1₁-1₂ 1₁ + 1₂ Adding eqs. (i) and (ii), we have, 1₁ + 1₂ On solving eqs. (iii) and (in = j5 1₁-1₂ = (200+j250) A 4 11 = TO = E₁-E₂ Z₁ -1242 +j1003 = E₁ + E₂ 2 Z+Z₁ (5686+j1003) + (6928 + j0) 2 (50+ j40) + j5 12614+j1003 12654 24.5° 100 + j 85 131-2240-4° 96-4 2-35-9° A = (78-1-j56-6) A (78-1-j56-6) A have, (5686+j1003) - (6928+ j0) j5 = (200+j250) A 169 A = = 1₂ = 164 A 1₂ How did he get the value inside the circuit or the value of the currents

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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..
and
Now
and
X₁ + RI
Z₁ = x₁ + 38 = 1·2+ 38 = 502,
Z₂ =
x₂ +41 = 0.9+4·1=552
Z₁
(j5) Ω ; Ζ = ( 5) Ω ; Ζ = (50 + j 40) Ω
E₁
..
=
E₂
Referring to Fig. 12.87, we have,
1, Z, + (1₁+1₂) Z
1₂ *Z₁ + (1₁ +1₂) Z = E₂
Subtracting eq. (ii) from eq. (i), we have,
1₁ - 1₂
-
=
=
=
1₁-1₂
Adding eqs. (i) and (ii), we have,
1₁ + 1₂
10,000
√√3
12000
=
210⁰ = 5774 210° volts (5686+j1003) volts
20° = - (6928+j0) volts
E₁-E₂
Z₁
-1242 +j1003
j5
(200+j250) A
E₁ + E₂
2 Z+Z₁
=
(5686+j1003) - (6928+ j0)
j5
= (200+j250) A
1₁ + 1₂
On solving eqs. (iii) and (iv)e nave, 67-
(5686+j1003) + (6928 + j 0)
2 (50+ j40) + j5
12654 24.5°
12614+j1003
100 + j 85
131-2240-4°
=
96-4 2-35-9° A = (78-1-j56-6) A
(78-1-j56-6) A
Į = 169 A ; I₂ = 164 A
597
...(iii)
How did he get the value inside the circuit
or the value of the currents
Transcribed Image Text:.. and Now and X₁ + RI Z₁ = x₁ + 38 = 1·2+ 38 = 502, Z₂ = x₂ +41 = 0.9+4·1=552 Z₁ (j5) Ω ; Ζ = ( 5) Ω ; Ζ = (50 + j 40) Ω E₁ .. = E₂ Referring to Fig. 12.87, we have, 1, Z, + (1₁+1₂) Z 1₂ *Z₁ + (1₁ +1₂) Z = E₂ Subtracting eq. (ii) from eq. (i), we have, 1₁ - 1₂ - = = = 1₁-1₂ Adding eqs. (i) and (ii), we have, 1₁ + 1₂ 10,000 √√3 12000 = 210⁰ = 5774 210° volts (5686+j1003) volts 20° = - (6928+j0) volts E₁-E₂ Z₁ -1242 +j1003 j5 (200+j250) A E₁ + E₂ 2 Z+Z₁ = (5686+j1003) - (6928+ j0) j5 = (200+j250) A 1₁ + 1₂ On solving eqs. (iii) and (iv)e nave, 67- (5686+j1003) + (6928 + j 0) 2 (50+ j40) + j5 12654 24.5° 12614+j1003 100 + j 85 131-2240-4° = 96-4 2-35-9° A = (78-1-j56-6) A (78-1-j56-6) A Į = 169 A ; I₂ = 164 A 597 ...(iii) How did he get the value inside the circuit or the value of the currents
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