Question

QI. Using Source Conversion, determine I in the circuit shown in figure 1.1. R38 7 E= 32 V A R$249 R 149 1 Figure 1.1 #) Find the equivalent resistance between A and B for the circuit shown in the Figure 1.2. 121 A R2 M 1.750 R5 1.50 w 1000 3.30 $1.50 RB w 7500

Answer 1

Sal?) 8-12 R$24-2 6A 5 9 R$ 140 E +'' = 32V Here the main problem is presence of both current and voltage Socerces. To solve this we have to converst eithere current Sococce into Voltage Socerece orc voltage Source into current scence. we preferr voltage to current socerce.conversion as current is required for our question. After conversion of voltage source to current source, the circult will be as follows. I A R=8-2 6 51 Ba{2422 R$ 1492 1, 32V 8224A 1,- 324 = 42 Now Herre two. current Socereces (1, X 12) are in parallel Therefore they can be replaced by one current source of 10A (say 13). As we need current through RL , so we choose parallel equivalence (R) of R&R2 .. ..Rg= R, 11R2 = RERUR. = 8X24 = 622 12 so our new diagram will be TIL I = I x R 1 RL 110A962 4 +1 = 10 x y = 31 .: Current through RL is 12-31 CS Scanned with CamScanner

(i) A R = 1.75KSL R = 100.5 Rr=1I5K2 Ry = 3.3k &R=1.5K R=22012 Reg1.75K2 www Mmmm MM B R3= 7500 R = 3.3ks For this type of circuit we generally start from close end. for reight end Ry & Rg are in series and Ro is pareallel with this combination R2 Ro Ro=1.5K 1.75ke 1082 - re, = R6 11 (Ryt Rq) {R, ISKEL = 220.12 115.05.12 = 220-2115050-12 R2= 750.2 Rg=3.3622 E 220 X 5050 220 +5050 = By same = 210.8252 Logic our next step is - R2² 1.75kr R2 = R, 11 (Ryt Rs try tRg) = 1:5K-2 11(100+1.5x18+ 210.82 R2=750 1. +3.3x103) = 1.5X10's2115410.82.52 - 1500 X 5110.82 1500+5110.82 Rz= 1159.6552 .. Therefore the equivalent reesistance beto A e B is RAR = R2+82 +R3 (1.75 X 10'+1159.65 + 750) 52 = 3659.65.2 - RAR 3. 66kA - Resistance beto na B is 3.66 ks. Scanned with CamScanner