detemine the currents I,I2 E, =18V Junction RuleI,4I,= Ig. Oloop rule: 18+0.5I2t 6エ,12.5エ2 6I,+3I2-18 = 0 エ 0.5 22 2.5513 =D0 エ2エ」 6.02 0.52 の-6エ,- 4.5エュー0.0Igt45 =0 -(6I,-213 +43=0 E2=45V エキエュI。→ エ。=工-I2 %3D I3 into (2 from (3) -6エ,-2 (Iーエ)+45:O F6エ-2エ,+ 2I,+45=0 - 8I,+ 2I 2+ 45:0円 Ox4 +4 x 3 (エ、+3エュ-18)×4 +(-8I、+2土z+4ら)×3 247, + 12エュース2%= 0 2イT.+ 6I t135 =0 18I, + 63 =O -63 エュニーろ.6A1 163 エエ,into の 本こ6I30(-3.5)18=0エュニ415-(-3.5)%30 I3=B.25A F2 into (D 2 エ.-28.5%30 +28.5 +28 5 エ28.5 6.0 H=4.15A
detemine the currents I,I2 E, =18V Junction RuleI,4I,= Ig. Oloop rule: 18+0.5I2t 6エ,12.5エ2 6I,+3I2-18 = 0 エ 0.5 22 2.5513 =D0 エ2エ」 6.02 0.52 の-6エ,- 4.5エュー0.0Igt45 =0 -(6I,-213 +43=0 E2=45V エキエュI。→ エ。=工-I2 %3D I3 into (2 from (3) -6エ,-2 (Iーエ)+45:O F6エ-2エ,+ 2I,+45=0 - 8I,+ 2I 2+ 45:0円 Ox4 +4 x 3 (エ、+3エュ-18)×4 +(-8I、+2土z+4ら)×3 247, + 12エュース2%= 0 2イT.+ 6I t135 =0 18I, + 63 =O -63 エュニーろ.6A1 163 エエ,into の 本こ6I30(-3.5)18=0エュニ415-(-3.5)%30 I3=B.25A F2 into (D 2 エ.-28.5%30 +28.5 +28 5 エ28.5 6.0 H=4.15A
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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I don’t understand why i’m getting 2 different answers for I3.
When plugging into junction rule: 4.75+(-3.5)= 1.25 BUT when i do 4.75-(-3.5)= 8.25
what am i doing wrong? :(
![determine the currents I,,Izr I
Dunction RuleI, I2 =Ig
loop rule :
-18+0.5I2t 6I,+2.5エ2=O
6I,+3I2-18 = 0
エ
22
2.52
0.50
エ2エ、 た、
Emi
ま。
1ら。
0.5 f
の-6エ,- 4.5Iュー0.5Igt45=0
-6I,-2I3 +43=0
E2=45V
工#Iュ+I I=II
る
Eる into 2 from (3)
-6I,-2 (I-エュ) +45:0
ト6エ-2エ,+2I,+45=0
エ,+ 2エ,+ 45=0(4)
Ox4 +(4) x 3
エ,+3I,-18)×4 +(-8I,+2エェリら)×3
24/
+12エ2-ユ2%30
2イT.+ 6 t135 =0
18I, + l63 =O
-63
Iュニーろ、白A
-63
エ,年エ2into ®
本に6ける0(-3.5-1830 エュニ4.75-(-3.5)%=0
エュ=8.25A
E2 into (D
6I,-28.5=0
728.5 +285
eI,=_28.5
6.0
H=4.15A
Shoes](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcb9f2fc8-72b4-4742-b563-2f7dbd26462e%2F724889f1-1a10-4800-bb79-1587a33d0c6b%2Fdhl923i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:determine the currents I,,Izr I
Dunction RuleI, I2 =Ig
loop rule :
-18+0.5I2t 6I,+2.5エ2=O
6I,+3I2-18 = 0
エ
22
2.52
0.50
エ2エ、 た、
Emi
ま。
1ら。
0.5 f
の-6エ,- 4.5Iュー0.5Igt45=0
-6I,-2I3 +43=0
E2=45V
工#Iュ+I I=II
る
Eる into 2 from (3)
-6I,-2 (I-エュ) +45:0
ト6エ-2エ,+2I,+45=0
エ,+ 2エ,+ 45=0(4)
Ox4 +(4) x 3
エ,+3I,-18)×4 +(-8I,+2エェリら)×3
24/
+12エ2-ユ2%30
2イT.+ 6 t135 =0
18I, + l63 =O
-63
Iュニーろ、白A
-63
エ,年エ2into ®
本に6ける0(-3.5-1830 エュニ4.75-(-3.5)%=0
エュ=8.25A
E2 into (D
6I,-28.5=0
728.5 +285
eI,=_28.5
6.0
H=4.15A
Shoes
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