ENGR-43_Lec-04a_1st_Order_Ckts

Engineering 43

RC & RL

1

st

Order Ckts

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu

Engineering-43: Engineering Circuit Analysis

1

Bruce Mayer, PE

BMayer@ChabotCollege.edu • ENGR-43_Lec-04a_1st_Order_Ckts.pptx

C&L Summary


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Bruce Mayer, PE


Introduction



Transient Circuits

 In Circuits Which Contain Inductors &

Capacitors, Currents & Voltages

CanNOT Change Instantaneously

 Even The Application, Or

Removal, Of Constant

Sources Creates

Transient (Time-

Dependent) Behavior


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Bruce Mayer, PE


1 st & 2 nd Order Circuits

 FIRST ORDER CIRCUITS

• Circuits That Contain

ONE Energy Storing Element

– Either a Capacitor or an Inductor

 SECOND ORDER CIRCUITS

• Circuits With TWO Energy Storing

Elements in ANY Combination


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Bruce Mayer, PE


Circuits with L’s and/or C’s

 Conventional DC Analysis Using

Mathematical Models Requires The

Determination of (a Set of) Equations

That Represent the Circuit Response

 Example; In Node Or Loop Analysis Of

Resistive Circuits One Represents The

Circuit By A Set Of Algebraic Equations

The Ckt


5

Analysis

The DC Math Model

G v

 i

Bruce Mayer, PE


Ckt w/ L’s & C’s cont.

 When The Circuit Includes Inductors Or

Capacitors The Models Become Linear

Ordinary Differential Equations (ODEs)

 Thus Need ODE Tools In Order To

Analyze Circuits With Energy Storing

Elements

• Recall ODEs from

ENGR25

6

• See Math-4 for

More Info on ODEs

Engineering-43: Engineering Circuit Analysis Bruce Mayer, PE


First Order Circuit Analysis

 A Method Based On Th évenin Will Be

Developed To Derive Mathematical

Models For Any Arbitrary Linear Circuit

With One Energy Storing Element

 This General Approach Can Be

Simplified In Some Special Cases

When The Form Of The Solution Can

Be Known BeforeHand

• Straight-Forward ParaMetric Solution


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Bruce Mayer, PE


Basic Concept

 Inductors & Capacitors

Can Store Energy

 Under Certain Conditions This

Energy Can Be Released

 RATE of Energy Storage/Release

Depends on the parameters Of The

Circuit Connected To The Terminals Of

The Energy Storing Element


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Bruce Mayer, PE


Example: Flash Circuit

 The Battery, V

S

,

Charges the Cap

To Prepare for a

Flash

 Moving the Switch to the Right

“Triggers” The

Flash

• i.e., The Cap

Releases its Stored

Energy to the Lamp


9

Say

“Cheese”

Bruce Mayer, PE


Flash Ckt Transient Response

 The Voltage Across the Flash-Ckt Storage

Cap as a Function of TIME

 Note That the Discharge Time (the Flash) is

Much Less Than the Charge-Time


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Bruce Mayer, PE


General Form of the Response

 Including the initial conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form

 dx dt

     

 This is the General Eqn

11

 Now By Linear

Differential Eqn Theorem

(SuperPosition) Let


 x p

(t)



ANY Solution to the

General ODE

• Called the “Particular”

Solution

 x c

(t)



The Solution to the

General Eqn with f(t) =0

• Called the “Complementary

Solution” or the “Natural”

(unforced) Response

• i.e., x c is the Soln to the

“Homogenous” Eqn

 dx dt

   



0

Bruce Mayer, PE


1 st Order Response Eqns

 Given x p and x c the Total Solution to the ODE x

 

 x p

 

 x c

 

 Consider the Case

Where the Forcing

Function is a Constant

• f(t) = A

 Now Solve the ODE in

Two Parts



 dx p dx c dt

 

 dt

 x p

 x c

 



0

A

 For the Particular Soln,

Notice that a

CONSTANT x p

 



K

1

Fits the Eqn: and dx p so

 



0 dt


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Bruce Mayer, PE


1 st Order Response Eqns cont

 Sub Into the General

(Particular) Eqn x p dx p

/dt

0



K

1



A and

K

1 or



A

13

 Next, Divide the

Homogeneous (RHS=0)

