In the given circuit the resistance R9 and R8 are Parallel with Resistance R10 therefore
(R9 + R8) || R10
= {(4 + 2) × 12} ⁄ {(4 + 2) + 12}
= (6 × 12) ⁄ (6 + 12)
RA = 4Ω
Now resistance RA and R7 is parallel with Resistance R6
∴ {(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RB = 4Ω
Now the circuit becomes as shown in the figure
Resistance RB and R7 is parallel with Resistance R4
{(4 + 8) × 6} ⁄ {(4 + 8) + 6}
= (12 × 6) ⁄ (12 + 6)
RC = 4Ω
Now resistance RC and R3 is parallel with Resistance R2
{(4 + 2) × 6} ⁄ {(4 + 2) + 6}
RD = 3
Now our final circuit becomes as shown in the figure
Therefore the equivalent resistance is
R1 + RD
Req = 4 + 3 = 7Ω
EXAMIANS
