104 11, V1=6 V. 11=40 A & K1-4.5, use the principle of superposition to find : R2 KIV, R3 V1 I1 R1 V01 (in volt) due to V1 only= O a. 0 b. -0.028190603132289 C. -0.018793735421526 O d. -0.0093968677107631

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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For R1-47 0, R2=72 Q, R3=104 Q, V1=6 V, 1=40 A & K1=4.5, use the principle of superposition to find :
R2
KIV,
R3
V1
I1
R1
V01 (in volt) due to V1 only=
a. 0
b. -0.028190603132289
C. -0.018793735421526
d. -0.0093968677107631
Transcribed Image Text:t of For R1-47 0, R2=72 Q, R3=104 Q, V1=6 V, 1=40 A & K1=4.5, use the principle of superposition to find : R2 KIV, R3 V1 I1 R1 V01 (in volt) due to V1 only= a. 0 b. -0.028190603132289 C. -0.018793735421526 d. -0.0093968677107631
d. -0.0093968677107631
(Note that V0=V01+V02, vwhere Vo1 is due to V1 and Vo2 is due to 1)
VO2 (in volt) due to 11 only=
a. -6.5151616127957
b. -13.030323225591
C. -16.939420193269
d. -3.9090969676774
(Note that VO=V01+V02, where Vo1 is due to V1 and Vo2 is due to 11)
then Vo (in volt)=
Power on R1 (in watt) =
Transcribed Image Text:d. -0.0093968677107631 (Note that V0=V01+V02, vwhere Vo1 is due to V1 and Vo2 is due to 1) VO2 (in volt) due to 11 only= a. -6.5151616127957 b. -13.030323225591 C. -16.939420193269 d. -3.9090969676774 (Note that VO=V01+V02, where Vo1 is due to V1 and Vo2 is due to 11) then Vo (in volt)= Power on R1 (in watt) =
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