Eqn by

 ∙x c

(t) to yield dx c x c

 

  dt

 



1


 Next Separate the

Variables & Integrate

 x c

1

   

  



1 d t

 Recognize LHS as a

Natural Log; so ln

 x c

  





 t where c



 const c

 Next Take “e” to

The Power of the

LHS & RHS

Bruce Mayer, PE


1 st Order Response Eqns cont

 Then x c

 

 e

 t

  c x c



K

2 e

 t



 e c e

 t



 Note that Units of TIME

CONSTANT,



, are Sec

 Thus the Solution for a

Constant Forcing Fcn x

 

 x p

 

 x c

  x

 



K

1



K

2 e

 t /




14

 For This Solution

Examine Extreme

Cases

• t =0 x 0

• t → ∞



K

1 x

 t

 







K

2

K

1



K

2 e

  

K

1

 The Latter Case (K

1

) is

Called the Steady-State

Response

• All Time-Dependent

Behavior has dissipated

Bruce Mayer, PE


Effect of the Time Constant

Tangent reaches x-axis in one time constant

Decreases 63.2% after One Time

Constant time


15

Drops to 1.8% after 4 Time

Constants e



4 

0 .

0183

Bruce Mayer, PE


Large vs Small Time Constants

 Larger Time Constants Result in Longer

Decay Times

• The Circuit has a Sluggish Response


16

Quick to Steady-State

Bruce Mayer, PE


1 st Order Ckt Solution Plan

1. Use some form of KVL/KCL, 𝑖 = 𝐶 𝑑𝑣 𝑑𝑡 𝑑𝑖 and 𝑣 = 𝐿 𝑑𝑡 to develop an ODE

2.

Isolate the “0 th” order term which reveals the

• Time Constant

• Forcing Function

3.

“EyeBall” the ODE to “Guesstimate the

Particular Solution, 𝑖 𝑝 𝑡 or 𝑣 𝑝 𝑡


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Bruce Mayer, PE



4. CHECK that the particular Solution

Satisfies the ODE

5. Separate the Variables in the

Homogeneous Equation and Integrate 𝑖 to obtain the Complementary Solution 𝑐 𝑡 or 𝑣 𝑐 𝑡

6. A the particular and complementary solutions to obtain the total Solution; 𝑥 𝑡𝑜𝑡 𝑡 = 𝑥 𝑐 𝑡 + 𝑥 𝑝 𝑡


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Bruce Mayer, PE



 The Solution Should include a Negative

Exponential with an UnKnown

PreFactor

7. Use the initial condition in the total solution to determine the Prefactor which completes the total solution

8. Check the total Solution for extreme cases:

• t = 0+ • t → ∞


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Bruce Mayer, PE


Example



1 st Order Ckt Soln

 For the Ckt Below Find 𝑖

𝐿 𝑡 for 𝑡 > 0


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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE



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Bruce Mayer, PE


Time Constant Example

 Charging a Cap v

C

 v

S

R

S a

R

S v

S





C

+ v

_ c

C dv

C dt b

 Use KCL at node-a

C dv dt

C

 v

C

R



S v

S



0

29

C dv

C

 v

C

 v

S dt R R

S


S

 Now let

• v

C

(t = 0 sec) = 0 V

• v

S

(t)= V

S

(a const)

 ReArrange the KCL Eqn

For the Homogenous

Case where V s

= 0 dv

C v

C dt

 

1

CR

S

 Thus the Time Constant

 

CR

S dx c x c

 

  dt

 



1

 x c



K

2 e

 t



Bruce Mayer, PE


Time Constant Example cont

 Charging a Cap

R

S a

+ v

S





C v c

_ b

 The Solution Can be shown to be v

C

Or v

C

( t

( t

)

) t



V

S

 e







V

S

1

V

S



 e

 t






30

 “Fully” Charged Criteria

• v

C

>0.99V

S

OR t e

 t





 





0 .

01 ln

 or

0 .

01





4 .

