Question How does Superposition Theorem differ from all the other network laws? What is/are its limitation/s in terms of applicability? find 16 using super position theorem 14 current source 9 ohms ( www DC (6 amps) 1 ohm current source DC (3 amps) #H 15 volts 3 ohms 8 ohms www 4 ohms) Is www 6 ohms +|+ 5 volts 2 ohms 10 volts 13 5 ohms ww 17 16 7 ohms

E8KETV The Asker · Electrical Engineering

How does Superposition Theorem differ from all the other network laws? What is/are its limitation/s in terms
of applicability?

Transcribed Image Text: find 16 using super position theorem 14 current source 9 ohms ( www DC (6 amps) 1 ohm current source DC (3 amps) #H 15 volts 3 ohms 8 ohms www 4 ohms) Is www 6 ohms +|+ 5 volts 2 ohms 10 volts 13 5 ohms ww 17 16 7 ohms
More
Transcribed Image Text: find 16 using super position theorem 14 current source 9 ohms ( www DC (6 amps) 1 ohm current source DC (3 amps) #H 15 volts 3 ohms 8 ohms www 4 ohms) Is www 6 ohms +|+ 5 volts 2 ohms 10 volts 13 5 ohms ww 17 16 7 ohms
Community Answer
PVGRWZ

Superposition thedem:-(1)It states that"The cursent through, or vottage acros, an element in a linear bilateral retwork equal to the algebric sum of the currents or voltages produced independently by each source".rarr The superposition theorem extends the use of ohnis law to circuits with multiple sources.Case (3) only 5V vortage Source acting alone, resnaining voctage sources will be chôted and current sources winl be opened. Now circuit islets i_(1)^('),i_(2)^('),i_(3)^(') and is are loop currents.By applying KVL to each 100 , we can write.{:[" Forloop "3:-0=-5i_(2)^(')+28i_(3)^(1)rarr(3)],[" Forlcop "4i-=>0=-4i_(1)^(')-6i_(2)^(1)+18i_(4)^(1)rarr]:}By Solving above four capitions, wo jed,B Now,weneed I_(6)^(')=i_(1)^(')-i_(2)^(')=0.22 q-0.296=-0.1204Ararr(a)Case {: hat(A)^(2)) enly 15vV (tage cource acting alove, I_(6)^(')=-0.00A4A.(I_(6)^(')=-0.004(A))0_(0)lets q_(1)^(2),i_(2)^(2),i_(3)^(2) and i_(4)^(2) are loop currenits. For loop 1:-=>15=12i_(1)^(2)-2i_(2)^(2)-5i_(3)^(2)-4i_(4)^(2)rarr(1)For 100P^(2):-7-15=-2i_(1)^(2)+18i_(2)^(2)-7i_(3)^(2)-6i_(4)^(2)rarr(2)forloop 3: =>0=-{i_(1)^(2)i_(1)^(2)-7_(1)^(12)+21i^(2)-T(3):}For 100p4:- =>0=-4i_(1)^(2)-6i_(2)^(2)+18 i4^(2)rarr (i)By colving above 4 equations, we get.i_(1)^(2)=1.175A;i_(2)^(2)=-0.667A;i_(3)^(2)=0.057A;i_(4)^(2)=0.038Awe need{:[I_(6)^(2)=i_(3)^(2)-i_(2)^(2)=0.057-(-0.667)=0.724A],[I_(C)^(2)=0.724Ararr" (b) "]:}Case (iii)? -10v voltage Source acting alones(3)circ cuit will be.For loopl:- 0=12i^(3)-2i_(2)^(3)-5i_(3)^(3)-4i_(4)^(3)rarr (1)For loop 2: 10=-2i_(1)^(3)+18i_(2)^(3)-7i_(3)^(3)-6ii_(4)^(3)rarr(2)For 100 p3:-0=-5i_(1)^(3)-7i_(2)^(3)+21i_(3)^(3)rarr(3)For loop4:- 0=-4i_(1)^(3)-6i_(2)^(3)+18i_(4)^(3)rarr (4)By solving, we geti_(beta)^(3)=0.452A;i_(2)^(3)=0.897A;i_(3)^(3)=0.406A;i_(4)^(3)=0.4Awe need I_(6)^(3)=i_(3)^(3)-i_(2)^(3)=0.406-0.897=-0.491AT_(6)^(3)=-0.491 ... See the full answer