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Exercise 2-Electrical Circuit Analysis-Lecture Slides, Exercises of Electrical Circuit Analysis

Bengal Engineering & Science UniversityElectrical Circuit Analysis

This lecture is part of lecture series on Electrical Circuit Analysis course. It was delivered by Prof. Mursleen Sayed at Bengal Engineering and Science University. It includes: Linear, Transformer, Parameters, Impedance, Voltage, Source, Frequency, Domain, Equivalent, Circuit

Typology: Exercises

2011/2012

Uploaded on 07/23/2012

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Exercise
Linear transformer parameters
R1= 200 ΩR2= 100 Ω
L1= 9 H L2= 4 H
k = 0.5
Transformer couples an impedance of R = 800 Ω
and C = 1μF to a voltage source
Voltage source = 300 Vrms has an internal
impedance of 500 + j100 and = 400 rad/s
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Download Exercise 2-Electrical Circuit Analysis-Lecture Slides and more Exercises Electrical Circuit Analysis in PDF only on Docsity!

Exercise ^ Linear transformer parameters ^ R

Ω^

R^2

^ L

= 9 H 1

L^2

= 4 H

^ k = 0.5 ^ Transformer couples an impedance of R = 800

and C = 1

μF to a voltage source

^ Voltage source = 300 V

rms

has an internal

impedance of 500 + j100 and

^

= 400 rad/s

Exercise ^ Frequency-Domain equivalent circuit^ 

jwL

= j(400)(9) = j3600 1

^ jwL

= j(400)(4) = j1600 2

^ M = k

L^1

L= 0.5^2

(9)(4) = 3 H

^ jwM = j(400)(3) = j

^ 1/jwC = 1/j(400)(1x

-6^ ) = -j

Exercise ^ Impedance reflected into primary winding ^ Scaling factor of the reflected impedance^ 

  • (^222222) 2 r

Z

Z

M

Z^

(^800) j 800 ) (^900) j

8900 ( 9 | (^900) j (^900) |

(^900) j (^900) () (^1200) ( Z^

2

2

r^

Exercise ^ Impedance looking into primary terminals of transformer^ 

Z

= 200 + j3600 + 800 + j800 = 1000 + j4400ab

Ω

^ rms value of primary and secondary current^ 

I s

=^ V

/(Zs

  • Zs

) = 300/(1500 + j4500) = 20 – j60ab

= 63.

-71.

0 mA (rms)

^ I =^2

I j^1

M/Z

= 59.63 22

63.

0 mA (rms)

Exploring Limiting values

22 22 L 2 L 2 22

jX

R

X

L(

j

R

R

Z^

22 22

2 2 1 1 ab^

jX R

M

Lj R Z^

2 22 2 22

22 22 2 2 1 1 ab^

X

R

jX R( M Lj R Z^

2 22 2 22

22 2 2 1 2 22 2 22

2 22 2 1 ab^

X

R

jX( M Lj X R

R

M

R

Z^

ab ab 2 22 2 22

2 22 2 1

2 22 2 22

2 22 2 1 ab^

jX R X R

X M L j X R

R

M

R

Z^

Exploring Limiting values

2 22 2 22

(^222)

1 2 22 2 22

(^222) 1 1 ab^

X

R

XL

L

X

R

XL

L

L

X

2 22 2 22

(^222)

2 22 2 22 1 ab^

X

R

XL

X

R

L

X

(^2) L 2 2 22

L 2 2 (^2) L 2 2 22 1 ab^

X

L(

R

X

L(

L

X

L(

R

L

X

(^2) L 2 2 22

(^2) L L 2 2 22 1 ab^

X

L(

R

X

XL

R

L

X

Exploring Limiting values ^ Similarly ^ Z

ab

^ With ideal transformer the secondary windingresistance and the load impedance are reflected tothe primary by a factor (N

/N 1

2 22 2 )^2

1 2 1 22 1 2 1 ab^

R

N N

R

R

L L

R

R^

^

L L (^212)

2 (^212)

1 ab^

jX R N N R N N R Z^

The Ideal Transformer ^ Consists of two magnetically coupled coils havingturns N

and N 1

2

^ Coefficient of coupling K =1 ^ Self-Inductance of each coil is infinite(L

= L 1

^ The coil losses, due to parasitic resistance arenegligible ^ Terminal behaviour of Ideal transformer

2 and^2

1 1

v N v^ N

^

2 2 (^11)

Ni Ni

Determining polarity of the voltageand current ratios

2 2 vv 1 NN 1

(^22) (^11)

iN iN ^

2 2 vv 1 NN 1

(^22) (^11)

iN iN 

2 2 vv 1 NN 1

(^22) (^11)

iN iN ^

2 2 vv 1 NN 1

(^22) (^11)

iN iN 

Quiz # 1 (Time = 10 min) ^ Linear transformer parameters ^ R

^200

Ω^

R^2

=^^100

^ L

9H

L^2

=^ 4H

k =

^ Transformer couples an impedance of

R^ L

Ω^ and

C^ L

μF

to a voltage source

^ Voltage source =

300V

rms

has an internal impedance of

500 + j

and

^

= 400rad/s

^ Find the Thevenin Equivalent Impedancew.r.t. load impedance?