6



 

R

S

C

Bruce Mayer, PE


Differential Eqn Approach

 Conditions for Using This Technique

• Circuit Contains ONE Energy Storing Device

• The Circuit Has Only CONSTANT,

INDEPENDENT Sources

• The Differential Equation For The Variable Of

Interest is SIMPLE To Obtain

– Normally by Using Basic Analysis Tools; e.g., KCL, KVL,

Thevenin, Norton, etc.

• The INITIAL CONDITION For The Differential

Equation is Known , Or Can Be Obtained Using

STEADY STATE Analysis Prior to Switching

– Based On: Cap is OPEN, Ind is SHORT


31

Bruce Mayer, PE


Example

 Given the RC Ckt At

Right with

• Initial Condition (IC):

– v(0 −

) = V

S

/2

 Find v(t) for t>0

 Looks Like a Single

E-Storage Ckt w/ a

Constant Forcing Fcn x (

• Assume Solution of Form t )



K

1



K

2 e



 t

, t



0

K

1

 x (



); K

1



K

2

 x ( 0



)


32

 Model t>0 using KCL at v(t) after switch is made v

 



R

V

S



C dv dt

 



0

 Find Time Constant; Put

Eqn into Std Form

• Multiply ODE by R

RC dv dt



V s

Bruce Mayer, PE


Example cont

 Compare Std-Form with

Model

• Const Force Fcn Model

 dx

 x



K

1 dt

• In This Case

RC dv

( t )

 v ( t )



V s dt

 Next Check Steady-

State (SS) Condition

• In SS the Time Derivative goes to ZERO

 Note: the SS condition is Often Called the

“Final” Condition (FC)

 In This Case the FC t

   confirms v ( t

: K

1

)



V s



V s

 Now Use IC to

Find K

2


33

Bruce Mayer, PE


Example cont.2

 At t = 0 +

• The Model Solution v ( 0 )



K

1



K

2 e



0



K

2

 v ( 0 )



K

1



34

 Recall The IC: v(0

−

) =

V

S

/2 = v(0 + ) for a Cap

• Then

K

K

K

2

2

2





 v ( 0 )

V

S

2



K

1



V

S



V

S

2


 The Total/General Soln v

  v ( t )





K

1

V

S



K

2 e



 t



 1



0 .

, t

5 e





0 t

RC



 or v v (

 Check

0



(



)

)





V

S

V

S



1



1





0 .

5 e

0 .

5 e

 



0









V s

0 .

5 V s



Bruce Mayer, PE




 v

R



Inductor Exmpl i

 Find i(t) Given

• i(0 − ) = 0

 Recognize Single

E-Storage Ckt w/ a

Constant Forcing Fcn x (

• Assume Solution of Form t )



K

1



K

2 e



 t

, t



0

K

1

 x (



); K

1



K

2

 x ( 0



)

 In This Case x→i

( t )



K

1



K

2 e



 t

; t



0

KVL i ( t )

 v

L

 i ( t )

 To Find the ODE Use

KVL for Single-Loop ckt

V

S

 v

R

 v

L



Ri

 Now Consider IC

( t )



L di dt

( t )

• By Physics, The Current

Thru an Inductor Can NOT

Change instantaneously

 i ( 0



)

 i ( 0



)



0


35

Bruce Mayer, PE


Inductor cont

 Casting ODE in

Standard form

L

R di

( t )

 i ( t ) dt



V

S

R

 Recognize Time Const

 

L R

 Also Note FC

 Next Using IC (t = 0 + )

K

2

 i ( 0



)



K

1

K

2



0



V

S

R

K

 

V

S

R

2

 Thus the ODE Solution t

   i ( t )



V s

 Thus

K

1



V s

R

R i ( t )



K

1



K

2 e



 t



V

R s



 1

 e



R t

L






36

Bruce Mayer, PE


i ( t )

Solution Process Summary

1. ReWrite ODE in Standard Form

• Yields The Time-Constant,



2. Analyze The Steady-State Behavior

• Finds The Final Condition Constant, K

1

3. Use the Initial Condition

• Gives The Exponential PreFactor, K

2

4. Check: Is The Solution Consistent

With the Extreme Cases

• t = 0+

• t → 


37

Bruce Mayer, PE


Solutions for f(t) ≠ Constant

1. Use KVL or KCL to Write the ODE

2. Perform Math Operations to Obtain a

CoEfficient of “1” for the “Zero th ” order

Term. This yields an Eqn of the form

 dx

 x

 f (yields



) dt

3. Find a Particular Solution, x p

(t) to the

FULL ODE above

38

• This depends on f(t), and may require

“Educated Guessing”

