary part

(sin ωt), sometimes written as;

t je

t e

t j

t j

sinIm

cosRe

A quantity with a real part and an imaginary part like e jωt is called a

complex number.



The RL Circuit

By KVL, the differential equation forthis circuit is,i

e = E m sin ωt v L

v R

L

t E Ridt

di L m sin

By Euler’s identity, Em sin ωt = Em Ime jωt ).

We can replace the sinusoidal source with a complex source

Em e jωt.

i

E = E m e j ωt v L

v R

L

This source now has two componentsources: the imaginary source

Emsin ωt which is our original source

and the real source Emcos ωt. Take

note e has been replace by E .



The RL Circuit

i

E = E m e j ωt v L

v R

L

The steady-state response (I ) of this RL

circuit using the complex source is also

complex of the form;

)(

0

t je I I

where φ is an arbitrary phase angle. Substituting this to the DEyields;

t j

m

t jt j e E e RI e I dt

d L )(

0

)(

0

Simplifying, yields;

L j R

E e I m j

0

Converting the denominator in terms of the complex exponential,



The RL Circuit

Thus;

)(tan

22)(tan220

1

1

)()(

R L

R L

jm j

m j e L R

E e L R

E e I

Substituting to the steady-state response, finally yields;

220

)( L R

E I m

)(tan 1

R L

)(tan

22

1

)(

R Lt jm e

L R

E I

)(tansin)(

)(tancos)(

1

22

1

22 R Lm

R Lm t

L R

E jt

L R

E I

Real part of the solution Imaginary part of the solution



The RL Circuit

)(tansin)(

)(tancos)(

1

22

1

22 R

Lm R Lm t

L R

E jt

L R

E I

Real part of the solution Imaginary part of the solution

Since our source is the imaginary part in Euler’s relation (since it

was stated as a sine function), our steady state solution must only

contain the imaginary part. Hence,

)(tansin)(tansin

)(

1

22

1

22 R

X

L

m

R Lm Lt

X R

E t

L R

E i

which is similar to the one derived in the RL circuit previously.



Assignment

Using complex source, derive the response for an RC circuit as

shown below.

i


v R sw.

Submit results in Edmodo. Deadline June 24, 2013.



The RC Circuit

By KVL, the differential equation forthis circuit is,

By Euler’s identity, Em sin ωt = Em Ime jωt ).

We can replace the sinusoidal source with a complex source

Em e jωt.

This source now has two component

sources: the imaginary source

Emsin ωt which is our original source

and the real source Emcos ωt.

t RC

E v

RC dt

dv mC

C sin1

i


v R sw.

E = E m e j ωt

i

v C C

v R sw.



The RC Circuit

The steady-state response (V C ) of this

RC circuit using the complex source is

also complex of the form;

)(

0

t j

C eV V

where φ is an arbitrary phase angle. Substituting this to the DEyields;

Simplifying, yields;

C

R j

E eV m j

/11

0

Converting the denominator in terms of the complex exponential,

E = E m e j ωt

i

v C C

v R sw.

t jmt jt j e RC

E eV

RC eV

dt

d )(

0

)(

0

1



The RC Circuit

Thus;

)(tan

2)(tan

20

1

/1

1

1/1

1

C X R

C R

j

C

m

j

m j e

X

R

E

eC

R

E eV

Substituting to the steady-state response, finally yields;

2

0

1

C

m

X

R

E V

)(tan

1

C X

R

))(tan(

2

1

1

C X Rt j

C

mC e

X

R

E V



The RC Circuit

)(tansin

1

)(tancos

1

1

2

1

2 C C X R

C

m

X R

C

mC t

X

R

E jt

X

R

E V

)(tansin

1

1

2 C X R

C

mC t

X

R

E v

which is equivalent to;

Since the actual source is the imaginary part of the complex source,

the time domain voltage will only contain the sine function. Thus,



The RC Circuit

C

C

mC X Rt X R

E dt dvC i /(tancos 1

22

C

o

C

m X Rt X R

E i /(tan90sin 1

22

Since tan-1(R/XC)+tan-1(XC/R) = 90o

C C C

C

m X R X R R X t X R

E i /(tan)/(tan)/(tansin 111

22

)/(tansin 1

22 R X t

X R

E i C

C

m



Thank You

module 2 - rlc circuits

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