Chapter 4 Inductance, Capacitance and 1st-order Source-Free RL and RC Circuits Dr. Wimol San-Um ENG-207: Electrical Engineering, Academic year 2/2555 Computer Engineering Program, Faculty of Engineering, Thai-Nichi Institute of Technology

Objectives 1. To understand inductance properties and inductance in combination configurations 2. To understand capacitance properties and capacitance in combination configurations 3. To understand source-Free RC circuits 4. To understand source-Free RL circuits

Introduction This chapter deals with some basic properties of inductor (L) and capacitor (C), which are basic energy storage elements in electric circuits. An inductor is a passive twoterminal electrical component that stores energy in its “magnetic field” while a capacitor is a passive two-terminal electrical component that stores energy in an “electric field”. Cascade and parallel combinations of inductor and capacitor are also investigated as for a simplification into simple circuit configuration in circuit analysis. The 1st-order sourcefree circuit that contains only an energy storage element, i.e. C or L, a resistor, and a switch, will be investigated in terms of a “natural response” and “time constant”. The study of the 1st-order source-free circuit will useful for modeling and describing the characteristics of electric networks found in all applications in the fields of electrical engineering.

ENG-207 Electrical Engineering

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Topic 1: Inductance Properties and Inductance Combination

v i

L

(a) Symbol of inductor

(b) Image of various inductors

Fig.1: The symbol of an inductor and the image of various types of inductors An “inductor” is a passive two-terminal electrical component that stores energy in its magnetic field as shown in Fig.1 (a). For comparison, capacitor stores energy in an electric field and a resistor does not store energy but rather dissipates energy as heat. As shown in Fig1 (b), the inductor is typically made of a wire or other conductor wound into a coil, to increase the magnetic field. The relationship between the time-varying voltage v(t) across an inductor with inductance and the time-varying current i(t) passing through it is described by

v(t)  L

di(t) dt

and

i(t) 

1 v(t)dt  I O L

(1)

It is seen in (1) that the effect of an inductor in a circuit is to oppose changes in current through it by developing a voltage across it proportional to the rate of change of the current. The absorbed power in an inductor is given by

di(t)  di(t)  PInductor  v(t )i (t )   L i (t )  Li (t ) dt  dt 

ENG-207 Electrical Engineering

(2)

Page 2

The energy accepted by the inductor is stored in the magnetic field around the coil and s expressed by the integral of the power over the desired time interval given by t

WInductor (t ) 

L

t 0

di(t) 1 1 dt  L i 2 (t) - i 2 (t 0 )  Li 2 (t) dt 2 2

(3)

Example 1: Given a simple RL circuit as shown below. Find the voltage across both R and L, the power dissipated in the resistor, the power absorbed by the inductor, and the energy stored in the inductor.

vR 0.1Ω

12sin(πt/6)

3H

vL

In general, the equivalent inductance of series-connected inductors is the sum of the individual inductances. However, the equivalent inductance of parallel-connected inductors is the sum of the reciprocal of individual inductances. (a) Cascade connection

L1

L2

(b) Parallel connection

Ln L1

L2

Ln

LEQ

LEQ

LEQ = L1 + L2 +…+ Ln

(1/LEQ ) = (1/L1) + (1/L2) +…+ (1/Ln)

Fig.2: Cascade and parallel connection of inductors.

ENG-207 Electrical Engineering

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Example 2: Find the equivalent inductance LEQ if all inductors are equal at 10mH.

L1

L4

L2 L3

L6 L5

L7

LEQ

Topic 2: Capacitance Properties and Inductance Combination

v i

C

(a) Symbol of capacitor

(b) Image of various capacitor types

Fig.3: The symbol of a capacitor and the image of various types of capacitors. A capacitor is a passive two-terminal electrical component used to store energy in an electric field. The forms of practical capacitors vary widely, but all contain at least two electrical conductors separated by a dielectric. Capacitors are widely used as parts of electrical circuits in many common electrical devices. The relationship between the timevarying voltage v(t) across a capacitor with capacitance and the time-varying current i(t) passing through it is described by

1 dv(t) and i(t)   i(t)dt  I O (4) C dt It is seen in (4) that the effect of an inductor in a circuit is to oppose changes in voltage through it by developing a current across it proportional to the rate of change of the voltage. The absorbed power in an inductor is given by i(t)  C

ENG-207 Electrical Engineering

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PCapacitor  Cv(t )

dv(t) dt

(5)

The energy accepted by the capacitor is stored in the electric field and is expressed by the integral of the power over the desired time interval given by t

WInductor (t ) 

C

t 0

dv(t) 1 1 dt  C v 2 (t) - v 2 (t 0 )  Cv 2 (t) dt 2 2

(6)

Example 3: Given a simple RC circuit as shown below. Find the voltage across both R and C, the power dissipated in the resistor, the power absorbed by the capacitor, and the energy stored in the capacitor.

100sin(2πt)

v

20uF

0.1MΩ

In general, the equivalent capacitance of series-connected inductors is the sum of the reciprocal of individual capacitance. However, the equivalent capacitance of seriesconnected inductors is the sum of the individual capacitances. (a) Cascade connection

C1

C2

(b) Parallel connection

Cn C1

C2

Cn

CEQ

CEQ (1/CEQ ) = (1/C1) + (1/C2) +…+ (1/Cn)

CEQ = C1 + C2 +…+Cn

Fig.4: Cascade and parallel connection of capacitors. ENG-207 Electrical Engineering

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Topic 3: Source-Free RL and RC Circuits and Natural Response

iC

iR VO

v

i

t=0

t=0

R

IO

C

(a) Simple RC circuit

vR

R

L

vL

(a) Simple RL circuit

Fig.5: Simple 1st-order RC and RL circuits A first-order circuit can contain only a single energy storage element, i.e. a capacitor or an inductor. The circuit will also contain resistance. Therefore, there are two possible types of first-order circuits, RC and RL circuits, as shown in Fig.5. A source-free circuit is one where all independent sources have been disconnected from the circuit after some switch action. The voltages and currents in the circuit typically will have some transient response due to initial conditions, i.e. initial capacitor voltages and initial inductor currents. In this topic, the use of exponential function will be useful and the expression is as follows; f (v)  VO exp(-(t/  )v) (7) where VO is a constant and τ is a time constant. The plot of Eqn.(7) is shown below;

Voltage

f(v) VO

τ increases

t Time

Fig.6: Plots of f(v) with different values of the time constant τ.

ENG-207 Electrical Engineering

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Simple RC circuit with disconnecting voltage source t=0

iC

iR VO

R

C

Case 1

Case 2

Case 3

t<0

t=0

t>0

iR VO

v

R

iC=0 VO

C

(a) v(t<0) = VO

v

R

C

VO

(b) v(0) = VO (Initial voltage)

v(t)

t<0

R

C

(c) v(t>0) = VOexp(-t/RC)

+

-

v(0 )

v(0 )

VO

iC

iR

iC

iR

t=0

t

t>0

Fig.3.7: Characteristics of the 1st-order RC circuit with disconnected voltage source 3.1

The natural response of 1st-Order RC circuit

Fig. 3.7 shows the 1st-Order RC circuit composed by a DC voltage source, R and C. There are three cases must be concerned, including circuit operations at time t<0, t=0, and t>0. Summary of the three cases are as follows; Case 1:

For t<0, the switch has been closed for a long time. Then, the circuit reaches a dc steady-state circuit where the capacitor is an open circuit. Therefore, v(t) = VO.

ENG-207 Electrical Engineering

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Case 2:

For t=0, the voltage in the capacitor remains unchanged in the first instant after the switch is opened, i.e. v(0-) = VO= v(0+).

Case 3:

For t>0, no source is connected and the energy stored in the capacitor will be released to the resistor. Applying KCL yields

IR  IC  0 or

dv v(t)  0 dt RC

(8)

Arranging Eqn.(8) results in  t   t v(t)  VO exp    VO exp  -   RC   τ

(9)

where τ=RC is a time constant which is time required for a circuit to reach its final state or steady state is 5τ as described in Fig.8 below. This characteristic is called “the natural response” of the RC circuit after the switch is disconnected.

v(t) VO Transient State

0.368VO

0.0067VO τ

t

Time Fig.3.8: Transient and steady states of the natural response f RC circuit.

ENG-207 Electrical Engineering

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Example 4: The switch in the circuit has been closed, and it is opened at t= 0. Find v(t) and i(t) for t≥0. Also draw the graph of v(t) and i(t). t=0

20V

i(t)

v(t)

20mF

Example 5: The switch in the circuit has been closed, and it is opened at t= 0. Find v(t) and i(t) for t≥0. Also draw the graph of v(t) and i(t).

24V

t=0

i(t) 12Ω

v(t)

(1/6)F

Example 6: The switch in the circuit has been closed, and it is opened at t= 0. Find v(t) and vO(t) for t≥0.

