Question

In the network determine the current flowing through the 4 ohm resistor

5 4 2 w 50230 j2 --j2 50/900

Answer 1

Que. In the Network determine the current flaving through the tr ar resistor 4 2 5 solgu -jz ī 50130 fig.lt) source transformation R &R I V=I v fig. (1) applying source transformation 4 25 50190 (ja j2 10 5 and + figla) resistance sn & inductance jan resistance and cofaci-tance (-42)n are connected in parallel For parallel connection - R,+R2 Reqk 'R, R2 & R2 RA Req = P, R. R,+ R2 I Rz Rea

after redrawing the the fig (2) 4^ (1) 25 10130 Z1 |z2 25 L90 Z1 = fig (3) 5 || ja Shop do 529 121= 10 5²+22 190 Z1 = 5x ja LZ go - 21080 5+j2 Itant / 0 za 1 = 10 68.198 L24 - 68.198 529 2. Z2 = 2||(-2) zz = 31.5) +43 +2(3-1) 21 (j-1) 1221 = 2 2 3 J1²+(-132 52 o [Z2 190 -45 = - 1359 1 96 (tant (1) SE L-45 of again applying sowice transformation 22 = figil (3)—

10 168 196 J2L-45 4r Zz 21 I 6 25 Zz L go = 25x52 1-45 & gó 17 2552 145 10x10 1684198+30 529 98.198 = 100 529 the 4 ohm resintor- current flowing through 2552 [45' I- 100 198 1989 529 Z, +4 + 22 25 52 1459 100 198.198 J29 10 168.198' + 4+ 521–45° 329 CO45 .: I - 525 ; we fenow that x caso +jxsino 100 casg8-198+jĐin 98498] -2552 [cos to a [cas68-19€ +join 68-196) + 4+ 52[ cas 45 – jsin 45"} 18.569[-0-142 + 2.142 +30-989] -2552 1856 [0.371 +30:928 ] +4 +52 [šą ju] 529 j - 216 367 + $18-36 4 - 25(1+i) 0.6885+ % 1.722 + 4+ (1-1)

I - 27.6367 -36-636 5.6885 +1 0.722 tant 6.636 2 +27:6367 I= 1627.6367 )+66-6362 115.6885)2+(0.722)? tant {018 00722 2 5-6885 1807.823 tant 0.2401 tant o.1269 907349 29.557 113.5 28.422 5.7341 | 7.233 6-267 amp 4.9567 16-267 = ty