Question

2 Using Kirchhoff's laws, calculate the branch currents in the network of fig. 1.41. www R ww 492 P 222 (Ans. : 1, = 0.714 A, 12 = 0.428 A, I, = 1.142 A) -10V 892 12V - S T U Fig. 1.41.


plz send full answer with step wise n proper solution ​

Answer 1

Answer:

613A, (22)/(13)A,2813A

Solution

According to kirchhoff's firse rule , at junction A

l1=l2+l3orl3=(1)−l2

Appliying kirchhoff's second rule to the closed loop BADCBwe have

3l2+2l1=−10+12=2or2l1+3l2=2…..(i)

to the loop AFEDA

4l1−3l2=10

or 4(l1−l2)−3l1=10or4l1−7l2=10....(ii)

Multipiying (i) by (2) and subratercting (ii) from it we get

6l1+7l2=4−10=−orl2=−613A

From (i) 2l1+3(−613)=2

or 2l1=2+1813=4413orl1=2213A

l3=l1−l2=2213−(−613)=2813A

Explanation: