Unit-2.pptx

ELE101: Fundamentals of Electrical and

Electronics Engineering

Unit II: Contents

 _{RL and RC transients in circuits with DC source}  Analysis of 2nd order circuit with DC source

 RMS values

 Use of phasors for constant frequency

sinusoidal sources

 _{Steady state AC analysis of series circuit}  Series & parallel combination of complex

impedances

Ohm for AC

 An AC circuit is made up with components.

 Power source  Resistors

 Capacitor  Inductors

 Kirchhoff’s laws apply just like DC.

IR V_R 

t V₀ sin

Inductor Current



The voltage across an inductor vL = Ldi/

dt

and at the moment of switching

(transient, t = 0) the voltage required to

induce a change in the inductor current is

∞, hence the

inductor current can not

change instantly

which gives the

statement: iL(0−) = iL(0+)



In the steady state condition in DC circuits

Capacitor Voltage

 The current flowing into the capacitor iC = C

dvC/dt and at the moment of switching

(transient, t = 0) the current required to induce a change in the capacitor voltage is ∞, hence

the capacitor voltage can not change

instantly which gives the statement: vC(0−) = vC(0+)

 In the steady state condition in DC circuits (i.e.

when t → ∞), the capacitor acts as an open circuit. (iC = C dvC/dt the capacitor is fully

Transients

The time-varying currents and voltages

resulting from the sudden application of

sources, usually due to switching, are

called

transients. By writing circuit

DC Steady State Response

The steps in determining the forced response

for

RLC circuits with dc sources are:

1. Replace capacitances with open

circuits.

2. Replace inductances with short

circuits.

First-Order Circuits

 A circuit that contains only sources, resistors and

an inductor is called an RL circuit.

 A circuit that contains only sources, resistors and

a capacitor is called an RC circuit.

 RL and RC circuits are called first-order circuits

because their voltages and currents are described by first-order differential equations.

– +

v

s L

R

– +

v _C

R

 _{In steady state (t = Infinity), an inductor behaves} like a short circuit



The

natural response

of an RL or RC

circuit is its behavior (

i.e.,

current and

voltage) when stored energy in the inductor

or capacitor is released to the resistive part

of the network (containing no independent

sources).



The

step response

of an RL or RC circuit is

Natural Response of an RL Circuit

is closed for t < 0, and then opened at t = 0. Find v for t>0

Notation:

0– is used to denote the time just prior to switching

0+ is used to denote the time immediately after

switching

 The current flowing in the inductor at t = 0– is I

L

R_o R

I_o

t = 0 _{i +}

v

Solving for the Current (

t



0)

 For t > 0, the circuit reduces to

 Applying KVL to the LR circuit yields first-order D.E.:

L di/dt + iR = 0

L[sI(s) – I(0-)] +RI(s) = 0 I(s) = I(0-)*L/(R+Ls)

 _Solution:

L

R_o R

I_o

i +

v

–

t L R

e

i

t

Solving for the Voltage (

t

> 0)

e

I

t

i

(

)



L

R_o R

I_o + v –

Re

I

iR

t

v

t

>





)

(

0,

Time Constant

t

In the example, we found that



Define the

time constant

 At t = t, the current has reduced to 1/_e

(~0.37) of its initial value.

 At t = 5t, the current has reduced to less

than 1% of its initial value.

R

L



t

t L R o

t L R

o

e

v

t

I

Re

I

t

i

(

)



and

(

)



Capacitors and Stored

Charge

 Current doesn’t really “flow through” a capacitor. No

electrons can go through the insulator.

 But, we say that current flows through a capacitor.

What we mean is that positive charge collects on one plate and leaves the other.

 _{A capacitor stores charge.}

 _{When a capacitor stores charge, it has nonzero}

voltage. In this case, we say the capacitor is

Capacitors in circuits



If you have a circuit with capacitors, you

can use KVL and KCL, nodal analysis, etc.

The voltage across the capacitor is related

to the current through it by a differential

equation instead of Ohm’s law.

dt

dV

C

CAPACITORS

C

i

dt

dV

So



capacitance is defined by

C

i(t)

| (V

+ 

dt

dV

C

Charging a Capacitor with a constant current

C

i

dt

dV(t)

_

C

i

dt

dV(t)





dt



dt

C

t

i

C

i

V(t)









dt

Voltage vs time for an RC discharge

0 0.2 0.4 0.6 0.8 1 1.2

0 1 2 3 4

Voltage

 _{Consider the following circuit, for which the switch is}

closed for t < 0, and then opened at t = 0:

Notation:

0– is used to denote the time just prior to switching

0+ is used to denote the time immediately after switching

 The voltage on the capacitor at t = 0– is V_o as cap is

open circuit

Natural Response of an RC Circuit

C R_o

R V_o

t = 0

+

 +_v

Solving for the Voltage (

t



0)

 _{For t > 0, the circuit reduces to}

Applying KCL to the RC circuit: V + RCdv/dt = 0

V(s) + RC[sV(s) – Vo] = 0

 _Solving:

+

v

–

RC t

e

v

t

v

(

)



(

0

)

C R_o

R V_o +_

Solving for the Current (

t

> 0)



Note that the current changes abruptly:

RC t o

e

V

t

v

(

)



R

V

i

e

R

V

R

v

t

i

t

i









>



)

0

(

)

(

0,

for

0

)

0

(

Time Constant

t



In the example, we found that



Define the

time constant

 _{At t =} t_{, the voltage has reduced to 1/e}

(~0.37) of its initial value.

 At t = 5t, the voltage has reduced to less

than 1% of its initial value.