10kΩ

240kΩ

100V 0.5uF

ENG-207 Electrical Engineering

32kΩ

t=0

v(t)

vO(t)

60kΩ

Page 9

The natural response of the 1st-Order LC circuit

3.2

Simple RL circuit with disconnecting voltage source i

t=0

IO

R

vR

vL

Case 1

Case 2

Case 3

t<0

t=0

t>0 i

i

IO

L

R

vR

i

L

vL

(a) i(t<0) = IO

IO

R

L

vR

vL

(b) i(0) = IO (Initial voltage)

i(t)

t<0

R

L

vL

(c) i(t>0) = IOexp(-Rt/L)

+

-

i(0 )

i(0 )

IO

vR

t=0

t

t>0

Fig.3.9: Characteristics of the 1st-order RL circuit with disconnected voltage source Fig. 3.9 also shows the simple 1st-Order RL circuit composed by a DC voltage source, R and L. In a similar manner to the simple RC circuit, there are three cases must be concerned, including circuit operations at time t<0, t=0, and t>0. Summary of the three cases are as follows; Case 1:

For t<0, the switch has been closed for a long time. Then, the circuit reaches a DC steady-state circuit where the inductor is a short circuit. Therefore, i(t) = IO.

ENG-207 Electrical Engineering

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Case 2:

For t=0, the current in the inductor remains unchanged in the first instant after the switch is opened, i.e. i(0-) = IO= i(0+).

Case 3:

For t>0, no source is connected and the energy stored in the inductor will be released to the resistor. Applying KVL yields

di Ri(t)  0 dt L

(9)

 R   t i(t)  I O exp  -   t   I O exp  -   τ  L 

(10)

VR  VL  0 or Arranging Eqn.(8) results in

where τ=L/R is a time constant which is time required for a circuit to reach its final state or steady state is 5τ in a similar manner to the RC circuit.

Example 7: The switch in the circuit has been closed, and it is opened at t= 0. Find i(t) and v(t) for t≥0.

t=0

i(t) 40V

ENG-207 Electrical Engineering

v(t)

12Ω

16Ω

2H

Page 11

Example 8: The switch in the circuit has been closed, and it is opened at t= 0. Find i(t) and v(t) for t≥0. 2Ω

i(t)

v(t) 10V

t=0

2H

Example 9: Assuming that i(0)=10A, calculate i(t) and ix(t) in the circuit.

i(t)

3i(t)

ix(t)

0.5H

Topic 4: Step-Response of RL and RC Circuits The response of a circuit to the sudden application of a constant voltage or current source is referred to as the step response of the circuit. This case presents the opposite conditions of the natural response. Now, in RL or RC circuits, the inductor or capacitor (assumed to be completely discharged) begins acquiring energy after a sudden application of an external power source. The voltages and currents that arise in the circuit under these conditions are discussed next. The calculations of step responses are similar to the source-free RL and RC circuits. Only summary will be described hereby. The unit-step input and output response is given by

ENG-207 Electrical Engineering

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VS

VO

Unit-Step Input

Unit-Step Response

Fig.3.10: Characteristics of t unit-step input and output response. 4.1

The step response of 1st-Order RC circuit In an RC circuit, the initial voltage across the capacitor is assumed to be Vo. The expression for the voltage in the capacitor after the current source is applied is

4.2

VO , t  0   v(t)    t VS  (VO  VS )exp   , t  0   τ  The step response of 1st-Order RL circuit

(11)

In an RL circuit, the initial conditions to determine the step response are assumed to be Io. The expression for the current the inductor after the voltage source is applied is IO , t  0   i(t)    t (12) IS  (I O  IS )exp   , t  0   τ  Example 10: The switch in the circuit below has been in position A for a long time. At t=0, the switch moves to B. Determine v(t) for t>0 and calculate its value at t =1s and 4s.

3kΩ

24V

ENG-207 Electrical Engineering

5kΩ

A t=0 B

0.5m F

4kΩ

240kΩ

v(t)

30V

Page 13

Example 11: The switch in the circuit has been in position x for long time. At t=0, the switch moves to position y. Answer the following questions. 1) Find the current through the inductor i(t) for t>0 2) What is the initial voltage across the inductor just after the switch has moved to y. 3) How long after the switch has been moved does the inductor voltage equals 24 V? 4) Plot both i(t) and v(t).

y

t=0

x

200mH 24V

8A

10Ω

v(t)

i(t)

Example 12: The switches in the circuit has been in position x for long time. At t=0, and t=35ms, the switches are changed their positions. Determine complete responses of v(t) and i(t). Also sketch the waveforms of vO(t) and v(t) on the time intervals [-6τ,6τ].

60V

12Ω

t=0

t=35ms

v(t)

i(t)

18Ω

150mH

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## chapter 4.pdf

ENG-207: Electrical Engineering, Academic year 2/2555. Computer Engineering Program, Faculty of Engineering, Thai-Nichi Institute of Technology. Objectives.

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