LCT4-SingleLoop.pdf

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Microsoft PowerPoint - LCT4-SingleLoop.ppt

Electrical

ENGINEERING

INTRODUCTION TO

CIRCUIT ANALYSIS

4. Single loop

Prof. dr hab. in Ň . Tadeusz Skoczkowski

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SINGLE NODE-PAIR CIRCUITS

PAIR CIRCUITS

THESE CIRCUITS ARE CHARACTERIZED BY ALL

THE ELMENTS HAVING THE SAME VOLTAGE

IN PRACTICE NODES MAY ASSUME STRANGE

FORMS

FORMS

THE ELMENTS HAVING THE SAME VOLTAGE

ACROSS THEM

ACROSS THEM - THEY ARE IN PARALLEL

THEY ARE IN PARALLEL

+

V

−

EXAMPLE OF SINGLE NODE-PAIR

+

LOW DISTORTION POWER AMPLIFIER

V

−

THIS ELEMENT IS INACTVE (SHORT-CIRCUITED)

CIRCUITED)

SINGLE NODE

IN PRACTICE NODES MAY ASSUME STRANGE

THESE CIRCUITS ARE CHARACTERIZED BY ALL

THE ELMENTS HAVING THE SAME VOLTAGE

ACROSS THEM

THIS ELEMENT IS INACTVE (SHORT

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LOW VOLTAGE POWER SUPPLY FOR CRT - PARTIAL VIEW

PARTIAL VIEW

SAMPLE PHYSICAL NODES

COMPONENT SIDE

CONNECTION SIDE

LOW VOLTAGE POWER SUPPLY FOR CRT

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BASIC CURRENT DIVIDER

R

p

THE CURRENT DIVISION

APPLY KCL

THE CURRENT i(t)

i(t) ENTERS THE NODE AND

ENTERS THE NODE AND

SPLITS - IT IS DIVIDED BETWEEN THE

IT IS DIVIDED BETWEEN THE

CURRENTS i1(t)

i1(t) AND

AND i2(t)

i2(t)

USE OHMÔS LAW TO REPLACE

CURRENTS

CURRENTS

DEFINE ÑPARALLEL RESISTANCE COMBINATIONÒ

I

=

1

(

=

mA

I

=

I

−

I

=

4

(

1

1

1

+

4

2

1

1

+

5

i

(

t

)

=

v

(

t

)

R

p

v

(

t

)

=

R

1

R

2

i

(

t

)

R

+

R

1

2

THE CURRENT

i(t)

SPLITS

CURRENTS

i1(t)

AND

USE OHMÔS LAW TO REPLACE

1

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FIND

I

1

I

2

V

O

WHEN IN DOUBTÈ REDRAW THE CIRCUIT TO

HIGHLIGHT ELECTRICAL CONNECTIONS!!

HIGHLIGHT ELECTRICAL CONNECTIONS!!

IS EASIER

TO SEE THE

TO SEE THE

DIVIDER

=

80 I

*

DIVIDER

2

24

,

,

WHEN IN DOUBTÈ REDRAW THE CIRCUIT TO

IS EASIER

TO SEE THE

k

V

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Plik z chomika:

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LCT2-OhmsLaw.pdf (270 KB)
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