RC



t

e

R

V

t

i

e

V

t

v

(

)



and

(

)



Natural Response Summary

RL Circuit

 Inductor current

cannot change instantaneously

RC Circuit

 Capacitor voltage

Basic RL and RC Circuits

About Calculation for The Initial Value

i_C i_L

i t= 0 + _ + -v_L(0+

v_C(0+)=4 V

i₍₀₊₎

i_C(0+) i_L(0+)





 

C

v    

 

 

2 2

i  

  

 

L

i    

  

 

 

C C

v _  v _

 

 

L L

Problem 1



_{Close the switch, and find I}

₁

_{, I}

₂

_{, I}

₃

_{, Q, and Vc}

Solution

The capacitor just starts to be charged

⇒ Q= 0.

Vc = Q/C = 0

Req=R1+R2R3/R2+R3

= 14.5kΩ

V2=V3=I1.(R2R3)/R2+R3

I2=V2/R2 = 0.1 mA

After the switch is closed for a length of time sufficiently long for the capacitor to become fully charged, find I₁,I₂,I₃,Q, andVc

I3 = 09 =I1(R1+R2) = 27.I1

I1=I2=9/27=0.33mA

V2 = V3 = 15/3 = 5

Q = C.Vc = 50uC

The switch is then reopened. Find I1,I2,I3,Q,and Vc immediately after the switch is reopened

Once the switch is open, the left-hand side circuit is open and therefore I1= 0.

Q = 50uC

Vc=Q/C = 5 V I2 = I3

Vc=I2R2+I3R3=I2(R2+R3) = 18*I2 I2 = 5/18 = 0.28 mA

After the switch is reopened for a length of

time sufficiently long for the capacitor to

become fully discharged, find

I1,I2,I3,Q,and Vc

I1 = 0

Q = Vc = 0 as cap is fully discharged

I2 = I3 = 0

Second-Order

Circuit

A second-order circuit is characterized by a second-order differential equation. It

consists of resistors and the equivalent of two energy storage elements.

Typical examples of second-order circuits:

C

R L

U_S

+

_

C U_S

+

_

Impedance of Series RLC

Circuit

 _{When R, L, and C are in a series circuit, the reactance of}

the inductor and reactance of the capacitor tend to offset each other, depending on the values. The total reactance is X_L-X_C



_The

_{total impedance}

_{for the RLC circuit is given}

by :

Z = R + jX

- jX



_In

_{polar form}

_{, this is written}





2 _tan 1 tot

L C

X

R X X

R

  

    _{ }

 

Vector Map

 Phase shifts are present in AC circuits.

 +90° for inductors

 -90° for capacitors VL=IXL

Phasor Diagram for Impedance in

Series RLC Circuit



A vector sum gives the total impedance.

X_L

X_C R

X_L X_C

R

Vector Sum

X_L X_C

R

Z

 The total impedance

is the magnitude of

Z.

 The phase between

the current and

voltage is the angle

f between Z and the x-axis. f





Z _{L C}

          _    R C L R X X_{L C}

R L C

V_S

What is the total impedance for the circuit?

330 mH

f = 100 kHz

470  2000

pF



 



2 2 100 kHz 330 H 207

L

X   fL   m  



 



1 1

796 2 2 100 kHz 2000 pF

C

X

fC

 

   

470 + 207 796 =470 589

L C

R jX jX j j

j

          

Z

In polar

form,

753 51.4

   

Z

Problem 5

X_L=X_C

There is a frequency at which X_C=X_L. This condition is called

series resonance.

Series resonance

At series

resonance, the

total impedance is a minimum.

X_L

f

X_C

The formula for resonance can be found by setting

X_C = X_L. The result is

1 2

r

f

LC





Problem 6

What is the resonant frequency for the circuit?

R = 470 ohm, L= 330 uH, C = 2000pF



 



1 2

1

2 330 μH 2000 pF r

f

LC









Problem 7:

What is

V

at resonance?

R L C

V_S 330 mH

470  2000

pF 5.0

Vrms

For parallel RLC circuits, the impedance is found using the reciprocal of the sum-of-reciprocals.

1 1 1 1

0 _L 90 _C 90

R X X

  

      

Z

R L C

A typical current phasor diagram for a parallel RLC circuit is

+90o

90o

I_C

I_R

I_L

The total current is given by:





2 _tan 1 CL

tot R C L

R

I

I I I

I

       _{ }  

I

Problem 8

What is

I

if

I

= 10 mA,

I

= 15 mA and

I

= 5 mA?

Problem 9 - What is the resonant frequency for the circuit?



 



1 2

1

2 680 μH 15 nF r

f

LC









 49.8 kHz

R L C

V_S

680 mH 1.0

k 15 nF

Series-parallel

RLC

circuits

V_S = R L

C

200 kHz

700 mH 1.0

k 2700

pF

1

2

0.880 k

0.295 k

2

L C

X

fL

j

X

j

fC













 





 



k 0.661 56.6 k 0.364 k 0.552 k 1.0 0.88

ybox Z j j        _{ }            

The impedance of the yellow box is

The total impedance is Z_tot  X_C  Z_ybox





0.295 0.364 0.552 k

tot

Z  _ j   j _ 

= 0.364 k + j0.257 k

0.445 35.2 k





 

Series Parallel RLC Circuit

RMS Value

where ΔVmax is the maximum output voltage of the AC source, or the voltage amplitude.

The instantaneous voltage across the resistor is

Power in an AC

Circuit

Problem 13 - Average Power in an RLC Series Circuit

Calculate the average power delivered to the series RLC