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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)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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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
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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