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Practice Problems Solution for Circuit Analysis | ECE 2040, Quizzes of Electrical Circuit Analysis

Georgia Institute of Technology - Main Campus Electrical Circuit Analysis

Prof. Jeffrey Davis

Material Type: Quiz; Professor: Davis; Class: Circuit Analysis; Subject: Electrical & Computer Engr; University: Georgia Institute of Technology-Main Campus; Term: Fall 2010;

Typology: Quizzes

Pre 2010

Uploaded on 12/10/2010

sunnypatel-125 🇺🇸

2 documents

1 / 10

This page cannot be seen from the preview

Don't miss anything!

Chapter 11 - Practice Problems Solutions

Problems

Section 11-3: Instantaneous Power and Average Power

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P11.3-1

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Discover Quizzes of Electrical Circuit Analysis Georgia Institute of Technology - Main Campus

Partial preview of the text

Download Practice Problems Solution for Circuit Analysis | ECE 2040 and more Quizzes Electrical Circuit Analysis in PDF only on Docsity!

Chapter 11 - Practice Problems Solutions

'

t

S

q q

q

P11.3-

j

q

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)

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&##

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!A! +&##, = $ = -./+ 0%" , !)%" "# !ȝ@!A!)"!>@

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q

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V I

P11.3-

.. ..

.

..

j j

j

j j

j

I I V

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V I V

I I V

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j j

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I

P11.3-

% D%+"#$%$&'()*"+-%

2 1B

j 7 I

q

V V

Section 11-5: Complex Power

P11.5-

0.6 1.2 1.342 63.43 A

j

q

q q

S

I

4 8.94 63.43 4 8 4 and 2 H

R j L j R L

q

P11.5-

3 1.5 3.35 26.56 A

j

q

q q

S

I

j j

R j L R L

q

R 4 :and L 2 H

P11.5-

Let

p

4 4 V

j

j j j

j

j j j

u

Z

1.342 26.6 A

4 4 8 j 8

q q

I

Z

q

Finally

7.2 3.6 VA

j

q q

S

The source current can be calculated from the apparent

power:

1

s s

s

2 50 cos 0.

5 36.9 A

s

2 20 0

q

s

) 5 36.9 q 4 j3 A

1

2 53.1 1.2 1.6 A

j

q

2 s 1

3.13 26.6 A

j j j

q

Finally,

s

2

q

(Using all rms values.)

(a)

2

2 2

P R = P R 500 20 100 Vrms

R

(b)

s L

5 5 5 2 45 A

j

j j

q q

q

(c)

s

j

q :

cos 45 leading

pf

q

Section 11-8: Maximum Power Transfer Theorem

P11.8-

t

L t

1.6 H

j j

j

R

R j L j

L

Z

Z Z

P11.8-

t

L t

20 k

100 H

j j

j

R

R j L j L

L

Z

Z Z

After selecting these values of R and L ,

2

3

max

1.4 mA and 20 10 19.5 mW

P

u

I

Since , yes, we can deliver 12 mW to the load.

max

P !12 mW

P11.8-

2

t L 2

800 1600 and

j

R

R j R C

C

j

RC

R

C

Z

Z Z

2

L t 2

j

R

R j R C

C

j

RC

R

C

Z

Z Z :

Equating the real parts gives

2 2

800 0.1 F

1 ( ) 1 [(5000)(4000) ]

R

C

RC C

P

Z

P11.9-

.. /

/ /. /

j j

q

I I I

.

I

. /

I !01.? 3@ !A!!(,>!!q I! !21B3 .? Aq!

C*,(::=!

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5D%#DE#>-!FGAHAI!?J/.J208!

P11.9-

P11.5-

Represent the circuit in the frequency domain and label the node voltages:

The node equations are:

1 1 2

1 2

j j

j

V V V

V V

2 1

2 1 2

1 2

j

j j

V V

V V V

V V 0

Using MATLAB:

1 1

2 2

1 10 5.33 36.9 V

2.75 67.2 V

j j

j

q ª º ª º ª º

q

¬ ¼ ¬ ¼ ¬ ¼

V V

then

1 2

2.66 126.9 A

V V

I q

Now the complex power can be calculated as

1

0.8889 VA

j j

q q

I V

S

Finally

0, VAR

S P jQ j P Q

(Checked using MATLAB and LNAP)

P11.4-

(a)

dc ac

v t 1 cos t v v

T

S

2 2

dc eff ac eff

0

1 0 1 V and V

T

T t T

v dt v

T T T

2

2 2 2 2

eff dc eff ac eff

1 1.225 V

v v v

(b) The period of the sinusoid is half the period of i ( t ). Let T be the period of i ( t ) and T /2 be

the period of the sinusoid. Then

10sin 0

10sin

T

t t

T

T T

i t t t

T

otherwise

S

/ 4 3 / 4

2 2

0 / 4

/ 4 2 / 4

0 /

sin sin

T T

RMS

T

T T

T

I t dt t d

T T T

t t

T T

T

T T

S S

2

t

rms rms

I = 25 I = 5 mA

P11.4-

t t

v t t t

t

d d

®

d d

2

0.1 0.

0.1 0.2 2 2 2 2 2

rms 0 0.

0 0.

2

18 V

V t dt t dt t dt t dt

rms

Practice Problems Solution for Circuit Analysis | ECE 2040, Quizzes of Electrical Circuit Analysis

Related documents

Partial preview of the text

Download Practice Problems Solution for Circuit Analysis | ECE 2040 and more Quizzes Electrical Circuit Analysis in PDF only on Docsity!

Chapter 11 - Practice Problems Solutions

t

t

S

S

P11.3-

j

j

  q

I

!A! +&##, = $ = -./+ 0%" , !)%" "# !ȝ@!A!)"!>@

P

T

u

q

 u

I

V I

P11.3-

I I V

I I

V I V

I I V

V

8"9:+;%)<$%+"#$%$&'()"+,%',+;%=>(?$>@,%>'9$%A$9#,%

302/ 4 30 3 B6C

/ 3 10C6 %D

270 604 2/0 1342/ 4 /3 36C

E<$+%

2/04 10 / 3 /00%H

P

I

P11.3-

% D%+"#$%$&'()*"+-%

2 1B

V V

Section 11-5: Complex Power

P11.5-

0.6 1.2 1.342 63.43 A

j

j

  q

 q  q

S

I

4 8.94 63.43 4 8 4 and 2 H

R j L j R L

 q

P11.5-

3 1.5 3.35 26.56 A

j

j

  q

 q  q

S

I

j j

R j L R L

  q

 q

 R 4 :and L 2 H

P11.5-

Let

4 4 V

j

j j j

j

j j j

u 

Z

Next

1.342 26.6 A

4 4 8 j 8

 q  q

I

Z

q

u

q

q q

q

q

q q

q

q

R 4 :and L 2 H

u

q q

q q

S