Kolok 1 SuperPozycja, Wzorzec K1a, /


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I62

R

I6

I7

I5

R1

J1

R4

R3

R5

E6

E7

R

R

I61

I71

Szacowane kierunki prądów

R1

J1

R4

R3

R5

Kolokwium 1 (wzorzec)

E7 = 7 V

Na końcu wg 1PK obliczymy I6:

I63 = I73 I53 = 2.454 mA

J1

8 kΩ

R1||R3

R5

Z obserwacji obwodu wnioskujemy,
że R oraz R5 są równoległe: R || R5.
To samo
: R1 || R3.
R4 jest równoległa do E6

Szukanie I52: Wg Ohma IL=6/3830=1.57mA. Wg dzielenia prądu: I52 = IL⋅R/(R5+R) = 1.5710-3⋅8/13 = 0.964mA,

Wg 1PK: I72 = I62 -I52 = 3.068 - 0.964 = 2.103mA

E6 = 6 V

Dla obliczenia I73 należy obliczyć opór zastępczy i skorzystać z prawa Ohma

J1

3/4 kΩ

5 kΩ

Z= (R1||R3||R5+R) || R4 = 8.652∗4 /12.652 = 2.735 kΩ.

Wg prawa Ohma:

I73 = 7 / 2735 = 2.559 mA

Uwaga! R4 || E7. Jednak nie należy usuwać R4, ponieważ interesuję nas prąd przez E7.

U

Źródło ↓

I5, mA

I6, mA

I7, mA

sprawdzenie

J1= 10 mA

I51 =1.206

I61 =1.96

I71 = 0.7538

1.96 = 1.206 + .7538

E6 = 6 V

I52 = 0.964

I62 = 3.068

I72 = 2.103

3.068 = .964 + 2.103

E7 = 7 V

I53 = 0.1055

I63 = 2.454

I73 = 2.559

2.559 = .1055 + 2.454

I5 = 0.347

I6 = +3.562

I7 = +3.908

3.908 = .347 + 3.562

1PK

- (wych)

(wych)

+ (wch)

R=8 kΩ

R4

0.652 kΩ

Dzielenie prądu J1:

G = 1/8+4/3+1/5 =1.658 mS.

I51 = J10.2/1.658 = 1.206 mA

I71 = J10.125/1.658 = 0.753 mA

wg 1PK: I61 = I51+I71= 1.96 mA

E7

R

I52

I62

R1

J1 = 10 mA

R4

R3

R5

E6

R4 usuwamy, bo IR4 = 0

Wg prawa Ohma

I62 = 6 / 1956 = 3.068mA

R4

R1||R3

R||R5

E6

I73

Zastępcza rezystancja dla E6 to

(R1||R3+ R||R5) || R4 = 3830 || 4000 =
= 1956 Ω

I72

E6

IL

1956 Ω

I62

R

R

R1

Realne kierunki prądów

R4

R3

R5

I73

E7

I73

I51

R4

R1||R3||R5= 0.652 kΩ

I7

E7

I51

Wg dzielenia napięcia:

U = 7∗0.652 / 8.652 = 0.528 V.

Wg prawa Ohma:

I53 = U / R5 = 0.528 / 5000 = 0.1055 mA

E7

I6

I5

J1 = 10mA, E6 = 6V, E7 = 7V,
R = 8 kΩ, R =1 kΩ, R3 = 3 kΩ, R4 = 4 kΩ, R5=5 kΩ.

I71

I63

I53

I73



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