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis


Solutions for f(t) ≠ Constant

4. Find the total solution by adding the

COMPLEMENTARY Solution, x c

(t) to previously determined x p

(t). x c

(t) takes the form:

• The Total Solution at this Point → x

  x c



 



Ke

 t

Ke





 t

 x p

5. Use the IC at t=0 + to find K; e.g.; IC = 7

7



Ke



0

  x p

 



7



K



M

• Where M is just a NUMBER


39

Bruce Mayer, PE


Capacitor Example

 For Ckt Below Find v o

(t) for t>0 (note f(t) = const; 12V)

C

R

1

R

2

 Assume a Solution of v

C the Form for v c

( t )



K

1



K

2 e



 t

, t

K

1

 v

C

(



); K

1



K

2



0

 v

C

( 0



)


40

 At t=0+ Apply KCL

C dv

C dt

( t )



R

1 v

C



R

2



0



( R

1



R

2

) C dv

C dt

( t )

 v c



Bruce Mayer, PE


0

Cap Exmp cont

 Step1: By Inspection of the ReGrouped KCL

Eqn Recognize









( R

1



R

2

) C

6 k

 

0 .

1 mF



0 .

6 s

 Now Examine the Reln

Between v o

• a V-Divider and v

C v

O

( t )



2

2



4 v

C

( t )



1

3 v

C

( t )


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 Step-2: Consider The

Steady-State

• In This Case After the

Switch Opens The

Energy Stored in the Cap

Will be Dissipated as

HEAT by the Resistors v o

( t



 With x p



)

 v

C

( t



= K

1

= SSsoln



)



0

K

1

 v

C

(



)



0

Bruce Mayer, PE


Cap Exmp cont

 Now The IC

 If the Switch is Closed for a Long Time before t =0, a

STEADY-STATE Condition Exists for NEGATIVE Times

Recall: Cap is

OPEN to DC

 v o

(0

−

) by V-Divider v

O

( 0



)



3



2

2



4

12 V



2

12 V

9 v

O

( 0



)



8

3

V


42 v

C

 Recall Reln Between v o

( t ) and v



3 v

O

C

( t ) for t ≥ 0

 v

C

( 0



)

 v

C

( 0



)



6

12 V

9



Bruce Mayer, PE

8 V


Cap Exmp cont

 Step-3: Apply The IC

K

1



K

2

 v

C

( 0



) and K

1

 v

C

(



)



0



K

2

 v

C

( 0



)



 Now have All the

8 V

Parameters needed To

Write The Solution v

C

( t )



K

1



K

2 e



 t

, t



0 v

C

( t )



0

 t

8 Ve



0 .

6 s , t



0 )

 Recall v v o

( t )

 o







= (1/3)v

8

3

V





 e



C t

0 .

6 s ,

 Note: For f(t)=Const, t

 puttingthe ODE in Std

Form yields

 and K

1 by

Inspection:

( R

1



R

2

) C dv

C dt

( t )

 v c



0



K

1

0


43

Bruce Mayer, PE


Thevenin/Norton Techniques

 Obtain The Thevenin Voltage Across The Capacitor,

Or The Norton Current Through The Inductor

Circuit with resistances and sources a

Inductor or

Capacitor b

Representation of an arbitrary circuit with one storage element

Th évenin

V

TH





R

TH a b

Inductor or

Capacitor

 With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find

• Time Constant using R

TH

• Steady-State Final Condition using v

TH

(if v

TH a constant)


