and (, we have, 1₁ + 1₂ E₁ + E₂ 2 Z+Z₁ 11

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ISBN:9780133923605
Author:Robert L. Boylestad
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and
Now
Z₁ = x₁ + 38 = 1·2+38 = 552
Z₂ = x₂+41 = 0.9+4·1= 552
Z₁ = (5) 2; Z₂ = (5) 2; Z= (50+ j40) 2
210⁰ = 5774 210° volts =
E₂
=
Referring to Fig. 12.87, we have,
1₁ Z₁ + (1₁ +1₂)Z =
1₁ + 1₂
=
1₁-1₂ =
Adding eqs. (i) and (), we have,
and
1₂ *Z₁ + (1₁ + 1₂) Z
Subtracting eq. (ii) from eq. (i), we have,
1₁ - 1₂
=
10,000
√√3
12000
√√3
=
E₁
E₂
E₁-E₂
Z₁
-1242 + j1003
j5
(200+j250) A
20⁰ = (6928+j0) volts
100
:.
1₁ + 1₂
On solving eqs. (iii) and (iv), we have,
=
E₁ + E₂
2 Z+Z₁
12614+j103
(5686+j1003) - (6928+ j0)
j5
= (200+j250) A
(5686+j1003) + (6928 + j0)
2 (50+ j40) + j5
12654 24.5°
85
131-2240-4°
96-4 2-35-9° A = (78-1-j56-6) A
(78-1-j56-6) A
(5686+j1003) volts
169 A; I, = 164 A
597
...(1)
.... (iii)
...(iv)
Hello expert, I want you to explain how
he got the law inside the circle. Can you
explain step by step and a clear line, please?
Transcribed Image Text:.. and Now Z₁ = x₁ + 38 = 1·2+38 = 552 Z₂ = x₂+41 = 0.9+4·1= 552 Z₁ = (5) 2; Z₂ = (5) 2; Z= (50+ j40) 2 210⁰ = 5774 210° volts = E₂ = Referring to Fig. 12.87, we have, 1₁ Z₁ + (1₁ +1₂)Z = 1₁ + 1₂ = 1₁-1₂ = Adding eqs. (i) and (), we have, and 1₂ *Z₁ + (1₁ + 1₂) Z Subtracting eq. (ii) from eq. (i), we have, 1₁ - 1₂ = 10,000 √√3 12000 √√3 = E₁ E₂ E₁-E₂ Z₁ -1242 + j1003 j5 (200+j250) A 20⁰ = (6928+j0) volts 100 :. 1₁ + 1₂ On solving eqs. (iii) and (iv), we have, = E₁ + E₂ 2 Z+Z₁ 12614+j103 (5686+j1003) - (6928+ j0) j5 = (200+j250) A (5686+j1003) + (6928 + j0) 2 (50+ j40) + j5 12654 24.5° 85 131-2240-4° 96-4 2-35-9° A = (78-1-j56-6) A (78-1-j56-6) A (5686+j1003) volts 169 A; I, = 164 A 597 ...(1) .... (iii) ...(iv) Hello expert, I want you to explain how he got the law inside the circle. Can you explain step by step and a clear line, please?
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