Question

Figure 4. (a) The equivalent resistance of the network shown in figure 4(A) is 2KΩ. (True, False)

(B) 1 ΚΩ 1 ΚΩ 1 ΚΩ - R2 w c 1.00 + 3100 im IR 3.00 7 Figure 4: 4. (a) The equivalent resistance of the network shown in figure 4(A) is 2K12 (True, False) (b) The equivalent resistance of the network shown in figure 4(B) is 272 [Hint see figure 4(B) and] Req = Rab = i1 + iz + i 1+ + + -i1 - (il - i3) + 3i2 = 0 -5(i1 - i3 + iz) + i3 - (i - iz) = 0 (c) The equivalent resistance of the network shown in figure 4(C) can be found from(same network as in figure 4(B)] -Riii - R3ii - i3) + R4i2 = 0 1- Ril - iz + i2) + R2iz - R3(i - is) = 0 i2 il (R1(R2 + R3 + Rs) + R2R3) R2R4 + R3(R4 + R5) + RARS iz i (R5(R1 + R4) + R3(R4 + Rs)) R2R4 + R3 (R4 + RS) + RARS ab R1 + R23 Req = Rab == R1 + Raiz ii + 2 R1 + R2 (R5 (R+RA)+R3 (R+R)), R21 R2 R4+R3 (R4+R3)+RARE (R1(R2+R3+R5)+R2 R3) 1+ RR4+R3 (RA+R3)+RAR (True, False) 1+

b) The equivalent resistance of the network shown in figure 4(B) is 27/17Ω [Hint see figure 4(B) and]

Answer 1

RI KR Rs=1ks Ilka R3 3 Rue Ro{ 2kas =)Ks - 2k2 ( FIG-A2 R = Rs +R6 = 2 kr R = R4R. – 2x2 ano. R1 = R2 + R" = 1+1 = 26 d. RW - R3xR" 3*3 1kn R – Rit RIV = (1 + 1) = 2Kr. ... The equiralent wristance of circuit shown in (FIG-A) is akr (prored). (True)

in 4 R iz & Rz apmom R2 R, =1.00 R2 =loor Rz=loon. R4 = 3.02. Rs 75.on v 2 -13tin d fLG,B equivalent resistance, Reg Req - Rab - Vab - stiz tin. Vooth, Aisyrer TReq= 14 + 111 Hi Now, from loopy from loop! tiek, it (2, -13) 23 - 22 R4 =0 iz R2 = (1, -iz+12) R5 or / Bi, Bitzen -11-13) R3 =0 or, bai, ti, iz-Zuz = 0 av, iz - (1, -ig +1)x5-(1,-) = 0 , 21, -ig - Biz :0 av aiz +5 i3 -51, -, -5i2 so or Fiz - 62,- 512=0 or; iz: 24 – 3 in so or + (21, - 312) - 61, - Sizzo or, 13 = 21, -3.41, or, 8i, – 2612 = 0 a, 1313 = 26 1,-121, or izli, 2826 = 4/13 trij av,isli, the- using equs ( 4 ) in equ is, we have. Req= 2 97 | Req=27/17 ore to True. 13

wana R, = R2 = R3 = 2002 &q=3.00h Rs = soon ib R4 Vri-igtl2 FI61.40 loopsis Looprug - 4;R, -R3 (1; 13) +R42=0 -P (1, -13 tip) A iz R₂ -R3 (1, -3) = 0 of, -i, (R1 + R 3) + 13 R3 4 12 R4 20 or,-i (R3+Rs) + 13 (P5 + he ths) =Roiz mr, -(R1+R2) +igR3 av, iz iz(Q5 + R24R3) - (B3+Rs) 1, + Ry z iz(85 +R3 +R3) – (R3 ts), LED ov, -(e+Ra) Rs - :,(R3+R$) R4 + izrais + iz R4CR2+R3 +Rs-) = 0 R5 (R1 + f₂) + R4 (R3tR;-) -R385 + R4 (Rat R3 + Rp) Rs from 12 1. R5 (R1 + R3) + R4 (R3 +Rs) I (R2 + R3+R-) —; (R372) - R5 R3 R5 + Rq (R2 + R3t R5) (R2 + R3 +Ps) 8ps (ft R3) + R4 (R3 +Rs)} - (Rat Rs) { Rahs + P4 (Rzthath) or, ir Ro- sli, 14- RI (R2 + R3 + 2) + R2 R3 RqR4 + R3 (R4 + Rs) +R4 R5 73 Req - Rab - Vab . R + R2 lali It bli

= R1 + R2. Ps (Rit R3) + R4 (R3 +R) RoRo a R4 (Ret R3 +R5) It RI (R2 + R3 485) + R2 R3 RzR4 + R3 (R4 +Rs) + R4 R5 BRIT R2 . RS (R1+R4) + R3(R4 +Rs) R2R4 + R3 (R4 + Rs) + R4 R5 - It RI (R2 + R3tRs) + R2 R3 Rg Ru + R3 (R4 +85) + Ru RS- Substituting vahus of R1, R2, R3, R4, Rs, we obtam, Req= lt1.5 (1+3) + 1.(3+5) It TX 3+ 1 (3+5) + 3x5 i(i+1+5) + 1x! ((x 3)+1-(3+5) + 3x5 28 It 26 I @ 3+1 26 - 54 - 27/172 T Req= 27/12e to (true)