44

Bruce Mayer, PE


Thevenin Models for ODE

R

TH a

R

TH a

V

TH



 i

R C i c

+ v c

_ b

Case 1.1

Voltage across capacitor

V

TH



 v

R



L



 i

L v



L

Case 1.2

b

Current through inductor i

R



 KCL at node a i c

 i

R



0 i c



C dv

C dt v

C

 v

TH

R

TH

C dv

C dt

 v

C



R

TH v

TH



0

 KVL for Single Loop v

R

 v

L

 v

TH v

R



R

TH i

L v

L



L di

L dt

L di

L dt



R

TH i

L

 v

TH





R

TH

C

 dv

C dt

 v

C

 v

TH


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



L

R

TH



 di

L dt

 i

L

 v

TH

R

TH



I

Bruce Mayer, PE


N

Find ODE Soln By Thevenin

 Break Out the Energy Storage Device

(C or L) as the “Load” for a

Driving Circuit

 Analyze the Driving Ckt to Arrive at it’s

Th évenin (or Norton) Equivalent

 ReAttach The C or L Load

 Use KCL or KVL to arrive at ODE

 Put The ODE in Standard Form


46

Bruce Mayer, PE


Find ODE Soln By Thevenin

 Recognize the Solution Parameters

 For Capacitor

• 

= R

TH

C

• K

1

(Final Condition) = v

TH

 For Inductor

= v

OC

= x p

47

• 

= L/R

TH

• K

1

(Final Condition) = v

TH

/R

TH

= i

SC

= i

N

 In Both Cases Use the v

C

(0

−

) or i

L

(0

−

)

Initial Condition to Find Parameter K

2

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis


= x p

Inductor Example

Find i

O

(t) ; t 

0  Since this Ckt has a

6



6



6



CONSTANT Forcing

Function of 0V (the 24V source is switched OUT

3 H i

O

( t )

24 V



 t



0 6

 at t = 0), Then The i

Solution Is Of The Form

 t

O

( t )



K

1



K

2 e



; t



0

 The Variable Of Interest

Is The Inductor Current

 The Th évenin model

L

R

TH di

O dt

 i

O

 v

TH

R

TH


48

 Next Construct the

Th évenin Equivalent for the Inductor

“Driving Circuit”

Bruce Mayer, PE


Inductor Example cont

Thevenin for t>0 at inductor terminals

 The f(t)=const ODE in

Standard Form

6



24 V





6

 t



0

6



6

 a b

 From This Ckt Observe v

TH

R

TH





0



6



6



6

 



6



10



0 .

3 s di

O dt

 i

O



0 ;

 The Solution t



0

Substituted into the

ODE at t > 0

0 .

3







K

2

0 .

3 t e



0 .

3









 K

1



K

2 e

 t

0 .

3







0

49

 

L



3 H

10





0 .

3 s

R

TH


 Combining the K

2 terms i shows shows K

1 t



O

( t )



K

2 e 0 .

3 ; t



= 0, thus

0

Bruce Mayer, PE


Inductor Example cont.2

Find i

O

(t) ; t 

0  Analyzing Ckt with 3H

Ind as Short Reveals

6



6



6

 i

O

( t )

 

3 H i

O

24 V

6



1

2





24 V



6



3









24 V



 t



0 6





16

A ; t



0

3

• The Reader Should

Verify the Above  Now Find K

2

 Assume Switch closed for a Long Time

Before t = 0

 Then the Entire Solution i

O

( t )





5 .

33 A

 e

 t

0 .

3 s ; t



0

• Inductor is SHORT to DC


50

Bruce Mayer, PE


Untangle to find i

O

(0 −)

 Be Faithful to Nodes


51

Bruce Mayer, PE


5.5

5.0

4.5

4.0

3.5

First Order Inductor Ciruit Transient Solution



0 .

3 s



 di

O dt di

O

 dt



3 s

 i

O i

O

10





0

 di

O dt di

O dt t



0







 



 i

O

0 .

3 s i

O

 

3 s 10

 di

O dt t



0





16 A

3 s 10

3



16



10

3



3

A s



17 .

78

A s i(t) (A)

3.0

2.5

2.0

1.5

1.0

0.5

0.0

-0.2

0.0

0.2

0.4

file = Engr44_Lec_06-1_Last_example_Fall03..xls


52 i

O

( t )

 t 

5 .

333 A

 e



0 .

