Question

please solve this que asap

Answer 1

Note : R(AB) ➡ EQ. Resistance b/w A and B
AF and AE are in series : 3+3=6
R(AE) = 6/2 = 3
AE and ED are in series : 3+3=6
R(AD) = 6/2 = 3
AD and CD are in series : 3+3=6
R(AC) = 6/2 = 3
AC and BC are in series : 3+3=6
R(AB) = (6*3)/(6+3) = 18/9 = 2ohm

Answer 2

Resistance AF and FE are in series so ,
R1 = 3 ohm + 3 ohm = 6 ohm

again, R1 and AE are in parallel so,
R2 = 6× 6/(6 + 6) = 3 ohm

again, R2 , ED are in series , so,
R3 = 3 ohm + 3 ohm = 6 ohm
R3 and AD are in parallel so,
R4 = 6 × 6 /( 6 + 6) = 3 ohm

again, R4 and DC are in series so,
R5 = 3 ohm + 3 ohm = 6 ohm
R5 and AC are in parallel so,
R6 = 6×6/(6 + 6) = 3 ohm

now, R6 and BC are in series so,
R7 = 3 ohm + 3 ohm = 6 ohm
again, R7 and AB are in Parallel so,
Req = 6×3/(6 +3) = 2 ohm

hence, Req = 2 ohm