3 s ; t



0

0.6

0.8

Time (s)

1.0

1.2

1.4

1.6

Bruce Mayer, PE


1.8

t=0

24V

WhiteBoard Work

 Let’s Work This

Problem

2



4



(1/4)H

6



4



8



+

V

C

(t)

-

 Well, Maybe

NEXT time…


53

Bruce Mayer, PE


WhiteBoard Work


54

Bruce Mayer, PE


Engineering 43

Appendix

DE Approach

Bruce Mayer, PE

Licensed Electrical & Mechanical Engineer

BMayer@ChabotCollege.edu


55

Bruce Mayer, PE


Differential Eqn Approach cont

 Math Property

• When all Independent Sources Are CONSTANT, then for ANY variable y(t); i.e., v(t) or i(t), in The

Circuit The Solution takes the Form y



K

1



K

2 e



 t

 The Solution Strategy

• Use The DIFFERENTIAL EQUATION And The

FINAL & INITIAL Conditions To Find The

Parameters K

1 and K

2


56

Bruce Mayer, PE



 If the ODE for y is

Known to Take This

Form a

1 dy y dt

( 0



)



 a

0 y

0 y



A

 We Can Use This

Structure to Find The

Unknowns. If: y ( t )



K

1



K

2 e



 t

, t



0

57

 dy dt

 

K

2

 e



 t


 Then Sub Into ODE a

1







K

2

 e



 t





 a

0



 K

1



K

2 e



 t







A

 a

1



 a

0

K

2 e



 t

 a

0

K

1



A

 Equating the TRANSIENT

(exponential) and

 a



CONSTANT Terms Find

1

 a

0



K

2 e

 t

 

0

   a

1 a

0

K

1



A a

0 Bruce Mayer, PE



 Up to Now y



A



K

2 e

 a

1 t a

0 a

0

 Next Use the Initial y

Condition

( 0



)

 y

0 y

K

( 0

1





)

K



2

K

1



 y

0

K

2 e

0  y

0

58 or

K

2

 y

0



K

1


 So Finally y



A a

0





 y

0

 a

A

0



 e

 a

1 t a

0

 If we Write the ODE in

Proper form We can

Determine By

Inspection

 and K

1 K

1 a

1 dy dt

 a

0 y



A

 a a

0

1

 dy

 dt y



A a

0

Bruce Mayer, PE


Inductor Example

 For The Ckt Shown

Find i

1

(t) for t>0

 Assume Solution i

1

( of the Form t )



K

1



K

2 e



 t

K

1

 i

1

(



); K

1



,

K

2 t



0

 i

1

( 0



)

 The Model for t>0 →

KVL on single-loop ckt

L di

1 dt





6



12

 i

1

( t )



0

59

 Rewrite In Std Form

Engineering-43: Engineering Circuit Analysis i

1

( t )

 v

L



2 di dt

1 ( t )

 i

1

( t )



0

18

 Recognize Time Const

 

1

9

S

Bruce Mayer, PE


L

Inductor Example cont

 Examine Std-Form Eqn to Find K

1

1 di

1

9

 dt

K

1

( t



)



0 i

1

( t )



0

 For Initial Conditions

Need the Inductor

Current for t<0

 Again Consider DC

(Steady-State)

Condition for t<0


60

 The SS Ckt Prior to

Switching

• Recall An Inductor is a

SHORT to DC i

1

( 0



)

 So i

1

( 0



)



12 V

12





1 A

Bruce Mayer, PE


Inductor Example cont.1

 Now Use Step-3 To

Find K

2 from IC

• Remember Current Thru an Inductor Must be

Time-Continuous i

1

( 0



)



1 [ A ]



 i

1

( 0



)

K

2



K

1



0

K

2

• Recall that K

1 was zero

 Construct From the

Parameters The ODE

Solution

 The Answer

 t

1 i

1

( t )

 e 9 [ A ], i

1

( t )

 e



9 t

[ A ], t t





0

0


61

Bruce Mayer, PE


ENGR-43_Lec-04a_1st_Order_Ckts

Engineering 43

RC & RL

1

Order Ckts

Engineering 43

Appendix

DE Approach

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ENGR-43_Lec-04a_1st_Order_Ckts

Engineering 43

RC & RL

1

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Engineering 43

Appendix

DE Approach

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